COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX...
Transcript of COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX...
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 1/39
COMPLEX ANALYSIS
Complex Numbers
Complex numbers are useful in the fields such as:
� Solutions of some types of linear differential equations
� Electrical circuit analyses
� Inverse transformations
� Solutions of field problems
� etc.
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 2/39
The general form of a complex number z, is z = x + i y or z = a + i b
where
x = Re (z) is a real number and called as the real part of z .
y = Im (z) is a real number and called as the imaginary part of z .
Another representation of a complex number is its polar form which is obtained by
applying usual rectangular to polar coordinate transformation as
x = r cosθ & y = r sinθ
z = r [cosθ + i sinθ] = r ei θ = r θ
i = -1
r = |z| = = mod(z) is called as the modulus (magnitude) of z22 yx +
xy
θ = arctan = arg(z) is called as the argument (angle) of z
Using EulerUsing Euler’’s identity s identity eeiiθ = = coscosθ + i + i sinsinθ
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 3/39
y
x
y
z
x 0
r
θ
z–plane
The figure below denotes the geometric representation of a complex number in a
plane called complex z-plane , or Argand diagram , defined by a Cartesian
coordinate system whose abscissa is used to represent the real part of z, and
ordinate is used to represent the imaginary part of z.
[ ]
2 2
1
z = x + i y
r = x + y x = r cos( )
y = tan y = r sin( )
x
z = r cos( ) + i r sin( )
z = r cos( ) + i sin( )
θ
θ θ
θ θθ θ
−
z = r eiθ
z = r θ
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 4/39
Note the following :
Geometrically, the argument θ is the directed angle measured in radians from
positive x–axis in counterclockwise direction. For z = 0, this angle is undefined.
For a given z ≠ 0, this angle is determined only up to integer multiples of 2π.
The value of θ that lies in the interval –π < θ ≤ π is called as the principle value
of the argument of z and is denoted by Arg(z) , with capital A. Thus, – π < Arg(z) ≤ + π
It becomes extremely important to consider the quadrant of the z–plane in which
the point z lies when determining the value of the argument by using the above
equation.
The modulus r, as the distance of the point z to the origin, is a non-negative quantity;
i.e., r ≥ 0
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 5/39
Example:
Determine the modulus and argument of the following complex numbers.
z = - 1 + i ⇒
z = 3 - 4 i ⇒
z = - 5 - 12 i ⇒
211r =+= )(135rad2.3564
3π1
1arctanθ o≅=
−=
5169r =+= )53.1(rad0.92734
arctanθ o−−=
−=
1314425r =+= )112.6(rad1.9655
12arctanθ o−−=
−−=
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 6/39
There are two fundamental rules for the manipulation of complex numbers:
1. A complex number z = x + i y is zero iff both its real and imaginary parts are zero;
i.e., z = 0 iff x = 0 & y = 0
it follows that two complex numbers z1 = x1 + i y1 and z2 = x2 + i y2 are equal
iff both their real and imaginary parts are equal;
i.e., z1 = z2 iff x1 = x2 & y1 = y2
Example :
What are the x and y values that satisfy the equation
(x2y – 2) + i (x + 2xy – 5) = 0
x2y – 2 = 0 and x + 2xy – 5 = 0 → x = 1 and y = 2x = 4 and y = 1/8
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 7/39
2. Complex numbers obey the ordinary rules of algebra
(a + i b) ± (c + i d) = (a ± c) + i (b ± d)
(a + i b) (c + i d) = (a c – b d) + i (a d + b c)
(a + i b)2 = a2 – b2 + i (2 a b)
with the addition that
i2 = – 1 , i3 = – i , i4 = +1 , i5 = i
Addition and Subtraction
