Equivalent Conditions for Differentiability - USM · 2010. 8. 30. · Cauchy-Riemann EquationsPolar...
Transcript of Equivalent Conditions for Differentiability - USM · 2010. 8. 30. · Cauchy-Riemann EquationsPolar...
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Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Equivalent Conditions for Differentiability
Bernd Schroder
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Introduction
1. Although all computations for differentiability work very wellwith complex numbers, it would be nice to have an idea whatcomplex differentiability means in terms of real functions of realnumbers.
2. The Cauchy-Riemann equations provide just that.3. As such, we obtain a new way to look at complex differentiable
functions and a new way to look at certain real differentiablefunctions.
4. This will allow us to prove the complex differentiability ofcertain functions (such as the exponential function and thelogarithm function). Proving these facts with differencequotients would be very painful with the tools we have at hand.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Introduction1. Although all computations for differentiability work very well
with complex numbers
, it would be nice to have an idea whatcomplex differentiability means in terms of real functions of realnumbers.
2. The Cauchy-Riemann equations provide just that.3. As such, we obtain a new way to look at complex differentiable
functions and a new way to look at certain real differentiablefunctions.
4. This will allow us to prove the complex differentiability ofcertain functions (such as the exponential function and thelogarithm function). Proving these facts with differencequotients would be very painful with the tools we have at hand.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Introduction1. Although all computations for differentiability work very well
with complex numbers, it would be nice to have an idea whatcomplex differentiability means in terms of real functions of realnumbers.
2. The Cauchy-Riemann equations provide just that.3. As such, we obtain a new way to look at complex differentiable
functions and a new way to look at certain real differentiablefunctions.
4. This will allow us to prove the complex differentiability ofcertain functions (such as the exponential function and thelogarithm function). Proving these facts with differencequotients would be very painful with the tools we have at hand.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Introduction1. Although all computations for differentiability work very well
with complex numbers, it would be nice to have an idea whatcomplex differentiability means in terms of real functions of realnumbers.
2. The Cauchy-Riemann equations provide just that.
3. As such, we obtain a new way to look at complex differentiablefunctions and a new way to look at certain real differentiablefunctions.
4. This will allow us to prove the complex differentiability ofcertain functions (such as the exponential function and thelogarithm function). Proving these facts with differencequotients would be very painful with the tools we have at hand.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Introduction1. Although all computations for differentiability work very well
with complex numbers, it would be nice to have an idea whatcomplex differentiability means in terms of real functions of realnumbers.
2. The Cauchy-Riemann equations provide just that.3. As such, we obtain a new way to look at complex differentiable
functions
and a new way to look at certain real differentiablefunctions.
4. This will allow us to prove the complex differentiability ofcertain functions (such as the exponential function and thelogarithm function). Proving these facts with differencequotients would be very painful with the tools we have at hand.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Introduction1. Although all computations for differentiability work very well
with complex numbers, it would be nice to have an idea whatcomplex differentiability means in terms of real functions of realnumbers.
2. The Cauchy-Riemann equations provide just that.3. As such, we obtain a new way to look at complex differentiable
functions and a new way to look at certain real differentiablefunctions.
4. This will allow us to prove the complex differentiability ofcertain functions (such as the exponential function and thelogarithm function). Proving these facts with differencequotients would be very painful with the tools we have at hand.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Introduction1. Although all computations for differentiability work very well
with complex numbers, it would be nice to have an idea whatcomplex differentiability means in terms of real functions of realnumbers.
2. The Cauchy-Riemann equations provide just that.3. As such, we obtain a new way to look at complex differentiable
functions and a new way to look at certain real differentiablefunctions.
4. This will allow us to prove the complex differentiability ofcertain functions (such as the exponential function and thelogarithm function).
Proving these facts with differencequotients would be very painful with the tools we have at hand.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Introduction1. Although all computations for differentiability work very well
with complex numbers, it would be nice to have an idea whatcomplex differentiability means in terms of real functions of realnumbers.
2. The Cauchy-Riemann equations provide just that.3. As such, we obtain a new way to look at complex differentiable
functions and a new way to look at certain real differentiablefunctions.
4. This will allow us to prove the complex differentiability ofcertain functions (such as the exponential function and thelogarithm function). Proving these facts with differencequotients would be very painful with the tools we have at hand.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem.
Cauchy-Riemann Equations. Let f (z) = u(x,y)+ iv(x,y)be a function on an open domain with continuous partial derivativesin the underlying real variables. Then f is differentiable at z = x+ iyif and only if
∂u∂x
(x,y) =∂v∂y
(x,y) and∂u∂y
(x,y) =−∂v∂x
(x,y).
These equations are called the Cauchy-Riemann Equations.Moreover, we have
f ′(z) =∂u∂x
(z)+ i∂v∂x
(z).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. Cauchy-Riemann Equations.
Let f (z) = u(x,y)+ iv(x,y)be a function on an open domain with continuous partial derivativesin the underlying real variables. Then f is differentiable at z = x+ iyif and only if
∂u∂x
(x,y) =∂v∂y
(x,y) and∂u∂y
(x,y) =−∂v∂x
(x,y).
These equations are called the Cauchy-Riemann Equations.Moreover, we have
f ′(z) =∂u∂x
(z)+ i∂v∂x
(z).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. Cauchy-Riemann Equations. Let f (z) = u(x,y)+ iv(x,y)be a function on an open domain with continuous partial derivativesin the underlying real variables.
Then f is differentiable at z = x+ iyif and only if
∂u∂x
(x,y) =∂v∂y
(x,y) and∂u∂y
(x,y) =−∂v∂x
(x,y).
These equations are called the Cauchy-Riemann Equations.Moreover, we have
f ′(z) =∂u∂x
(z)+ i∂v∂x
(z).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. Cauchy-Riemann Equations. Let f (z) = u(x,y)+ iv(x,y)be a function on an open domain with continuous partial derivativesin the underlying real variables. Then f is differentiable at z = x+ iy
if and only if
∂u∂x
(x,y) =∂v∂y
(x,y) and∂u∂y
(x,y) =−∂v∂x
(x,y).
These equations are called the Cauchy-Riemann Equations.Moreover, we have
f ′(z) =∂u∂x
(z)+ i∂v∂x
(z).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. Cauchy-Riemann Equations. Let f (z) = u(x,y)+ iv(x,y)be a function on an open domain with continuous partial derivativesin the underlying real variables. Then f is differentiable at z = x+ iyif and only if
∂u∂x
(x,y) =∂v∂y
(x,y)
and∂u∂y
(x,y) =−∂v∂x
(x,y).
These equations are called the Cauchy-Riemann Equations.Moreover, we have
f ′(z) =∂u∂x
(z)+ i∂v∂x
(z).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. Cauchy-Riemann Equations. Let f (z) = u(x,y)+ iv(x,y)be a function on an open domain with continuous partial derivativesin the underlying real variables. Then f is differentiable at z = x+ iyif and only if
∂u∂x
(x,y) =∂v∂y
(x,y) and∂u∂y
(x,y) =−∂v∂x
(x,y).
These equations are called the Cauchy-Riemann Equations.Moreover, we have
f ′(z) =∂u∂x
(z)+ i∂v∂x
(z).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. Cauchy-Riemann Equations. Let f (z) = u(x,y)+ iv(x,y)be a function on an open domain with continuous partial derivativesin the underlying real variables. Then f is differentiable at z = x+ iyif and only if
∂u∂x
(x,y) =∂v∂y
(x,y) and∂u∂y
(x,y) =−∂v∂x
(x,y).
These equations are called the Cauchy-Riemann Equations.
Moreover, we have
f ′(z) =∂u∂x
(z)+ i∂v∂x
(z).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. Cauchy-Riemann Equations. Let f (z) = u(x,y)+ iv(x,y)be a function on an open domain with continuous partial derivativesin the underlying real variables. Then f is differentiable at z = x+ iyif and only if
∂u∂x
(x,y) =∂v∂y
(x,y) and∂u∂y
(x,y) =−∂v∂x
(x,y).
These equations are called the Cauchy-Riemann Equations.Moreover, we have
f ′(z) =∂u∂x
(z)+ i∂v∂x
(z).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇒.”