z = z1 ± z2 = (x1 ± x2) + i (y1 ± y2)
� Easier to perform in Cartesian form
� Similar to addition and subtraction of vectors in a plane
Triangle Inequality:
Length of any one side of a triangle is less than or equal to the sum of the lengths
of other two sides. That is r ≤ r1 + r2
, i6 = – 1 , i7 = – i , ..... modularity
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 8/39
Minus, Conjugate, Addition, and Subtraction in Polar Coordinates
x
y
z = a + b i
_
z = a - i b- z = - a - b i
x
y
z1
z2z1 + z2
- z2
z1 - z2
conjugate
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 9/39
Multiplication
z = z1 z2 = (x1 x2 – y1 y2) + i (x1 y2 + y1 x2) = r1r2 θ1 + θ2
r = r1 r2
θ = θ1 + θ2
� Easier to perform in polar form
� Not similar to multiplication of vectors in plane
( )( )
( )
1 1 1 1
2 2 2 2
1 2 1 2 1 2 1 2
z = r cos( ) + i sin( )
z = r cos( ) + i sin( )
z z = r r cos( + ) + i sin( + )
θ θ
θ θ
θ θ θ θ
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 10/39
Division
1
2
1 2 1 2 1 2 1 22 2 2 22 2 2 2
z x x + y y y x - x yz = = + i
z x + y x + y
r = r1 / r2
θ = θ1 – θ2
� Easier to perform in polar form
� Not similar to any operation of vectors in plane
[ ][ ]
[ ]
1 1 1 1
2 2 2 2
1 11 2 1 2 2
2 2
z = r cos( ) + i sin( )
z = r cos( ) + i sin( )
z r = cos( - ) + i sin( - ) , r 0
z r
θ θ
θ θ
θ θ θ θ ≠
212
1 θθ rr −=
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 11/39
Complex conjugate
The complex conjugate of a complex number z = x + i y is defined as_
z = x - i y
Another practical way to obtain division of two complex numbers is to use the
conjugate of the denominator:
Example:
2 2 2 2
2 2
a + i b a + i b c - i d a c + b d b c - a d = = + i
c + i d c + i d c - i d c + d c + d
c + d 0
≠
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 12/39
Note the following:
Letz = x + iy → = x – iy
then(x + iy)2 + 2(x – iy) = x2 – y2 + i2xy + 2x – i2y = –1 + i6
which gives the following two real equationsx2 – y2 + 2x = –1 → y = ± (x+1)
2xy – 2y = 6 → y(x–1) = 3whose solutions are found as
x = 2 & y = 3 and x = –2 & y = –1Hence, the required solution for z is found as
z1 = 2 + i3 and z2 = – 2 – i
2 z + z
Re(z)x2x = z + z ==→i2
z zIm(z)yi2y = z z
−==→−
222 |z| = y+ x=z z 2121 zz)z(z ±=± 2121 zz)z(z =2
1
2
1
zz
zz =
Example :
Solve the following (find z): i6 1 z2 z2 +−=+
z
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 13/39
Integer powers of complex numbers (de Moivre’s formu la)
The idea of product of two complex numbers can be extended to the product of n complex numbers
If all the complex numbers multiplied are the same, the above expression gives an important result
zn = rn einθ = rn [cosθ + i sinθ]n = rn [cos(nθ) + i sin(nθ)]
where n is either an integer or a rational number (that is, n=p/q where p and q areintegers) with the condition that z ≠ 0 for n = –1.
For r = 1, the expression reduces the form
(cosθ + i sinθ)n = cos(nθ) + i sin(nθ)
called as the de Moivre’s formula named after Abraham de Moivre (1667-1754)
∑
=…=…= =∏∏
=
+++
=
n
1kk
n21
θin
1k
k)θ...θi(θ
n21n21
n
1k
k erer rr z zzz
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 14/39
Example: Find z5 for z = 1 + i 3
Solution 1 : Taking the power of a binomial
( )55 2 3 4 5z = 1 + i 3 = 1 + 5 ( i 3) + 10 ( i 3) + 10 ( i 3) + 5 ( i 3) + ( i 3)
= 1 + i 5 3 - 30 - i 30 3 + 45 + i 9 3
= 16 - i 16 3
Solution 2 : Using de Moivre’s formula
( )5 55
5 i 5 3
5i 5 5 3
1 3z = 1 + i 3 = 2 + i = 2 cos + i sin = 2 e
2 2 3 3
5 5 = 2 cos + i sin = 2 e
3 3
1 3 = 32 - i = 16 - i 16 3
2 2
π
π
π π
π π
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 15/39
Let
z = wn where w = R eiφ = ?
thenthen
z = Rn (cos nφ + i sin nφ) = Rn einφ
Hence
r (cos θ + i sin θ) = Rn (cos nφ + i sin nφ) or r eiθ = Rn einφ
De Moivre’s formula may also be employed to evaluate the nth root of a complex
number z = r (cos θ + i sin θ)
n 1/n nr = R or R = r = r
Therefore
1/n 1/n θ + 2 k θ + 2 k z = r cos + i sin , k = 0, 1, ..., n-1
n nπ π
and cos θ = cos nφ & sin θ = sin nφ
1n , 1, 0, k,n2kπθ
or2kπθn −…=+=φ+=φ
w =w =
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 16/39
Example: Find 3 8
Let z = 8 + i 0
1/3 1/3 0 + 2 k 0 + 2 k z = 8 cos + i sin , k = 0, 1, 2
3 3π π
1/31z = 8 cos(0) + i sin(0) = 2
1/32
0 + 2 0 + 2 1 3z = 8 cos + i sin = 2 - + i = - 1 + i 3
3 3 2 2π π
1/33
0 + 4 0 + 4 1 3z = 8 cos + i sin = 2 - - i = - 1 - i 3
3 3 2 2π π
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 17/39
Note that all solutions have a common modulus of 2, but their arguments differ
from each other, and they are located on a circle of radius of 2 about the origin of
the z–plane, equally spaced around it with an incremental angle of 2π/3 as
illustrated
(81/3)2
Im
Re
(81/3)1
i2
0 2
2π/3
z–plane
(81/3)3
4π/3
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 18/39
Example: Find i
Let i /2z = 0 + i = 1 e π
1/2 1/2 /2 + 2 k /2 + 2 k z = 1 cos + i sin , k = 0, 1
2 2π π π π
i /41
2 2z = 1 cos + i sin = e = + i
4 4 2 2ππ π
i 5 /42
5 5 2 2z = 1 cos + i sin = e = - - i
4 4 2 2ππ π
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 19/39
Complex Functions
Let z and w are two complex variables defined as z = x + i y and w = u + i v
where x, y, u, and v are real variables.
If, for each value of z in some portion of the complex z–plane, one or more value(s)
of w are defined, then w is said to be a complex function of z.
w = f(z) = u + i v = f(x + i y) = u(x,y) + i v(x,y)
This complex functional relationship between z and w may be regarded as a
complex mapping or complex transformation of points P within a region in
the z–plane (called as the Domain ) to corresponding image point(s) Q within a
region in the w–plane (called as the Range ).
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 20/39
w = f(z) mapping
y
x
domain
P
z–plane v
u
range
Q
w–plane
Complex Functions as Complex Mapping
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 21/39
Example:
Find the ranges, R, in the complex w–plane of the following complex functions
corresponding to their domains, D, in z–plane. Plot D and R regions.
Use the standard notation, z = x + i y = r eiθ & w = u + i v
(a) w = f(z) = i z , D: Re(z) ≥ 0
w = i z = i (x + i y) = – y + i x
u(x,y) = – y & v(x,y) = x
Re(z) = x ≥ 0 => v(x,y) ≥ 0
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 22/39
(b) w = f(z) = 3 z – π , D: – π ≤ Re(z) ≤ π
w = 3 z – π = (3x – π) + i 3 y
u(x,y) = 3 x – π & v(x,y) = 3 y
–π ≤ x ≤ π => – 4 π ≤ u(x,y) ≤ 2 π
(c) w = f(z) = z2 , D: |z| ≤ 1 and 0 ≤ Arg(z) ≤ π /4
w = z2 = (r ei θ)2 = r2 ei 2θ
|w| = r2 = |z|2 & Arg(w) = 2 Arg(z)
|z| ≤ 1 => |w| ≤ 1
0 ≤ Arg(z) ≤ π /4 => 0 ≤ Arg(w) ≤ π /2
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 23/39
In dealing with complex functions, it is possible to distinguish the following two cases.
� Complex valued functions of a real variable
� Complex valued functions of a complex variable
Complex valued functions of a real variable
In this case, to each value of a real variable, t (a ≤ t ≤ b), one or more complex
value(s) of z are assigned, which can be shown as z(t) = x(t) + i y(t) = f(t)
Some examples for this type of complex function are:
z = ei t (0 ≤ t ≤ 2 π) => z = cos(t) + i sin(t) => x(t) = cos(t) & y(t) = sin(t)
z = t + i t2 (for all t) => x(t) = t & y(t) = t2
z = (1 – i ) (1 + i 2 ) (t ≥ 0) => x(t) = (1 + 2 t) & y(t) = tt t t t
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 24/39
Example:
For the slider-crank mechanism shown in the figure, it is desired to represent the
position of the point Q as a (complex valued) function of the horizontal position t
(a real variable) of the slider P as z(t) = x(t) + i y(t) (h – r ≤ t ≤ h + r)
y
x
P
Q
θ
r h
β
t
This form can conveniently be used to represent the parametric equations of planar
curves in complex z–plane.
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 25/39
Note that as P moves on a straight line, Q moves on a circle.
For every position of P denoted by t, it is possible to find two positions for Q denoted by z.
Therefore, the relationship between the positions of P and Q can be considered as
a mapping of points lying on a straight line to points lying on a curve (circle).