∂u∂x
(x,y) = limt→x
u(t,y)−u(x,y)t− x
= ℜ
(lim
t+iy→x+iy
f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)
)= ℜ
(f ′(z)
)= ℑ
(if ′(z)
)= ℑ
(i lim
x+it→x+iy
f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)
)= ℑ
(lim
x+it→x+iyi
f (x+ it)− f (x+ iy)i(t− y)
)= ℑ
(limt→y
f (x+ it)− f (x+ iy)t− y
)= lim
t→y
v(x, t)− v(x,y)t− y
=∂v∂y
(x,y)
The other equation is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇒.”∂u∂x
(x,y)
= limt→x
u(t,y)−u(x,y)t− x
= ℜ
(lim
t+iy→x+iy
f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)
)= ℜ
(f ′(z)
)= ℑ
(if ′(z)
)= ℑ
(i lim
x+it→x+iy
f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)
)= ℑ
(lim
x+it→x+iyi
f (x+ it)− f (x+ iy)i(t− y)
)= ℑ
(limt→y
f (x+ it)− f (x+ iy)t− y
)= lim
t→y
v(x, t)− v(x,y)t− y
=∂v∂y
(x,y)
The other equation is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇒.”∂u∂x
(x,y) = limt→x
u(t,y)−u(x,y)t− x
= ℜ
(lim
t+iy→x+iy
f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)
)= ℜ
(f ′(z)
)= ℑ
(if ′(z)
)= ℑ
(i lim
x+it→x+iy
f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)
)= ℑ
(lim
x+it→x+iyi
f (x+ it)− f (x+ iy)i(t− y)
)= ℑ
(limt→y
f (x+ it)− f (x+ iy)t− y
)= lim
t→y
v(x, t)− v(x,y)t− y
=∂v∂y
(x,y)
The other equation is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇒.”∂u∂x
(x,y) = limt→x
u(t,y)−u(x,y)t− x
= ℜ
(lim
t+iy→x+iy
f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)
)
= ℜ(f ′(z)
)= ℑ
(if ′(z)
)= ℑ
(i lim
x+it→x+iy
f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)
)= ℑ
(lim
x+it→x+iyi
f (x+ it)− f (x+ iy)i(t− y)
)= ℑ
(limt→y
f (x+ it)− f (x+ iy)t− y
)= lim
t→y
v(x, t)− v(x,y)t− y
=∂v∂y
(x,y)
The other equation is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇒.”∂u∂x
(x,y) = limt→x
u(t,y)−u(x,y)t− x
= ℜ
(lim
t+iy→x+iy
f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)
)= ℜ
(f ′(z)
)
= ℑ(if ′(z)
)= ℑ
(i lim
x+it→x+iy
f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)
)= ℑ
(lim
x+it→x+iyi
f (x+ it)− f (x+ iy)i(t− y)
)= ℑ
(limt→y
f (x+ it)− f (x+ iy)t− y
)= lim
t→y
v(x, t)− v(x,y)t− y
=∂v∂y
(x,y)
The other equation is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇒.”∂u∂x
(x,y) = limt→x
u(t,y)−u(x,y)t− x
= ℜ
(lim
t+iy→x+iy
f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)
)= ℜ
(f ′(z)
)= ℑ
(if ′(z)
)
= ℑ
(i lim
x+it→x+iy
f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)
)= ℑ
(lim
x+it→x+iyi
f (x+ it)− f (x+ iy)i(t− y)
)= ℑ
(limt→y
f (x+ it)− f (x+ iy)t− y
)= lim
t→y
v(x, t)− v(x,y)t− y
=∂v∂y
(x,y)
The other equation is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇒.”∂u∂x
(x,y) = limt→x
u(t,y)−u(x,y)t− x
= ℜ
(lim
t+iy→x+iy
f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)
)= ℜ
(f ′(z)
)= ℑ
(if ′(z)
)= ℑ
(i lim
x+it→x+iy
f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)
)
= ℑ
(lim
x+it→x+iyi
f (x+ it)− f (x+ iy)i(t− y)
)= ℑ
(limt→y
f (x+ it)− f (x+ iy)t− y
)= lim
t→y
v(x, t)− v(x,y)t− y
=∂v∂y
(x,y)
The other equation is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇒.”∂u∂x
(x,y) = limt→x
u(t,y)−u(x,y)t− x
= ℜ
(lim
t+iy→x+iy
f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)
)= ℜ
(f ′(z)
)= ℑ
(if ′(z)
)= ℑ
(i lim
x+it→x+iy
f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)
)= ℑ
(lim
x+it→x+iyi
f (x+ it)− f (x+ iy)i(t− y)
)
= ℑ
(limt→y
f (x+ it)− f (x+ iy)t− y
)= lim
t→y
v(x, t)− v(x,y)t− y
=∂v∂y
(x,y)
The other equation is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇒.”∂u∂x
(x,y) = limt→x
u(t,y)−u(x,y)t− x
= ℜ
(lim
t+iy→x+iy
f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)
)= ℜ
(f ′(z)
)= ℑ
(if ′(z)
)= ℑ
(i lim
x+it→x+iy
f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)
)= ℑ
(lim
x+it→x+iyi
f (x+ it)− f (x+ iy)i(t− y)
)= ℑ
(limt→y
f (x+ it)− f (x+ iy)t− y
)
= limt→y
v(x, t)− v(x,y)t− y
=∂v∂y
(x,y)
The other equation is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇒.”∂u∂x
(x,y) = limt→x
u(t,y)−u(x,y)t− x
= ℜ
(lim
t+iy→x+iy
f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)
)= ℜ
(f ′(z)
)= ℑ
(if ′(z)
)= ℑ
(i lim
x+it→x+iy
f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)
)= ℑ
(lim
x+it→x+iyi
f (x+ it)− f (x+ iy)i(t− y)
)= ℑ
(limt→y
f (x+ it)− f (x+ iy)t− y
)= lim
t→y
v(x, t)− v(x,y)t− y
=∂v∂y
(x,y)
The other equation is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇒.”∂u∂x
(x,y) = limt→x
u(t,y)−u(x,y)t− x
= ℜ
(lim
t+iy→x+iy
f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)
)= ℜ
(f ′(z)
)= ℑ
(if ′(z)
)= ℑ
(i lim
x+it→x+iy
f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)
)= ℑ
(lim
x+it→x+iyi
f (x+ it)− f (x+ iy)i(t− y)
)= ℑ
(limt→y
f (x+ it)− f (x+ iy)t− y
)= lim
t→y
v(x, t)− v(x,y)t− y
=∂v∂y
(x,y)
The other equation is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇒.”∂u∂x
(x,y) = limt→x
u(t,y)−u(x,y)t− x
= ℜ
(lim
t+iy→x+iy
f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)
)= ℜ
(f ′(z)
)= ℑ
(if ′(z)
)= ℑ
(i lim
x+it→x+iy
f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)
)= ℑ
(lim
x+it→x+iyi
f (x+ it)− f (x+ iy)i(t− y)
)= ℑ
(limt→y
f (x+ it)− f (x+ iy)t− y
)= lim
t→y
v(x, t)− v(x,y)t− y
=∂v∂y
(x,y)
The other equation is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇐.”
lim∆z→0
f (z+∆z)− f (z)∆z
= lim∆x,∆y→0
f (x+∆x,y+∆y)− f (x,y)∆x+ i∆y
= lim∆x,∆y→0
f (x+∆x,y+∆y)− f (x,y+∆y)∆x+ i∆y
+f (x,y+∆y)− f (x,y)
∆x+ i∆y
= lim∆x,∆y→0
u(x+∆x,y+∆y)−u(x,y+∆y)∆x+ i∆y
+ iv(x+∆x,y+∆y)− v(x,y+∆y)
∆x+ i∆y
+u(x,y+∆y)−u(x,y)
∆x+ i∆y+ i
v(x,y+∆y)− v(x,y)∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y
+∂u∂y (x,y+ηu)∆y
∆x+ i∆y+ i
∂v∂y (x,y+ηv)∆y
∆x+ i∆y
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇐.”
lim∆z→0
f (z+∆z)− f (z)∆z
= lim∆x,∆y→0
f (x+∆x,y+∆y)− f (x,y)∆x+ i∆y
= lim∆x,∆y→0
f (x+∆x,y+∆y)− f (x,y+∆y)∆x+ i∆y
+f (x,y+∆y)− f (x,y)
∆x+ i∆y
= lim∆x,∆y→0
u(x+∆x,y+∆y)−u(x,y+∆y)∆x+ i∆y
+ iv(x+∆x,y+∆y)− v(x,y+∆y)
∆x+ i∆y
+u(x,y+∆y)−u(x,y)
∆x+ i∆y+ i
v(x,y+∆y)− v(x,y)∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y
+∂u∂y (x,y+ηu)∆y
∆x+ i∆y+ i
∂v∂y (x,y+ηv)∆y
∆x+ i∆y
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇐.”
lim∆z→0
f (z+∆z)− f (z)∆z
= lim∆x,∆y→0
f (x+∆x,y+∆y)− f (x,y)∆x+ i∆y
= lim∆x,∆y→0
f (x+∆x,y+∆y)− f (x,y+∆y)∆x+ i∆y
+f (x,y+∆y)− f (x,y)
∆x+ i∆y
= lim∆x,∆y→0
u(x+∆x,y+∆y)−u(x,y+∆y)∆x+ i∆y
+ iv(x+∆x,y+∆y)− v(x,y+∆y)
∆x+ i∆y
+u(x,y+∆y)−u(x,y)
∆x+ i∆y+ i
v(x,y+∆y)− v(x,y)∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y
+∂u∂y (x,y+ηu)∆y
∆x+ i∆y+ i
∂v∂y (x,y+ηv)∆y
∆x+ i∆y
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇐.”
lim∆z→0
f (z+∆z)− f (z)∆z
= lim∆x,∆y→0
f (x+∆x,y+∆y)− f (x,y)∆x+ i∆y
= lim∆x,∆y→0
f (x+∆x,y+∆y)− f (x,y+∆y)∆x+ i∆y
+f (x,y+∆y)− f (x,y)
∆x+ i∆y
= lim∆x,∆y→0
u(x+∆x,y+∆y)−u(x,y+∆y)∆x+ i∆y
+ iv(x+∆x,y+∆y)− v(x,y+∆y)
∆x+ i∆y
+u(x,y+∆y)−u(x,y)
∆x+ i∆y+ i
v(x,y+∆y)− v(x,y)∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y
+∂u∂y (x,y+ηu)∆y
∆x+ i∆y+ i
∂v∂y (x,y+ηv)∆y
∆x+ i∆y
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇐.”
lim∆z→0
f (z+∆z)− f (z)∆z
= lim∆x,∆y→0
f (x+∆x,y+∆y)− f (x,y)∆x+ i∆y
= lim∆x,∆y→0
f (x+∆x,y+∆y)− f (x,y+∆y)∆x+ i∆y
+f (x,y+∆y)− f (x,y)
∆x+ i∆y
= lim∆x,∆y→0
u(x+∆x,y+∆y)−u(x,y+∆y)∆x+ i∆y
+ iv(x+∆x,y+∆y)− v(x,y+∆y)
∆x+ i∆y
+u(x,y+∆y)−u(x,y)
∆x+ i∆y+ i
v(x,y+∆y)− v(x,y)∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y
+∂u∂y (x,y+ηu)∆y
∆x+ i∆y+ i
∂v∂y (x,y+ηv)∆y
∆x+ i∆y
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇐.”
lim∆z→0
f (z+∆z)− f (z)∆z
= lim∆x,∆y→0
f (x+∆x,y+∆y)− f (x,y)∆x+ i∆y
= lim∆x,∆y→0
f (x+∆x,y+∆y)− f (x,y+∆y)∆x+ i∆y
+f (x,y+∆y)− f (x,y)
∆x+ i∆y
= lim∆x,∆y→0
u(x+∆x,y+∆y)−u(x,y+∆y)∆x+ i∆y
+ iv(x+∆x,y+∆y)− v(x,y+∆y)
∆x+ i∆y
+u(x,y+∆y)−u(x,y)
∆x+ i∆y+ i
v(x,y+∆y)− v(x,y)∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y
+∂u∂y (x,y+ηu)∆y
∆x+ i∆y+ i
∂v∂y (x,y+ηv)∆y
∆x+ i∆y
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇐” (concl.)
lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y+
∂u∂y (x,y+ηu)∆y
∆x+ i∆y+ i
∂v∂y (x,y+ηv)∆y
∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y
+− ∂v
∂x (x,y+ηu)∆y∆x+ i∆y
+ i∂u∂x (x,y+ηv)∆y
∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x+ ∂u
∂x (x,y+ηv)i∆y∆x+ i∆y
+i∂v∂x (x+ξv,y+∆y)∆x+ ∂v
∂x (x,y+ηu)i∆y∆x+ i∆y
=∂u∂x
(x,y)+ i∂v∂x
(x,y)
(Technical argument at end omitted.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇐” (concl.)
lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y+
∂u∂y (x,y+ηu)∆y
∆x+ i∆y+ i
∂v∂y (x,y+ηv)∆y
∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y
+− ∂v
∂x (x,y+ηu)∆y∆x+ i∆y
+ i∂u∂x (x,y+ηv)∆y
∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x+ ∂u
∂x (x,y+ηv)i∆y∆x+ i∆y
+i∂v∂x (x+ξv,y+∆y)∆x+ ∂v
∂x (x,y+ηu)i∆y∆x+ i∆y
=∂u∂x
(x,y)+ i∂v∂x
(x,y)
(Technical argument at end omitted.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇐” (concl.)
lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y+
∂u∂y (x,y+ηu)∆y
∆x+ i∆y+ i
∂v∂y (x,y+ηv)∆y
∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y
+− ∂v
∂x (x,y+ηu)∆y∆x+ i∆y
+ i∂u∂x (x,y+ηv)∆y
∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x+ ∂u
∂x (x,y+ηv)i∆y∆x+ i∆y
+i∂v∂x (x+ξv,y+∆y)∆x+ ∂v
∂x (x,y+ηu)i∆y∆x+ i∆y
=∂u∂x
(x,y)+ i∂v∂x
(x,y)
(Technical argument at end omitted.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇐” (concl.)
lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y+
∂u∂y (x,y+ηu)∆y
∆x+ i∆y+ i
∂v∂y (x,y+ηv)∆y
∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y
+− ∂v
∂x (x,y+ηu)∆y∆x+ i∆y
+ i∂u∂x (x,y+ηv)∆y
∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x+ ∂u
∂x (x,y+ηv)i∆y∆x+ i∆y
+i∂v∂x (x+ξv,y+∆y)∆x+ ∂v
∂x (x,y+ηu)i∆y∆x+ i∆y
=∂u∂x
(x,y)+ i∂v∂x
(x,y)
(Technical argument at end omitted.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇐” (concl.)
lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y+
∂u∂y (x,y+ηu)∆y
∆x+ i∆y+ i
∂v∂y (x,y+ηv)∆y
∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y
+− ∂v
∂x (x,y+ηu)∆y∆x+ i∆y
+ i∂u∂x (x,y+ηv)∆y
∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x+ ∂u
∂x (x,y+ηv)i∆y∆x+ i∆y
+i∂v∂x (x+ξv,y+∆y)∆x+ ∂v
∂x (x,y+ηu)i∆y∆x+ i∆y
=∂u∂x
(x,y)+ i∂v∂x
(x,y)
(Technical argument at end omitted.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof of “⇐” (concl.)
lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y+
∂u∂y (x,y+ηu)∆y
∆x+ i∆y+ i
∂v∂y (x,y+ηv)∆y
∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x
∆x+ i∆y+ i
∂v∂x (x+ξv,y+∆y)∆x
∆x+ i∆y
+− ∂v
∂x (x,y+ηu)∆y∆x+ i∆y
+ i∂u∂x (x,y+ηv)∆y
∆x+ i∆y
= lim∆x,∆y→0
∂u∂x (x+ξu,y+∆y)∆x+ ∂u
∂x (x,y+ηv)i∆y∆x+ i∆y
+i∂v∂x (x+ξv,y+∆y)∆x+ ∂v
∂x (x,y+ηu)i∆y∆x+ i∆y
=∂u∂x
(x,y)+ i∂v∂x
(x,y)
(Technical argument at end omitted.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example.
The function f (z) = z3 is differentiable on C.
f (z) = z3 = (x+ iy)3
= x3 +3ix2y−3xy2− iy3
= x3−3xy2 + i(3x2y− y3)
∂u∂x
= 3x2−3y2 =∂v∂y
∂u∂y
= −6xy =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = z3 is differentiable on C.
f (z) = z3 = (x+ iy)3
= x3 +3ix2y−3xy2− iy3
= x3−3xy2 + i(3x2y− y3)
∂u∂x
= 3x2−3y2 =∂v∂y
∂u∂y
= −6xy =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = z3 is differentiable on C.
f (z) = z3
= (x+ iy)3
= x3 +3ix2y−3xy2− iy3
= x3−3xy2 + i(3x2y− y3)
∂u∂x
= 3x2−3y2 =∂v∂y
∂u∂y
= −6xy =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = z3 is differentiable on C.
f (z) = z3 = (x+ iy)3
= x3 +3ix2y−3xy2− iy3
= x3−3xy2 + i(3x2y− y3)
∂u∂x
= 3x2−3y2 =∂v∂y
∂u∂y
= −6xy =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = z3 is differentiable on C.
f (z) = z3 = (x+ iy)3
= x3 +3ix2y−3xy2− iy3
= x3−3xy2 + i(3x2y− y3)
∂u∂x
= 3x2−3y2 =∂v∂y
∂u∂y
= −6xy =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = z3 is differentiable on C.
f (z) = z3 = (x+ iy)3
= x3 +3ix2y−3xy2− iy3
= x3−3xy2 + i(3x2y− y3)
∂u∂x
= 3x2−3y2 =∂v∂y
∂u∂y
= −6xy =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = z3 is differentiable on C.
f (z) = z3 = (x+ iy)3
= x3 +3ix2y−3xy2− iy3
= x3−3xy2 + i(3x2y− y3)
∂u∂x
= 3x2−3y2 =∂v∂y
∂u∂y
= −6xy =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = z3 is differentiable on C.
f (z) = z3 = (x+ iy)3
= x3 +3ix2y−3xy2− iy3
= x3−3xy2 + i(3x2y− y3)
∂u∂x
= 3x2−3y2
=∂v∂y
∂u∂y
= −6xy =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = z3 is differentiable on C.
f (z) = z3 = (x+ iy)3
= x3 +3ix2y−3xy2− iy3
= x3−3xy2 + i(3x2y− y3)
∂u∂x
= 3x2−3y2 =∂v∂y
∂u∂y
= −6xy =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = z3 is differentiable on C.
f (z) = z3 = (x+ iy)3
= x3 +3ix2y−3xy2− iy3
= x3−3xy2 + i(3x2y− y3)
∂u∂x
= 3x2−3y2 =∂v∂y
∂u∂y
= −6xy =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = z3 is differentiable on C.
f (z) = z3 = (x+ iy)3
= x3 +3ix2y−3xy2− iy3
= x3−3xy2 + i(3x2y− y3)
∂u∂x
= 3x2−3y2 =∂v∂y
∂u∂y
= −6xy
=−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = z3 is differentiable on C.
f (z) = z3 = (x+ iy)3
= x3 +3ix2y−3xy2− iy3
= x3−3xy2 + i(3x2y− y3)
∂u∂x
= 3x2−3y2 =∂v∂y
∂u∂y
= −6xy =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = z3 is differentiable on C.
f (z) = z3 = (x+ iy)3
= x3 +3ix2y−3xy2− iy3
= x3−3xy2 + i(3x2y− y3)
∂u∂x
= 3x2−3y2 =∂v∂y
∂u∂y
= −6xy =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example.
The function f (z) = ez is differentiable on C.
f (z) = ez = ex+iy = exeiy = ex cos(y)+ iex sin(y)∂u∂x
= ex cos(y) =∂v∂y
∂u∂y
= −ex sin(y) =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = ez is differentiable on C.
f (z) = ez = ex+iy = exeiy = ex cos(y)+ iex sin(y)∂u∂x
= ex cos(y) =∂v∂y
∂u∂y
= −ex sin(y) =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = ez is differentiable on C.
f (z)
= ez = ex+iy = exeiy = ex cos(y)+ iex sin(y)∂u∂x
= ex cos(y) =∂v∂y
∂u∂y
= −ex sin(y) =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = ez is differentiable on C.
f (z) = ez
= ex+iy = exeiy = ex cos(y)+ iex sin(y)∂u∂x
= ex cos(y) =∂v∂y
∂u∂y
= −ex sin(y) =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = ez is differentiable on C.
f (z) = ez = ex+iy
= exeiy = ex cos(y)+ iex sin(y)∂u∂x
= ex cos(y) =∂v∂y
∂u∂y
= −ex sin(y) =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = ez is differentiable on C.
f (z) = ez = ex+iy = exeiy
= ex cos(y)+ iex sin(y)∂u∂x
= ex cos(y) =∂v∂y
∂u∂y
= −ex sin(y) =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = ez is differentiable on C.
f (z) = ez = ex+iy = exeiy = ex cos(y)+ iex sin(y)
∂u∂x
= ex cos(y) =∂v∂y
∂u∂y
= −ex sin(y) =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = ez is differentiable on C.
f (z) = ez = ex+iy = exeiy = ex cos(y)+ iex sin(y)∂u∂x
= ex cos(y) =∂v∂y
∂u∂y
= −ex sin(y) =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = ez is differentiable on C.
f (z) = ez = ex+iy = exeiy = ex cos(y)+ iex sin(y)∂u∂x
= ex cos(y)
=∂v∂y
∂u∂y
= −ex sin(y) =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = ez is differentiable on C.
f (z) = ez = ex+iy = exeiy = ex cos(y)+ iex sin(y)∂u∂x
= ex cos(y) =∂v∂y
∂u∂y
= −ex sin(y) =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = ez is differentiable on C.
f (z) = ez = ex+iy = exeiy = ex cos(y)+ iex sin(y)∂u∂x
= ex cos(y) =∂v∂y
∂u∂y
= −ex sin(y) =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = ez is differentiable on C.
f (z) = ez = ex+iy = exeiy = ex cos(y)+ iex sin(y)∂u∂x
= ex cos(y) =∂v∂y
∂u∂y
= −ex sin(y)
=−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = ez is differentiable on C.
f (z) = ez = ex+iy = exeiy = ex cos(y)+ iex sin(y)∂u∂x
= ex cos(y) =∂v∂y
∂u∂y
= −ex sin(y) =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = ez is differentiable on C.
f (z) = ez = ex+iy = exeiy = ex cos(y)+ iex sin(y)∂u∂x
= ex cos(y) =∂v∂y
∂u∂y
= −ex sin(y) =−∂v∂x
So by the Cauchy Riemann Equations, f is differentiable on C.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example.
The function f (z) = |z| is not differentiable at any z ∈ C.
f (z) = |z|=√
x2 + y2 + i ·0∂u∂x
=x√
x2 + y26= 0 (x 6= 0)
∂u∂y
=y√
x2 + y26= 0 (y 6= 0)
So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = |z| is not differentiable at any z ∈ C.
f (z) = |z|=√
x2 + y2 + i ·0∂u∂x
=x√
x2 + y26= 0 (x 6= 0)
∂u∂y
=y√
x2 + y26= 0 (y 6= 0)
So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = |z| is not differentiable at any z ∈ C.
f (z)
= |z|=√
x2 + y2 + i ·0∂u∂x
=x√
x2 + y26= 0 (x 6= 0)
∂u∂y
=y√
x2 + y26= 0 (y 6= 0)
So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = |z| is not differentiable at any z ∈ C.
f (z) = |z|
=√
x2 + y2 + i ·0∂u∂x
=x√
x2 + y26= 0 (x 6= 0)
∂u∂y
=y√
x2 + y26= 0 (y 6= 0)
So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = |z| is not differentiable at any z ∈ C.
f (z) = |z|=√
x2 + y2
+ i ·0∂u∂x
=x√
x2 + y26= 0 (x 6= 0)
∂u∂y
=y√
x2 + y26= 0 (y 6= 0)
So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = |z| is not differentiable at any z ∈ C.
f (z) = |z|=√
x2 + y2 + i ·0
∂u∂x
=x√
x2 + y26= 0 (x 6= 0)
∂u∂y
=y√
x2 + y26= 0 (y 6= 0)
So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = |z| is not differentiable at any z ∈ C.
f (z) = |z|=√
x2 + y2 + i ·0∂u∂x
=x√
x2 + y26= 0 (x 6= 0)
∂u∂y
=y√
x2 + y26= 0 (y 6= 0)
So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = |z| is not differentiable at any z ∈ C.
f (z) = |z|=√
x2 + y2 + i ·0∂u∂x
=x√
x2 + y2
6= 0 (x 6= 0)
∂u∂y
=y√
x2 + y26= 0 (y 6= 0)
So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = |z| is not differentiable at any z ∈ C.
f (z) = |z|=√
x2 + y2 + i ·0∂u∂x
=x√
x2 + y26= 0
(x 6= 0)
∂u∂y
=y√
x2 + y26= 0 (y 6= 0)
So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = |z| is not differentiable at any z ∈ C.
f (z) = |z|=√
x2 + y2 + i ·0∂u∂x
=x√
x2 + y26= 0 (x 6= 0)
∂u∂y
=y√
x2 + y26= 0 (y 6= 0)
So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = |z| is not differentiable at any z ∈ C.
f (z) = |z|=√
x2 + y2 + i ·0∂u∂x
=x√
x2 + y26= 0 (x 6= 0)
∂u∂y
=y√
x2 + y26= 0 (y 6= 0)
So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = |z| is not differentiable at any z ∈ C.
f (z) = |z|=√
x2 + y2 + i ·0∂u∂x
=x√
x2 + y26= 0 (x 6= 0)
∂u∂y
=y√
x2 + y2
6= 0 (y 6= 0)
So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = |z| is not differentiable at any z ∈ C.
f (z) = |z|=√
x2 + y2 + i ·0∂u∂x
=x√
x2 + y26= 0 (x 6= 0)
∂u∂y
=y√
x2 + y26= 0
(y 6= 0)
So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = |z| is not differentiable at any z ∈ C.
f (z) = |z|=√
x2 + y2 + i ·0∂u∂x
=x√
x2 + y26= 0 (x 6= 0)
∂u∂y
=y√
x2 + y26= 0 (y 6= 0)
So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = |z| is not differentiable at any z ∈ C.
f (z) = |z|=√
x2 + y2 + i ·0∂u∂x
=x√
x2 + y26= 0 (x 6= 0)
∂u∂y
=y√
x2 + y26= 0 (y 6= 0)
So f is not differentiable for z 6= 0
and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = |z| is not differentiable at any z ∈ C.
f (z) = |z|=√
x2 + y2 + i ·0∂u∂x
=x√
x2 + y26= 0 (x 6= 0)
∂u∂y
=y√
x2 + y26= 0 (y 6= 0)
So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem.