By using either θ or β angle, the coordinates x and y of Q can be related to the
position t of P.
y
x
P
Q
θ
r h
β
t
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 26/39
y
x
P
Q
θ
r h
β
t
x(t) = r cos(θ) = t – h cos(β)
y(t) = r sin(θ) = h sin(β)
Since x2 + y2 = r2 => [t – h cos(β)]2 + [h sin(β)]2 = r2
2 2 2t + h - rcos(β) =
2 h t
22 2 2t + h - rsin( ) = 1 -
2 h tβ
±
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 27/39
The parametric representation of the position of Q in the complex z–plane becomes
( )
−+−±
+−=
−+−±
−+−=
t2
rhtth4i
t2
rht
ht2
rht1hi
t2
rhtt)t(z
222222222
2222222
y
x
P
Q
θ
r h
β
t
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 28/39
For r = 1 and h = 4, x(t) and y(t) are shown for 3 ≤ t ≤ 5 in the Figure.
-1.00
-0.75
-0.50
-0.25
0.00
0.25
0.50
0.75
1.00
3.0 3.5 4.0 4.5 5.0Real variable t
y(t)
x(t)
Position of Q as a function of Slider Position
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 29/39
Complex valued functions of a complex variable
In this case, to each value of a complex variable z in a domain D of the z-plane,
one or more complex values w are assigned, which can be shown as w = f(z)
For a given complex variable z, a complex value w may be obtained regardless
of the function, f, itself being a real function or a complex function.
Some examples for this type of complex function are:
w = i z → i (x + i y) = – y + i x → u = – y & v = x
w = z2 → (x + i y)2 = (x2 – y2) + i 2 x y → u = x2 – y2 & v = 2 x y
w = z1/4 → (r eiθ)1/4 = R eiΦ → R = r1/4 & Φ = (θ + 2 k π) / 4
w = ez → ex + iy = ex (cos y + i sin y) → u = ex cos y & v = ex sin y
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 30/39
Note that once w = f(z) is known, it is a straightforward algebra to obtain u & v (or R
& f) in terms of x & y (or r & q).
However, if u(x,y) & v(x,y) are given in turn, to express w as a function of z is not a
trivial problem at all.
� In fact, it may even not have a solution.
� If there is a solution, then a suitable manipulation must be carried out in order
to come up with the correct w = f(z) expression.
Example :
Given w = u(x,y) + i v(x,y) = (x2 + x – y2 + 1) + i y (2x + 1) express w as a function
of z = x + i y
Rearranging gives
w = x2 + x – y2 + 1 + i 2 x y + i y
= x2 + i 2 x y – y2 + x + i y + 1
= (x + i y)2 + (x + i y) + 1
= z2 + z + 1
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 31/39
Some elementary functions of z
Exponential function: Using the basic definition of exponential function for real
variables, one gets
( )1
n 2 3 4z z z z z
f z = e = = 1 + z + + + +…n! 2! 3! 4!
∞
=∑k
u(x,y) = ex cos(y) & v(x,y) = ex sin(y) Mod(ez) = |ez| = ex & Arg(ez) = y
Note that ( )1
n 2 3 4- z (-z) z z z
f z = e = = 1 - z + - + - …n! 2! 3! 4!
∞
=∑k
( )1
n 2 3 4i z (i z) z i z z
f z = e = = 1 + i z - - + - …n! 2! 3! 4!
∞
=∑k
( )1
n 2 3 4- i z (- i z) z i z z
f z = e = = 1 - i z - + + - …n! 2! 3! 4!