Cauchy-Riemann Equations revisited. Letf (z) = f
(reiθ)
= u(r,θ)+ iv(r,θ) be a function on an open domainthat does not contain zero and with continuous partial derivatives inthe underlying real variables. Then f is differentiable at z = reiθ ifand only if
r∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
.
The latter equations are called the Cauchy-Riemann Equations inpolar form.The value of the derivative is
f ′(z) = e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. Cauchy-Riemann Equations revisited.
Letf (z) = f
(reiθ)
= u(r,θ)+ iv(r,θ) be a function on an open domainthat does not contain zero and with continuous partial derivatives inthe underlying real variables. Then f is differentiable at z = reiθ ifand only if
r∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
.
The latter equations are called the Cauchy-Riemann Equations inpolar form.The value of the derivative is
f ′(z) = e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. Cauchy-Riemann Equations revisited. Letf (z) = f
(reiθ)
= u(r,θ)+ iv(r,θ) be a function on an open domainthat does not contain zero and with continuous partial derivatives inthe underlying real variables.
Then f is differentiable at z = reiθ ifand only if
r∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
.
The latter equations are called the Cauchy-Riemann Equations inpolar form.The value of the derivative is
f ′(z) = e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. Cauchy-Riemann Equations revisited. Letf (z) = f
(reiθ)
= u(r,θ)+ iv(r,θ) be a function on an open domainthat does not contain zero and with continuous partial derivatives inthe underlying real variables. Then f is differentiable at z = reiθ
ifand only if
r∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
.
The latter equations are called the Cauchy-Riemann Equations inpolar form.The value of the derivative is
f ′(z) = e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. Cauchy-Riemann Equations revisited. Letf (z) = f
(reiθ)
= u(r,θ)+ iv(r,θ) be a function on an open domainthat does not contain zero and with continuous partial derivatives inthe underlying real variables. Then f is differentiable at z = reiθ ifand only if
r∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
.
The latter equations are called the Cauchy-Riemann Equations inpolar form.The value of the derivative is
f ′(z) = e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. Cauchy-Riemann Equations revisited. Letf (z) = f
(reiθ)
= u(r,θ)+ iv(r,θ) be a function on an open domainthat does not contain zero and with continuous partial derivatives inthe underlying real variables. Then f is differentiable at z = reiθ ifand only if
r∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
.
The latter equations are called the Cauchy-Riemann Equations inpolar form.The value of the derivative is
f ′(z) = e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. Cauchy-Riemann Equations revisited. Letf (z) = f
(reiθ)
= u(r,θ)+ iv(r,θ) be a function on an open domainthat does not contain zero and with continuous partial derivatives inthe underlying real variables. Then f is differentiable at z = reiθ ifand only if
r∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
.
The latter equations are called the Cauchy-Riemann Equations inpolar form.
The value of the derivative is
f ′(z) = e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. Cauchy-Riemann Equations revisited. Letf (z) = f
(reiθ)
= u(r,θ)+ iv(r,θ) be a function on an open domainthat does not contain zero and with continuous partial derivatives inthe underlying real variables. Then f is differentiable at z = reiθ ifand only if
r∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
.
The latter equations are called the Cauchy-Riemann Equations inpolar form.The value of the derivative is
f ′(z) = e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps).
Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x
=x√
x2 + y2= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y
=y√
x2 + y2= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x
=∂ arctan
( yx
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)
=− yx2 + y2 =−sin(θ)
r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2
=−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y
=∂ arctan
( yx
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2
=cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.
∂ r∂x
=∂√
x2 + y2
∂x=
x√x2 + y2
= cos(θ)
∂ r∂y
=∂√
x2 + y2
∂y=
y√x2 + y2
= sin(θ)
∂θ
∂x=
∂ arctan( y
x
)∂x
=1
1+( y
x
)2
(− y
x2
)=− y
x2 + y2 =−sin(θ)r
∂θ
∂y=
∂ arctan( y
x
)∂y
=1
1+( y
x
)21x
=x
x2 + y2 =cos(θ)
r
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x
=∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)
∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y
=∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y
=∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x
=∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)
If r∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
.
Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂x
=∂u∂ r
∂ r∂x
+∂u∂θ
∂θ
∂x=
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)∂v∂y
=∂v∂ r
∂ r∂y
+∂v∂θ
∂θ
∂y=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂y
=∂u∂ r
∂ r∂y
+∂u∂θ
∂θ
∂y=
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
∂v∂x
=∂v∂ r
∂ r∂x
+∂v∂θ
∂θ
∂x=
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
)If r
∂u∂ r
=∂v∂θ
and∂u∂θ
=−r∂v∂ r
, then by direct substitution∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
. Conversely, if∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
, setting the
above equal to each other, we obtain
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
= −∂v∂ r
cos(θ)+∂v∂θ
sin(θ)r
Multiplying the first with r cos(θ), the second with r sin(θ) and
adding gives r∂u∂ r
=∂v∂θ
. Multiplying the first with −r sin(θ), the
second with r cos(θ) and adding gives∂u∂θ
=−r∂v∂ r
.The above shows that f is differentiable if and only if theCauchy-Riemann Equations in polar form are satisfied. (We omittedthe technical proof that continuity of the partial derivatives in onecoordinate system gives their continuity in the other.)Now we turn to the formula for the derivative.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)
=∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
= −∂v∂ r
cos(θ)+∂v∂θ
sin(θ)r
Multiplying the first with r cos(θ), the second with r sin(θ) and
adding gives r∂u∂ r
=∂v∂θ
. Multiplying the first with −r sin(θ), the
second with r cos(θ) and adding gives∂u∂θ
=−r∂v∂ r
.The above shows that f is differentiable if and only if theCauchy-Riemann Equations in polar form are satisfied. (We omittedthe technical proof that continuity of the partial derivatives in onecoordinate system gives their continuity in the other.)Now we turn to the formula for the derivative.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
= −∂v∂ r
cos(θ)+∂v∂θ
sin(θ)r
Multiplying the first with r cos(θ), the second with r sin(θ) and
adding gives r∂u∂ r
=∂v∂θ
. Multiplying the first with −r sin(θ), the
second with r cos(θ) and adding gives∂u∂θ
=−r∂v∂ r
.The above shows that f is differentiable if and only if theCauchy-Riemann Equations in polar form are satisfied. (We omittedthe technical proof that continuity of the partial derivatives in onecoordinate system gives their continuity in the other.)Now we turn to the formula for the derivative.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
= −∂v∂ r
cos(θ)+∂v∂θ
sin(θ)r
Multiplying the first with r cos(θ), the second with r sin(θ) and
adding gives r∂u∂ r
=∂v∂θ
. Multiplying the first with −r sin(θ), the
second with r cos(θ) and adding gives∂u∂θ
=−r∂v∂ r
.The above shows that f is differentiable if and only if theCauchy-Riemann Equations in polar form are satisfied. (We omittedthe technical proof that continuity of the partial derivatives in onecoordinate system gives their continuity in the other.)Now we turn to the formula for the derivative.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
= −∂v∂ r
cos(θ)+∂v∂θ
sin(θ)r
Multiplying the first with r cos(θ), the second with r sin(θ) and
adding gives r∂u∂ r
=∂v∂θ
. Multiplying the first with −r sin(θ), the
second with r cos(θ) and adding gives∂u∂θ
=−r∂v∂ r
.The above shows that f is differentiable if and only if theCauchy-Riemann Equations in polar form are satisfied. (We omittedthe technical proof that continuity of the partial derivatives in onecoordinate system gives their continuity in the other.)Now we turn to the formula for the derivative.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
= −∂v∂ r
cos(θ)+∂v∂θ
sin(θ)r
Multiplying the first with r cos(θ), the second with r sin(θ) and
adding gives r∂u∂ r
=∂v∂θ
.
Multiplying the first with −r sin(θ), the
second with r cos(θ) and adding gives∂u∂θ
=−r∂v∂ r
.The above shows that f is differentiable if and only if theCauchy-Riemann Equations in polar form are satisfied. (We omittedthe technical proof that continuity of the partial derivatives in onecoordinate system gives their continuity in the other.)Now we turn to the formula for the derivative.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
= −∂v∂ r
cos(θ)+∂v∂θ
sin(θ)r
Multiplying the first with r cos(θ), the second with r sin(θ) and
adding gives r∂u∂ r
=∂v∂θ
. Multiplying the first with −r sin(θ), the
second with r cos(θ) and adding gives∂u∂θ
=−r∂v∂ r
.
The above shows that f is differentiable if and only if theCauchy-Riemann Equations in polar form are satisfied. (We omittedthe technical proof that continuity of the partial derivatives in onecoordinate system gives their continuity in the other.)Now we turn to the formula for the derivative.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
= −∂v∂ r
cos(θ)+∂v∂θ
sin(θ)r
Multiplying the first with r cos(θ), the second with r sin(θ) and
adding gives r∂u∂ r
=∂v∂θ
. Multiplying the first with −r sin(θ), the
second with r cos(θ) and adding gives∂u∂θ
=−r∂v∂ r
.The above shows that f is differentiable if and only if theCauchy-Riemann Equations in polar form are satisfied.
(We omittedthe technical proof that continuity of the partial derivatives in onecoordinate system gives their continuity in the other.)Now we turn to the formula for the derivative.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
= −∂v∂ r
cos(θ)+∂v∂θ
sin(θ)r
Multiplying the first with r cos(θ), the second with r sin(θ) and
adding gives r∂u∂ r
=∂v∂θ
. Multiplying the first with −r sin(θ), the
second with r cos(θ) and adding gives∂u∂θ
=−r∂v∂ r
.The above shows that f is differentiable if and only if theCauchy-Riemann Equations in polar form are satisfied. (We omittedthe technical proof that continuity of the partial derivatives in onecoordinate system gives their continuity in the other.)
Now we turn to the formula for the derivative.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)=
∂v∂ r
sin(θ)+∂v∂θ
cos(θ)r
∂u∂ r
sin(θ)+∂u∂θ
cos(θ)r
= −∂v∂ r
cos(θ)+∂v∂θ
sin(θ)r
Multiplying the first with r cos(θ), the second with r sin(θ) and
adding gives r∂u∂ r
=∂v∂θ
. Multiplying the first with −r sin(θ), the
second with r cos(θ) and adding gives∂u∂θ
=−r∂v∂ r
.The above shows that f is differentiable if and only if theCauchy-Riemann Equations in polar form are satisfied. (We omittedthe technical proof that continuity of the partial derivatives in onecoordinate system gives their continuity in the other.)Now we turn to the formula for the derivative.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
f ′(z) =∂u∂x
+ i∂v∂x
=∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)+i(
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
))=
∂u∂ r
cos(θ)+∂v∂ r
sin(θ)+ i(
∂v∂ r
cos(θ)+∂u∂ r
(−sin(θ)))
=∂u∂ r
cos(−θ)+∂u∂ r
isin(−θ)+ i∂v∂ r
isin(−θ)+ i∂v∂ r
cos(−θ)
= e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
f ′(z)
=∂u∂x
+ i∂v∂x
=∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)+i(
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
))=
∂u∂ r
cos(θ)+∂v∂ r
sin(θ)+ i(
∂v∂ r
cos(θ)+∂u∂ r
(−sin(θ)))
=∂u∂ r
cos(−θ)+∂u∂ r
isin(−θ)+ i∂v∂ r
isin(−θ)+ i∂v∂ r
cos(−θ)
= e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
f ′(z) =∂u∂x
+ i∂v∂x
=∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)+i(
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
))=
∂u∂ r
cos(θ)+∂v∂ r
sin(θ)+ i(
∂v∂ r
cos(θ)+∂u∂ r
(−sin(θ)))
=∂u∂ r
cos(−θ)+∂u∂ r
isin(−θ)+ i∂v∂ r
isin(−θ)+ i∂v∂ r
cos(−θ)
= e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
f ′(z) =∂u∂x
+ i∂v∂x
=∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)
+i(
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
))=
∂u∂ r
cos(θ)+∂v∂ r
sin(θ)+ i(
∂v∂ r
cos(θ)+∂u∂ r
(−sin(θ)))
=∂u∂ r
cos(−θ)+∂u∂ r
isin(−θ)+ i∂v∂ r
isin(−θ)+ i∂v∂ r
cos(−θ)
= e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
f ′(z) =∂u∂x
+ i∂v∂x
=∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)+i(
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
))
=∂u∂ r
cos(θ)+∂v∂ r
sin(θ)+ i(
∂v∂ r
cos(θ)+∂u∂ r
(−sin(θ)))
=∂u∂ r
cos(−θ)+∂u∂ r
isin(−θ)+ i∂v∂ r
isin(−θ)+ i∂v∂ r
cos(−θ)
= e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
f ′(z) =∂u∂x
+ i∂v∂x
=∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)+i(
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
))=
∂u∂ r
cos(θ)+∂v∂ r
sin(θ)
+ i(
∂v∂ r
cos(θ)+∂u∂ r
(−sin(θ)))
=∂u∂ r
cos(−θ)+∂u∂ r
isin(−θ)+ i∂v∂ r
isin(−θ)+ i∂v∂ r
cos(−θ)
= e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
f ′(z) =∂u∂x
+ i∂v∂x
=∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)+i(
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
))=
∂u∂ r
cos(θ)+∂v∂ r
sin(θ)+ i(
∂v∂ r
cos(θ)+∂u∂ r
(−sin(θ)))
=∂u∂ r
cos(−θ)+∂u∂ r
isin(−θ)+ i∂v∂ r
isin(−θ)+ i∂v∂ r
cos(−θ)
= e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
f ′(z) =∂u∂x
+ i∂v∂x
=∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)+i(
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
))=
∂u∂ r
cos(θ)+∂v∂ r
sin(θ)+ i(
∂v∂ r
cos(θ)+∂u∂ r
(−sin(θ)))
=∂u∂ r
cos(−θ)+∂u∂ r
isin(−θ)+ i∂v∂ r
isin(−θ)+ i∂v∂ r
cos(−θ)
= e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
f ′(z) =∂u∂x
+ i∂v∂x
=∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)+i(
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
))=
∂u∂ r
cos(θ)+∂v∂ r
sin(θ)+ i(
∂v∂ r
cos(θ)+∂u∂ r
(−sin(θ)))
=∂u∂ r
cos(−θ)+∂u∂ r
isin(−θ)+ i∂v∂ r
isin(−θ)+ i∂v∂ r
cos(−θ)
= e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
f ′(z) =∂u∂x
+ i∂v∂x
=∂u∂ r
cos(θ)+∂u∂θ
(−sin(θ)
r
)+i(
∂v∂ r
cos(θ)+∂v∂θ
(−sin(θ)
r
))=
∂u∂ r
cos(θ)+∂v∂ r
sin(θ)+ i(
∂v∂ r
cos(θ)+∂u∂ r
(−sin(θ)))
=∂u∂ r
cos(−θ)+∂u∂ r
isin(−θ)+ i∂v∂ r
isin(−θ)+ i∂v∂ r
cos(−θ)
= e−iθ(
∂u∂ r
(r,θ)+ i∂v∂ r
(r,θ))
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example.
The function f (z) = f(reiθ)
= r12 ei θ
2 is differentiable onany set r > δ ≥ 0, α < θ < α +2π .
f (z) = r12 ei θ
2
= r12 cos
(θ
2
)+ ir
12 sin
(θ
2
)r
∂u∂ r
= r12
r−12 cos
(θ
2
)=
12
r12 cos
(θ
2
)=
∂v∂θ
− r∂v∂ r
= −r12
r−12 sin
(θ
2
)= −1
2r
12 sin
(θ
2
)=
∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(reiθ)
= r12 ei θ
2 is differentiable onany set r > δ ≥ 0, α < θ < α +2π .
f (z) = r12 ei θ
2
= r12 cos
(θ
2
)+ ir
12 sin
(θ
2
)r
∂u∂ r
= r12
r−12 cos
(θ
2
)=
12
r12 cos
(θ
2
)=
∂v∂θ
− r∂v∂ r
= −r12
r−12 sin
(θ
2
)= −1
2r
12 sin
(θ
2
)=
∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(reiθ)
= r12 ei θ
2 is differentiable onany set r > δ ≥ 0, α < θ < α +2π .
f (z) = r12 ei θ
2
= r12 cos
(θ
2
)+ ir
12 sin
(θ
2
)r
∂u∂ r
= r12
r−12 cos
(θ
2
)=
12
r12 cos
(θ
2
)=
∂v∂θ
− r∂v∂ r
= −r12
r−12 sin
(θ
2
)= −1
2r
12 sin
(θ
2
)=
∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(reiθ)
= r12 ei θ
2 is differentiable onany set r > δ ≥ 0, α < θ < α +2π .
f (z) = r12 ei θ
2
= r12 cos
(θ
2
)+ ir
12 sin
(θ
2
)
r∂u∂ r
= r12
r−12 cos
(θ
2
)=
12
r12 cos
(θ
2
)=
∂v∂θ
− r∂v∂ r
= −r12
r−12 sin
(θ
2
)= −1
2r
12 sin
(θ
2
)=
∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(reiθ)
= r12 ei θ
2 is differentiable onany set r > δ ≥ 0, α < θ < α +2π .
f (z) = r12 ei θ
2
= r12 cos
(θ
2
)+ ir
12 sin
(θ
2
)r
∂u∂ r
= r12
r−12 cos
(θ
2
)
=12
r12 cos
(θ
2
)=
∂v∂θ
− r∂v∂ r
= −r12
r−12 sin
(θ
2
)= −1
2r
12 sin
(θ
2
)=
∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(reiθ)
= r12 ei θ
2 is differentiable onany set r > δ ≥ 0, α < θ < α +2π .
f (z) = r12 ei θ
2
= r12 cos
(θ
2
)+ ir
12 sin
(θ
2
)r
∂u∂ r
= r12
r−12 cos
(θ
2
)=
12
r12 cos
(θ
2
)
=∂v∂θ
− r∂v∂ r
= −r12
r−12 sin
(θ
2
)= −1
2r
12 sin
(θ
2
)=
∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(reiθ)
= r12 ei θ
2 is differentiable onany set r > δ ≥ 0, α < θ < α +2π .
f (z) = r12 ei θ
2
= r12 cos
(θ
2
)+ ir
12 sin
(θ
2
)r
∂u∂ r
= r12
r−12 cos
(θ
2
)=
12
r12 cos
(θ
2
)=
∂v∂θ
− r∂v∂ r
= −r12
r−12 sin
(θ
2
)= −1
2r
12 sin
(θ
2
)=
∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(reiθ)
= r12 ei θ
2 is differentiable onany set r > δ ≥ 0, α < θ < α +2π .
f (z) = r12 ei θ
2
= r12 cos
(θ
2
)+ ir
12 sin
(θ
2
)r
∂u∂ r
= r12
r−12 cos
(θ
2
)=
12
r12 cos
(θ
2
)=
∂v∂θ
− r∂v∂ r
= −r12
r−12 sin
(θ
2
)
= −12
r12 sin
(θ
2
)=
∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(reiθ)
= r12 ei θ
2 is differentiable onany set r > δ ≥ 0, α < θ < α +2π .
f (z) = r12 ei θ
2
= r12 cos
(θ
2
)+ ir
12 sin
(θ
2
)r
∂u∂ r
= r12
r−12 cos
(θ
2
)=
12
r12 cos
(θ
2
)=
∂v∂θ
− r∂v∂ r
= −r12
r−12 sin
(θ
2
)= −1
2r
12 sin
(θ
2
)
=∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(reiθ)
= r12 ei θ
2 is differentiable onany set r > δ ≥ 0, α < θ < α +2π .
f (z) = r12 ei θ
2
= r12 cos
(θ
2
)+ ir
12 sin
(θ
2
)r
∂u∂ r
= r12
r−12 cos
(θ
2
)=
12
r12 cos
(θ
2
)=
∂v∂θ
− r∂v∂ r
= −r12
r−12 sin
(θ
2
)= −1
2r
12 sin
(θ
2
)=
∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example.
The function f (z) = f(
reiθ)
= ln(r)+ iθ is differentiableon any set r > δ ≥ 0, α < θ < α +2π .
f (z) = ln(r)+ iθ
r∂u∂ r
= r1r
= 1 =∂v∂θ
−r∂v∂ r
= 0 =∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(
reiθ)
= ln(r)+ iθ is differentiableon any set r > δ ≥ 0, α < θ < α +2π .
f (z) = ln(r)+ iθ
r∂u∂ r
= r1r
= 1 =∂v∂θ
−r∂v∂ r
= 0 =∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(
reiθ)
= ln(r)+ iθ is differentiableon any set r > δ ≥ 0, α < θ < α +2π .
f (z) = ln(r)+ iθ
r∂u∂ r
= r1r
= 1 =∂v∂θ
−r∂v∂ r
= 0 =∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(
reiθ)
= ln(r)+ iθ is differentiableon any set r > δ ≥ 0, α < θ < α +2π .
f (z) = ln(r)+ iθ
r∂u∂ r
= r1r
= 1 =∂v∂θ
−r∂v∂ r
= 0 =∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(
reiθ)
= ln(r)+ iθ is differentiableon any set r > δ ≥ 0, α < θ < α +2π .
f (z) = ln(r)+ iθ
r∂u∂ r
= r1r
= 1 =∂v∂θ
−r∂v∂ r
= 0 =∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(
reiθ)
= ln(r)+ iθ is differentiableon any set r > δ ≥ 0, α < θ < α +2π .
f (z) = ln(r)+ iθ
r∂u∂ r
= r1r
= 1
=∂v∂θ
−r∂v∂ r
= 0 =∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(
reiθ)
= ln(r)+ iθ is differentiableon any set r > δ ≥ 0, α < θ < α +2π .
f (z) = ln(r)+ iθ
r∂u∂ r
= r1r
= 1 =∂v∂θ
−r∂v∂ r
= 0 =∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(
reiθ)
= ln(r)+ iθ is differentiableon any set r > δ ≥ 0, α < θ < α +2π .
f (z) = ln(r)+ iθ
r∂u∂ r
= r1r
= 1 =∂v∂θ
−r∂v∂ r
= 0 =∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(
reiθ)
= ln(r)+ iθ is differentiableon any set r > δ ≥ 0, α < θ < α +2π .
f (z) = ln(r)+ iθ
r∂u∂ r
= r1r
= 1 =∂v∂θ
−r∂v∂ r
= 0
=∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The function f (z) = f(
reiθ)
= ln(r)+ iθ is differentiableon any set r > δ ≥ 0, α < θ < α +2π .
f (z) = ln(r)+ iθ
r∂u∂ r
= r1r
= 1 =∂v∂θ
−r∂v∂ r
= 0 =∂u∂θ
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem.