∞
=∑k
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 32/39
Hyperbolic functions:
cosh(z) = (ez + e–z)/2 = cosh(x) cos(y) + i sinh(x) sin(y)
sinh(z) = (ez – e–z)/2 = sinh(x) cos(y) + i cosh(x) sin(y)
If z = 0 + iy => cosh(iy) = cos(y) sinh(iy) = i sin(y)
Familiar laws for the hyperbolic functions:
cosh2(z) – sinh2(z) = 1
cosh(z1 + z2) = cosh(z1) cosh(z2) + sinh(z1) sinh(z2)
sinh(z1 + z2) = sinh(z1) cosh(z2) + cosh(z1) sinh(z2)
cosh(2 z) = cosh2(z) + sinh2(z) = 1 + 2 sinh2(z) = 2 cosh2(z) – 1
sinh(2 z) = 2 sinh(z) cosh(z)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 33/39
Trigonometric functions:
cos(z) = (eiz + e–iz)/2 = cos(x) cosh(y) – i sin(x) sinh(y)
sin(z) = (eiz – e–iz)/2i = sin(x) cosh(y) + i cos(x) sinh(y)
If z = 0 + iy => cos(iy) = cosh(y) sin(iy) = i sinh(y)
Familiar laws for the trigonometric functions:
cos2(z) + sin2(z) = 1
cos(z1 + z2) = cos(z1) cos(z2) – sin(z1) sin(z2)
sin(z1 + z2) = sin(z1) cos(z2) + cos(z1) sin(z2)
cos(2 z) = cos2(z) – sin2(z) = 1 – 2 sin2(z) = 2 cos2(z) – 1
sin(2 z) = 2 sin(z) cos(z)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 34/39
Logarithmic function:
The logarithm of z = r eiθ that is defined implicitly as the function w = ln(z) which
satisfies the equation z = ew or r eiθ = eu + i v = eu eiv
Hence, eu = r or u = ln(r), and v = θ
Thus, w = u + i v = ln(r) + i θ = ln |z| + i arg(z)
If the principal argument of z is denoted by Arg(z), then this equation can
be rewritten as ln(z) = ln |z| + i [Arg(z) + 2 k π] k = 0, ±1, ±2, …
This indicates that complex logarithmic function is infinitely multi–valued.
For k = 0, the part (branch) of the logarithmic function is called as the principal
value .
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 35/39
Familiar laws for the logarithms of real quantities all hold for the logarithms of
complex quantities in the following sense:
ln(z1z2) = ln(z1) + ln (z2)
ln(z1/z2) = ln(z1) – ln (z2)
ln(zk) = k ln(z) k = 0, ±1, ±2, …
Example:
Compute w(z) = ln(1 – i)
ln(z) = ln |z| + i [Arg(z) + 2 k π] k = 0, ±1, ±2, …
( ) L2,1,0,k,2kπ4π
i2lni)ln(1 ±±=
+−+=−
( ) ( ) ( ) L,4
9πi2ln,
47π
i2ln,4π
i2lni)ln(1 −+−=−
–iπ/4
ln( )2
i7π/4
i15π/4
–i9π/4
i23π/4
–i17π/4
w-plane
Re(w)
Im(w)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 36/39
General Powers of z
w(z) = zc , c is any number, real or complex
( )0c ln(r) + i + 2 k c ln(z)z = e = e , k = 0, 1, 2, ...c θ π ± ±
If c = n:( ) ( )0 0 0
ln rn ln(r) + i i n i n n ln(z)
i 2 n k
z = e = e = e e = r e
e = 1
nn nθ θ θ
π
Because for all k’s
If c = m/n: ( )
( ) ( )
0i m/n + 2 k / /
/0 0
z = r e
m m = r cos + 2 k + i sin + 2 k
n n
k = 0, 1, 2, ...
m n m n
m n
θ π
θ π θ π
, n, n--11
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 37/39
If c is an irrational number, not expressible in the form m/n
( )
( )( ) ( )( )0i c + 2 k
0 0
z = r e
= r cos c + 2 k + i sin c + 2 k
k = 0, 1, 2, ...
c c
c
θ π
θ π θ π
± ±
If c is complex, i.e., c = a + i b
( )
( ) ( )
( ) ( )( )( )( )
0
00
0
(a + i b) ln(r) + i + 2 k
i b ln(r) 4 a + 2 k a ln(r) - b + 2 k
0a ln(r) - b + 2 k
0
z = e
= e e
cos b ln(r) + a + 2 k = e
+ i sin b ln(r) + a + 2 k
k =
c θ π
θ πθ π
θ πθ π
θ π
0, 1, 2, ... ± ±
+
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 38/39
Example:
Find all possible values of (1)i
Summary: zc is
single valued if c is an integer
n-valued if c = 1/n
n-valued if c = m/n
infinite valued if c is real and irrational
infinite valued if Im(c) ≠ 0
→→ r = 1 , r = 1 , θθoo = 0 , c = i = 0 , c = i
Example:
Find all possible values of (i)i →→ r = 1 , r = 1 , θθoo = = ππ/2/2 , c = i , c = i
...2,1,0,k,eee(1) 2kπ)]2ki(0i[ln(1)ln(1)ii ±±==== −π++
...2,1,0,k,eee(i) 21)4k)]2k2
i(i[ln(1)ln(i)ii ±±==== π+−π+π+ /(
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 39/39
END OF WEEKEND OF WEEKEND OF WEEKEND OF WEEKEND OF WEEKEND OF WEEKEND OF WEEKEND OF WEEK 1313131313131313