If f is analytic on a connected domain D and f ′(z) = 0 forevery z in the domain, then f must be constant.
Proof. Via the Fundamental Theorem of Calculus, we obtain
u(x1,y0)−u(x0,y0) =∫ x1
x0
∂u∂x
(x,y0) dx
u(x0,y1)−u(x0,y0) =∫ y1
y0
∂u∂y
(x0,y) dy
So, because∂u∂x
=∂u∂y
= 0, we conclude that u does not change along
horizontal or vertical lines. Similarly v does not change alonghorizontal or vertical lines. Hence f does not change along horizontalor vertical lines.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If f is analytic on a connected domain D and f ′(z) = 0 forevery z in the domain, then f must be constant.
Proof. Via the Fundamental Theorem of Calculus, we obtain
u(x1,y0)−u(x0,y0) =∫ x1
x0
∂u∂x
(x,y0) dx
u(x0,y1)−u(x0,y0) =∫ y1
y0
∂u∂y
(x0,y) dy
So, because∂u∂x
=∂u∂y
= 0, we conclude that u does not change along
horizontal or vertical lines. Similarly v does not change alonghorizontal or vertical lines. Hence f does not change along horizontalor vertical lines.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If f is analytic on a connected domain D and f ′(z) = 0 forevery z in the domain, then f must be constant.
Proof.
Via the Fundamental Theorem of Calculus, we obtain
u(x1,y0)−u(x0,y0) =∫ x1
x0
∂u∂x
(x,y0) dx
u(x0,y1)−u(x0,y0) =∫ y1
y0
∂u∂y
(x0,y) dy
So, because∂u∂x
=∂u∂y
= 0, we conclude that u does not change along
horizontal or vertical lines. Similarly v does not change alonghorizontal or vertical lines. Hence f does not change along horizontalor vertical lines.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If f is analytic on a connected domain D and f ′(z) = 0 forevery z in the domain, then f must be constant.
Proof. Via the Fundamental Theorem of Calculus, we obtain
u(x1,y0)−u(x0,y0) =∫ x1
x0
∂u∂x
(x,y0) dx
u(x0,y1)−u(x0,y0) =∫ y1
y0
∂u∂y
(x0,y) dy
So, because∂u∂x
=∂u∂y
= 0, we conclude that u does not change along
horizontal or vertical lines. Similarly v does not change alonghorizontal or vertical lines. Hence f does not change along horizontalor vertical lines.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If f is analytic on a connected domain D and f ′(z) = 0 forevery z in the domain, then f must be constant.
Proof. Via the Fundamental Theorem of Calculus, we obtain
u(x1,y0)−u(x0,y0) =∫ x1
x0
∂u∂x
(x,y0) dx
u(x0,y1)−u(x0,y0) =∫ y1
y0
∂u∂y
(x0,y) dy
So, because∂u∂x
=∂u∂y
= 0, we conclude that u does not change along
horizontal or vertical lines. Similarly v does not change alonghorizontal or vertical lines. Hence f does not change along horizontalor vertical lines.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If f is analytic on a connected domain D and f ′(z) = 0 forevery z in the domain, then f must be constant.
Proof. Via the Fundamental Theorem of Calculus, we obtain
u(x1,y0)−u(x0,y0) =∫ x1
x0
∂u∂x
(x,y0) dx
u(x0,y1)−u(x0,y0) =∫ y1
y0
∂u∂y
(x0,y) dy
So, because∂u∂x
=∂u∂y
= 0, we conclude that u does not change along
horizontal or vertical lines. Similarly v does not change alonghorizontal or vertical lines. Hence f does not change along horizontalor vertical lines.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If f is analytic on a connected domain D and f ′(z) = 0 forevery z in the domain, then f must be constant.
Proof. Via the Fundamental Theorem of Calculus, we obtain
u(x1,y0)−u(x0,y0) =∫ x1
x0
∂u∂x
(x,y0) dx
u(x0,y1)−u(x0,y0) =∫ y1
y0
∂u∂y
(x0,y) dy
So, because∂u∂x
=∂u∂y
= 0
, we conclude that u does not change along
horizontal or vertical lines. Similarly v does not change alonghorizontal or vertical lines. Hence f does not change along horizontalor vertical lines.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If f is analytic on a connected domain D and f ′(z) = 0 forevery z in the domain, then f must be constant.
Proof. Via the Fundamental Theorem of Calculus, we obtain
u(x1,y0)−u(x0,y0) =∫ x1
x0
∂u∂x
(x,y0) dx
u(x0,y1)−u(x0,y0) =∫ y1
y0
∂u∂y
(x0,y) dy
So, because∂u∂x
=∂u∂y
= 0, we conclude that u does not change along
horizontal or vertical lines.
Similarly v does not change alonghorizontal or vertical lines. Hence f does not change along horizontalor vertical lines.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If f is analytic on a connected domain D and f ′(z) = 0 forevery z in the domain, then f must be constant.
Proof. Via the Fundamental Theorem of Calculus, we obtain
u(x1,y0)−u(x0,y0) =∫ x1
x0
∂u∂x
(x,y0) dx
u(x0,y1)−u(x0,y0) =∫ y1
y0
∂u∂y
(x0,y) dy
So, because∂u∂x
=∂u∂y
= 0, we conclude that u does not change along
horizontal or vertical lines. Similarly v does not change alonghorizontal or vertical lines.
Hence f does not change along horizontalor vertical lines.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If f is analytic on a connected domain D and f ′(z) = 0 forevery z in the domain, then f must be constant.
Proof. Via the Fundamental Theorem of Calculus, we obtain
u(x1,y0)−u(x0,y0) =∫ x1
x0
∂u∂x
(x,y0) dx
u(x0,y1)−u(x0,y0) =∫ y1
y0
∂u∂y
(x0,y) dy
So, because∂u∂x
=∂u∂y
= 0, we conclude that u does not change along
horizontal or vertical lines. Similarly v does not change alonghorizontal or vertical lines. Hence f does not change along horizontalor vertical lines.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
�
C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines
, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proof.
up
uq
� C
Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example.
Determine where f (x) =3z2 +5
(z3−2)2 (z4 +16)3 is analytic.
Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3
√2, 3√
2ei 2π
3 , 3√
2ei 4π
3 andz = 2ei π
4 ,2ei 3π
4 ,2ei 5π
4 ,2ei 7π
4 .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Determine where f (x) =3z2 +5
(z3−2)2 (z4 +16)3 is analytic.
Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3
√2, 3√
2ei 2π
3 , 3√
2ei 4π
3 andz = 2ei π
4 ,2ei 3π
4 ,2ei 5π
4 ,2ei 7π
4 .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Determine where f (x) =3z2 +5
(z3−2)2 (z4 +16)3 is analytic.
Note that there is no need to compute the derivative
or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3
√2, 3√
2ei 2π
3 , 3√
2ei 4π
3 andz = 2ei π
4 ,2ei 3π
4 ,2ei 5π
4 ,2ei 7π
4 .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Determine where f (x) =3z2 +5
(z3−2)2 (z4 +16)3 is analytic.
Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations.
(Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3
√2, 3√
2ei 2π
3 , 3√
2ei 4π
3 andz = 2ei π
4 ,2ei 3π
4 ,2ei 5π
4 ,2ei 7π
4 .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Determine where f (x) =3z2 +5
(z3−2)2 (z4 +16)3 is analytic.
Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.)
By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3
√2, 3√
2ei 2π
3 , 3√
2ei 4π
3 andz = 2ei π
4 ,2ei 3π
4 ,2ei 5π
4 ,2ei 7π
4 .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Determine where f (x) =3z2 +5
(z3−2)2 (z4 +16)3 is analytic.
Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain.
So itis analytic except at z = 3
√2, 3√
2ei 2π
3 , 3√
2ei 4π
3 andz = 2ei π
4 ,2ei 3π
4 ,2ei 5π
4 ,2ei 7π
4 .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Determine where f (x) =3z2 +5
(z3−2)2 (z4 +16)3 is analytic.
Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z =
3√
2, 3√
2ei 2π
3 , 3√
2ei 4π
3 andz = 2ei π
4 ,2ei 3π
4 ,2ei 5π
4 ,2ei 7π
4 .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Determine where f (x) =3z2 +5
(z3−2)2 (z4 +16)3 is analytic.
Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3
√2
, 3√
2ei 2π
3 , 3√
2ei 4π
3 andz = 2ei π
4 ,2ei 3π
4 ,2ei 5π
4 ,2ei 7π
4 .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Determine where f (x) =3z2 +5
(z3−2)2 (z4 +16)3 is analytic.
Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3
√2, 3√
2ei 2π
3
, 3√
2ei 4π
3 andz = 2ei π
4 ,2ei 3π
4 ,2ei 5π
4 ,2ei 7π
4 .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Determine where f (x) =3z2 +5
(z3−2)2 (z4 +16)3 is analytic.
Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3
√2, 3√
2ei 2π
3 , 3√
2ei 4π
3
andz = 2ei π
4 ,2ei 3π
4 ,2ei 5π
4 ,2ei 7π
4 .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Determine where f (x) =3z2 +5
(z3−2)2 (z4 +16)3 is analytic.
Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3
√2, 3√
2ei 2π
3 , 3√
2ei 4π
3 andz =
2ei π
4 ,2ei 3π
4 ,2ei 5π
4 ,2ei 7π
4 .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Determine where f (x) =3z2 +5
(z3−2)2 (z4 +16)3 is analytic.
Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3
√2, 3√
2ei 2π
3 , 3√
2ei 4π
3 andz = 2ei π
4
,2ei 3π
4 ,2ei 5π
4 ,2ei 7π
4 .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Determine where f (x) =3z2 +5
(z3−2)2 (z4 +16)3 is analytic.
Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3
√2, 3√
2ei 2π
3 , 3√
2ei 4π
3 andz = 2ei π
4 ,2ei 3π
4
,2ei 5π
4 ,2ei 7π
4 .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Determine where f (x) =3z2 +5
(z3−2)2 (z4 +16)3 is analytic.
Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3
√2, 3√
2ei 2π
3 , 3√
2ei 4π
3 andz = 2ei π
4 ,2ei 3π
4 ,2ei 5π
4
,2ei 7π
4 .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Determine where f (x) =3z2 +5
(z3−2)2 (z4 +16)3 is analytic.
Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3
√2, 3√
2ei 2π
3 , 3√
2ei 4π
3 andz = 2ei π
4 ,2ei 3π
4 ,2ei 5π
4 ,2ei 7π
4 .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition.
Let f be an analytic function on a connected domain sothat f is analytic, too. Then f is constant.
Proof. Applying the Cauchy Riemann Equations to f = u+ iv and
f = u− iv, we obtain∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
as well as∂u∂x
=−∂v∂y
and∂u∂y
=∂v∂x
. But then all partial derivatives are zero. Hence f ′ is
zero and f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat f is analytic, too.
Then f is constant.
Proof. Applying the Cauchy Riemann Equations to f = u+ iv and
f = u− iv, we obtain∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
as well as∂u∂x
=−∂v∂y
and∂u∂y
=∂v∂x
. But then all partial derivatives are zero. Hence f ′ is
zero and f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat f is analytic, too. Then f is constant.
Proof. Applying the Cauchy Riemann Equations to f = u+ iv and
f = u− iv, we obtain∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
as well as∂u∂x
=−∂v∂y
and∂u∂y
=∂v∂x
. But then all partial derivatives are zero. Hence f ′ is
zero and f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat f is analytic, too. Then f is constant.
Proof.
Applying the Cauchy Riemann Equations to f = u+ iv and
f = u− iv, we obtain∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
as well as∂u∂x
=−∂v∂y
and∂u∂y
=∂v∂x
. But then all partial derivatives are zero. Hence f ′ is
zero and f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat f is analytic, too. Then f is constant.
Proof. Applying the Cauchy Riemann Equations to f = u+ iv and
f = u− iv, we obtain∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
as well as∂u∂x
=−∂v∂y
and∂u∂y
=∂v∂x
. But then all partial derivatives are zero. Hence f ′ is
zero and f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat f is analytic, too. Then f is constant.
Proof. Applying the Cauchy Riemann Equations to f = u+ iv and
f = u− iv, we obtain∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
as well as∂u∂x
=−∂v∂y
and∂u∂y
=∂v∂x
. But then all partial derivatives are zero. Hence f ′ is
zero and f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat f is analytic, too. Then f is constant.
Proof. Applying the Cauchy Riemann Equations to f = u+ iv and
f = u− iv, we obtain∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
as well as∂u∂x
=−∂v∂y
and∂u∂y
=∂v∂x
. But then all partial derivatives are zero. Hence f ′ is
zero and f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat f is analytic, too. Then f is constant.
Proof. Applying the Cauchy Riemann Equations to f = u+ iv and
f = u− iv, we obtain∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
as well as∂u∂x
=−∂v∂y
and∂u∂y
=∂v∂x
.
But then all partial derivatives are zero. Hence f ′ is
zero and f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat f is analytic, too. Then f is constant.
Proof. Applying the Cauchy Riemann Equations to f = u+ iv and
f = u− iv, we obtain∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
as well as∂u∂x
=−∂v∂y
and∂u∂y
=∂v∂x
. But then all partial derivatives are zero.
Hence f ′ is
zero and f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat f is analytic, too. Then f is constant.
Proof. Applying the Cauchy Riemann Equations to f = u+ iv and
f = u− iv, we obtain∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
as well as∂u∂x
=−∂v∂y
and∂u∂y
=∂v∂x
. But then all partial derivatives are zero. Hence f ′ is
zero
and f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat f is analytic, too. Then f is constant.
Proof. Applying the Cauchy Riemann Equations to f = u+ iv and
f = u− iv, we obtain∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
as well as∂u∂x
=−∂v∂y
and∂u∂y
=∂v∂x
. But then all partial derivatives are zero. Hence f ′ is
zero and f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat f is analytic, too. Then f is constant.
Proof. Applying the Cauchy Riemann Equations to f = u+ iv and
f = u− iv, we obtain∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
as well as∂u∂x
=−∂v∂y
and∂u∂y
=∂v∂x
. But then all partial derivatives are zero. Hence f ′ is
zero and f is constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition.
Let f be an analytic function on a connected domain sothat |f | is constant. Then f is constant.
Proof. Because f f = |f |2 we have that f =|f |2
fis analytic. By the
preceding result, f must be constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat |f | is constant.
Then f is constant.
Proof. Because f f = |f |2 we have that f =|f |2
fis analytic. By the
preceding result, f must be constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat |f | is constant. Then f is constant.
Proof. Because f f = |f |2 we have that f =|f |2
fis analytic. By the
preceding result, f must be constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat |f | is constant. Then f is constant.
Proof.
Because f f = |f |2 we have that f =|f |2
fis analytic. By the
preceding result, f must be constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat |f | is constant. Then f is constant.
Proof. Because f f = |f |2
we have that f =|f |2
fis analytic. By the
preceding result, f must be constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat |f | is constant. Then f is constant.
Proof. Because f f = |f |2 we have that f =|f |2
fis analytic.
By the
preceding result, f must be constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat |f | is constant. Then f is constant.
Proof. Because f f = |f |2 we have that f =|f |2
fis analytic. By the
preceding result, f must be constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Proposition. Let f be an analytic function on a connected domain sothat |f | is constant. Then f is constant.
Proof. Because f f = |f |2 we have that f =|f |2
fis analytic. By the
preceding result, f must be constant.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition.
A real valued function u of two variables is calledharmonic if and only if it has continuous first and second order
partial derivatives and satisfies the equation∂ 2u∂x2 +
∂ 2u∂y2 = 0. The
above equation is called Laplace’s equation.
Laplace’s equation describes steady state heat transfer as well as wavemotion, because the (two-dimensional) wave equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂ 2u∂ t2
and the (two-dimensional) heat equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂u∂ t
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. A real valued function u of two variables is calledharmonic
if and only if it has continuous first and second order
partial derivatives and satisfies the equation∂ 2u∂x2 +
∂ 2u∂y2 = 0. The
above equation is called Laplace’s equation.
Laplace’s equation describes steady state heat transfer as well as wavemotion, because the (two-dimensional) wave equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂ 2u∂ t2
and the (two-dimensional) heat equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂u∂ t
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. A real valued function u of two variables is calledharmonic if and only if it has continuous first and second order
partial derivatives
and satisfies the equation∂ 2u∂x2 +
∂ 2u∂y2 = 0. The
above equation is called Laplace’s equation.
Laplace’s equation describes steady state heat transfer as well as wavemotion, because the (two-dimensional) wave equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂ 2u∂ t2
and the (two-dimensional) heat equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂u∂ t
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. A real valued function u of two variables is calledharmonic if and only if it has continuous first and second order
partial derivatives and satisfies the equation∂ 2u∂x2 +
∂ 2u∂y2 = 0.
The
above equation is called Laplace’s equation.
Laplace’s equation describes steady state heat transfer as well as wavemotion, because the (two-dimensional) wave equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂ 2u∂ t2
and the (two-dimensional) heat equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂u∂ t
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. A real valued function u of two variables is calledharmonic if and only if it has continuous first and second order
partial derivatives and satisfies the equation∂ 2u∂x2 +
∂ 2u∂y2 = 0. The
above equation is called Laplace’s equation.
Laplace’s equation describes steady state heat transfer as well as wavemotion, because the (two-dimensional) wave equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂ 2u∂ t2
and the (two-dimensional) heat equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂u∂ t
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. A real valued function u of two variables is calledharmonic if and only if it has continuous first and second order
partial derivatives and satisfies the equation∂ 2u∂x2 +
∂ 2u∂y2 = 0. The
above equation is called Laplace’s equation.
Laplace’s equation describes steady state heat transfer as well as wavemotion
, because the (two-dimensional) wave equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂ 2u∂ t2
and the (two-dimensional) heat equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂u∂ t
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. A real valued function u of two variables is calledharmonic if and only if it has continuous first and second order
partial derivatives and satisfies the equation∂ 2u∂x2 +
∂ 2u∂y2 = 0. The
above equation is called Laplace’s equation.
Laplace’s equation describes steady state heat transfer as well as wavemotion, because the (two-dimensional) wave equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂ 2u∂ t2
and the (two-dimensional) heat equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂u∂ t
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. A real valued function u of two variables is calledharmonic if and only if it has continuous first and second order
partial derivatives and satisfies the equation∂ 2u∂x2 +
∂ 2u∂y2 = 0. The
above equation is called Laplace’s equation.
Laplace’s equation describes steady state heat transfer as well as wavemotion, because the (two-dimensional) wave equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂ 2u∂ t2
and the (two-dimensional) heat equation is
∂ 2u∂x2 +
∂ 2u∂y2 = k
∂u∂ t
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example.
Show that the function u(x,y) = e−y sin(x) is harmonic for0 < x < π and y > 0.
∂ 2u∂x2 +
∂ 2u∂y2 = e−y(− sin(x)
)+ e−y sin(x)
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Show that the function u(x,y) = e−y sin(x) is harmonic for0 < x < π and y > 0.
∂ 2u∂x2 +
∂ 2u∂y2 = e−y(− sin(x)
)+ e−y sin(x)
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Show that the function u(x,y) = e−y sin(x) is harmonic for0 < x < π and y > 0.
∂ 2u∂x2 +
∂ 2u∂y2
= e−y(− sin(x))+ e−y sin(x)
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Show that the function u(x,y) = e−y sin(x) is harmonic for0 < x < π and y > 0.
∂ 2u∂x2 +
∂ 2u∂y2 = e−y(− sin(x)
)
+ e−y sin(x)
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Show that the function u(x,y) = e−y sin(x) is harmonic for0 < x < π and y > 0.
∂ 2u∂x2 +
∂ 2u∂y2 = e−y(− sin(x)
)+ e−y sin(x)
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Show that the function u(x,y) = e−y sin(x) is harmonic for0 < x < π and y > 0.
∂ 2u∂x2 +
∂ 2u∂y2 = e−y(− sin(x)
)+ e−y sin(x)
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem.
If the complex function f (z) = u(x,y)+ iv(x,y) is analyticon its domain D, then its real and imaginary parts are harmonic on D.
Proof. (Existence of the second derivatives will follow from latertheorems.)
∂ 2u∂x2 +
∂ 2u∂y2 =
∂
∂x∂u∂x
+∂
∂y∂u∂y
=∂
∂x∂v∂y
+∂
∂y
(−∂v
∂x
)=
∂ 2v∂x∂y
− ∂ 2v∂y∂x
= 0
Harmonicity of v is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If the complex function f (z) = u(x,y)+ iv(x,y) is analyticon its domain D, then its real and imaginary parts are harmonic on D.
Proof. (Existence of the second derivatives will follow from latertheorems.)
∂ 2u∂x2 +
∂ 2u∂y2 =
∂
∂x∂u∂x
+∂
∂y∂u∂y
=∂
∂x∂v∂y
+∂
∂y
(−∂v
∂x
)=
∂ 2v∂x∂y
− ∂ 2v∂y∂x
= 0
Harmonicity of v is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If the complex function f (z) = u(x,y)+ iv(x,y) is analyticon its domain D, then its real and imaginary parts are harmonic on D.
Proof.
(Existence of the second derivatives will follow from latertheorems.)
∂ 2u∂x2 +
∂ 2u∂y2 =
∂
∂x∂u∂x
+∂
∂y∂u∂y
=∂
∂x∂v∂y
+∂
∂y
(−∂v
∂x
)=
∂ 2v∂x∂y
− ∂ 2v∂y∂x
= 0
Harmonicity of v is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If the complex function f (z) = u(x,y)+ iv(x,y) is analyticon its domain D, then its real and imaginary parts are harmonic on D.
Proof. (Existence of the second derivatives will follow from latertheorems.)
∂ 2u∂x2 +
∂ 2u∂y2 =
∂
∂x∂u∂x
+∂
∂y∂u∂y
=∂
∂x∂v∂y
+∂
∂y
(−∂v
∂x
)=
∂ 2v∂x∂y
− ∂ 2v∂y∂x
= 0
Harmonicity of v is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If the complex function f (z) = u(x,y)+ iv(x,y) is analyticon its domain D, then its real and imaginary parts are harmonic on D.
Proof. (Existence of the second derivatives will follow from latertheorems.)
∂ 2u∂x2 +
∂ 2u∂y2
=∂
∂x∂u∂x
+∂
∂y∂u∂y
=∂
∂x∂v∂y
+∂
∂y
(−∂v
∂x
)=
∂ 2v∂x∂y
− ∂ 2v∂y∂x
= 0
Harmonicity of v is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If the complex function f (z) = u(x,y)+ iv(x,y) is analyticon its domain D, then its real and imaginary parts are harmonic on D.
Proof. (Existence of the second derivatives will follow from latertheorems.)
∂ 2u∂x2 +
∂ 2u∂y2 =
∂
∂x∂u∂x
+∂
∂y∂u∂y
=∂
∂x∂v∂y
+∂
∂y
(−∂v
∂x
)=
∂ 2v∂x∂y
− ∂ 2v∂y∂x
= 0
Harmonicity of v is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If the complex function f (z) = u(x,y)+ iv(x,y) is analyticon its domain D, then its real and imaginary parts are harmonic on D.
Proof. (Existence of the second derivatives will follow from latertheorems.)
∂ 2u∂x2 +
∂ 2u∂y2 =
∂
∂x∂u∂x
+∂
∂y∂u∂y
=∂
∂x∂v∂y
+∂
∂y
(−∂v
∂x
)
=∂ 2v
∂x∂y− ∂ 2v
∂y∂x= 0
Harmonicity of v is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If the complex function f (z) = u(x,y)+ iv(x,y) is analyticon its domain D, then its real and imaginary parts are harmonic on D.
Proof. (Existence of the second derivatives will follow from latertheorems.)
∂ 2u∂x2 +
∂ 2u∂y2 =
∂
∂x∂u∂x
+∂
∂y∂u∂y
=∂
∂x∂v∂y
+∂
∂y
(−∂v
∂x
)=
∂ 2v∂x∂y
− ∂ 2v∂y∂x
= 0
Harmonicity of v is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If the complex function f (z) = u(x,y)+ iv(x,y) is analyticon its domain D, then its real and imaginary parts are harmonic on D.
Proof. (Existence of the second derivatives will follow from latertheorems.)
∂ 2u∂x2 +
∂ 2u∂y2 =
∂
∂x∂u∂x
+∂
∂y∂u∂y
=∂
∂x∂v∂y
+∂
∂y
(−∂v
∂x
)=
∂ 2v∂x∂y
− ∂ 2v∂y∂x
= 0
Harmonicity of v is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If the complex function f (z) = u(x,y)+ iv(x,y) is analyticon its domain D, then its real and imaginary parts are harmonic on D.
Proof. (Existence of the second derivatives will follow from latertheorems.)
∂ 2u∂x2 +
∂ 2u∂y2 =
∂
∂x∂u∂x
+∂
∂y∂u∂y
=∂
∂x∂v∂y
+∂
∂y
(−∂v
∂x
)=
∂ 2v∂x∂y
− ∂ 2v∂y∂x
= 0
Harmonicity of v is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Theorem. If the complex function f (z) = u(x,y)+ iv(x,y) is analyticon its domain D, then its real and imaginary parts are harmonic on D.
Proof. (Existence of the second derivatives will follow from latertheorems.)
∂ 2u∂x2 +
∂ 2u∂y2 =
∂
∂x∂u∂x
+∂
∂y∂u∂y
=∂
∂x∂v∂y
+∂
∂y
(−∂v
∂x
)=
∂ 2v∂x∂y
− ∂ 2v∂y∂x
= 0
Harmonicity of v is proved similarly.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example.
The real and imaginary parts of f (z) =1z3 are harmonic
for (x,y) 6= (0,0).
Proof. f is analytic away from the origin!
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The real and imaginary parts of f (z) =1z3 are harmonic
for (x,y) 6= (0,0).
Proof. f is analytic away from the origin!
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The real and imaginary parts of f (z) =1z3 are harmonic
for (x,y) 6= (0,0).
Proof.
f is analytic away from the origin!
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The real and imaginary parts of f (z) =1z3 are harmonic
for (x,y) 6= (0,0).
Proof. f is analytic away from the origin!
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. The real and imaginary parts of f (z) =1z3 are harmonic
for (x,y) 6= (0,0).
Proof. f is analytic away from the origin!
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition.
Let u and v be harmonic on the domain D. The function vis the called a harmonic conjugate of u if and only if
∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
.
Theorem. The complex function f (z) = u(x,y)+ iv(x,y) is analytic onits domain D if and only if u and v are harmonic and v is a harmonicconjugate of u.
Proof. The proof that analyticity of f implies that u and v areharmonic and harmonic conjugates follows from the precedingtheorem and from the Cauchy-Riemann Equations. The proof of theconverse follows from the fact that the Cauchy-Riemann Equationsand continuous partial derivatives implied analyticity.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. Let u and v be harmonic on the domain D. The function vis the called a harmonic conjugate of u if and only if
∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
.
Theorem. The complex function f (z) = u(x,y)+ iv(x,y) is analytic onits domain D if and only if u and v are harmonic and v is a harmonicconjugate of u.
Proof. The proof that analyticity of f implies that u and v areharmonic and harmonic conjugates follows from the precedingtheorem and from the Cauchy-Riemann Equations. The proof of theconverse follows from the fact that the Cauchy-Riemann Equationsand continuous partial derivatives implied analyticity.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. Let u and v be harmonic on the domain D. The function vis the called a harmonic conjugate of u if and only if
∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
.
Theorem. The complex function f (z) = u(x,y)+ iv(x,y) is analytic onits domain D if and only if u and v are harmonic and v is a harmonicconjugate of u.
Proof. The proof that analyticity of f implies that u and v areharmonic and harmonic conjugates follows from the precedingtheorem and from the Cauchy-Riemann Equations. The proof of theconverse follows from the fact that the Cauchy-Riemann Equationsand continuous partial derivatives implied analyticity.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. Let u and v be harmonic on the domain D. The function vis the called a harmonic conjugate of u if and only if
∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
.
Theorem. The complex function f (z) = u(x,y)+ iv(x,y) is analytic onits domain D if and only if u and v are harmonic and v is a harmonicconjugate of u.
Proof. The proof that analyticity of f implies that u and v areharmonic and harmonic conjugates follows from the precedingtheorem and from the Cauchy-Riemann Equations. The proof of theconverse follows from the fact that the Cauchy-Riemann Equationsand continuous partial derivatives implied analyticity.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. Let u and v be harmonic on the domain D. The function vis the called a harmonic conjugate of u if and only if
∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
.
Theorem.
The complex function f (z) = u(x,y)+ iv(x,y) is analytic onits domain D if and only if u and v are harmonic and v is a harmonicconjugate of u.
Proof. The proof that analyticity of f implies that u and v areharmonic and harmonic conjugates follows from the precedingtheorem and from the Cauchy-Riemann Equations. The proof of theconverse follows from the fact that the Cauchy-Riemann Equationsand continuous partial derivatives implied analyticity.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. Let u and v be harmonic on the domain D. The function vis the called a harmonic conjugate of u if and only if
∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
.
Theorem. The complex function f (z) = u(x,y)+ iv(x,y) is analytic onits domain D if and only if u and v are harmonic and v is a harmonicconjugate of u.
Proof. The proof that analyticity of f implies that u and v areharmonic and harmonic conjugates follows from the precedingtheorem and from the Cauchy-Riemann Equations. The proof of theconverse follows from the fact that the Cauchy-Riemann Equationsand continuous partial derivatives implied analyticity.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. Let u and v be harmonic on the domain D. The function vis the called a harmonic conjugate of u if and only if
∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
.
Theorem. The complex function f (z) = u(x,y)+ iv(x,y) is analytic onits domain D if and only if u and v are harmonic and v is a harmonicconjugate of u.
Proof.
The proof that analyticity of f implies that u and v areharmonic and harmonic conjugates follows from the precedingtheorem and from the Cauchy-Riemann Equations. The proof of theconverse follows from the fact that the Cauchy-Riemann Equationsand continuous partial derivatives implied analyticity.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. Let u and v be harmonic on the domain D. The function vis the called a harmonic conjugate of u if and only if
∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
.
Theorem. The complex function f (z) = u(x,y)+ iv(x,y) is analytic onits domain D if and only if u and v are harmonic and v is a harmonicconjugate of u.
Proof. The proof that analyticity of f implies that u and v areharmonic and harmonic conjugates follows from the precedingtheorem and from the Cauchy-Riemann Equations.
The proof of theconverse follows from the fact that the Cauchy-Riemann Equationsand continuous partial derivatives implied analyticity.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. Let u and v be harmonic on the domain D. The function vis the called a harmonic conjugate of u if and only if
∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
.
Theorem. The complex function f (z) = u(x,y)+ iv(x,y) is analytic onits domain D if and only if u and v are harmonic and v is a harmonicconjugate of u.
Proof. The proof that analyticity of f implies that u and v areharmonic and harmonic conjugates follows from the precedingtheorem and from the Cauchy-Riemann Equations. The proof of theconverse follows from the fact that the Cauchy-Riemann Equationsand continuous partial derivatives implied analyticity.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Definition. Let u and v be harmonic on the domain D. The function vis the called a harmonic conjugate of u if and only if
∂u∂x
=∂v∂y
and∂u∂y
=−∂v∂x
.
Theorem. The complex function f (z) = u(x,y)+ iv(x,y) is analytic onits domain D if and only if u and v are harmonic and v is a harmonicconjugate of u.
Proof. The proof that analyticity of f implies that u and v areharmonic and harmonic conjugates follows from the precedingtheorem and from the Cauchy-Riemann Equations. The proof of theconverse follows from the fact that the Cauchy-Riemann Equationsand continuous partial derivatives implied analyticity.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example.
Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy =−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx =−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy =−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx =−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy =−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx =−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy =−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx =−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy =−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx =−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v
=∫
e−y cos(x) dy =−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx =−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy
=−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx =−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy =−e−y cos(x)
+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx =−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy =−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx =−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy =−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx =−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy =−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx =−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy =−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx =−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy =−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v
=∫
e−y sin(x) dx =−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy =−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx
=−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy =−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx =−e−y cos(x)
+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy =−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx =−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability
logo1
Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions
Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.
∂v∂y
=∂u∂x
= e−y cos(x)
v =∫
e−y cos(x) dy =−e−y cos(x)+ cy
∂v∂x
= −∂u∂y
= e−y sin(x)
v =∫
e−y sin(x) dx =−e−y cos(x)+ cx
So v(x,y) =−e−y cos(x).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Equivalent Conditions for Differentiability