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1 Complex Numbers
1.1 Sums and Products
Definition: The complex plane, denoted C is the set of all ordered pairs (x, y) with x, y ∈ R,where Re z = x is called the real part and Imz = y is called the imaginary part.
• We call the x-axis the real axis.
• We call the y-axis the imaginary axis.
• Usually we write z = (x, y) or z = x+ iy, this is called rectangular form.
• For two complex numbers, z1 = (x1, y1) and z2 = (x2, y2) we have z1 = z2 if and only if
x1 = x2 and y1 = y2
For z1 = (x1, y1) and z2 = (x2, y2), we define addition as
z1 + z2 = (x1, y1) + (x2, y2) = (x1 + x2, y1 + y2)
and multiplication as
z1z2 = (x1, y1)(x2, y2) = (x1x2 − y1y2, y1x2 + x1y2)
How do the real numbers interact here?
• For any r ∈ R we have (r, 0) ∈ C, and
(r, 0) + (s, 0) = (r + s, 0)
and(r, 0)(s, 0) = (rs, 0)
• For i = (0, 1) ∈ C we havei2 = (0, 1)(0, 1) = (−1, 0)
Now we can see that multiplication is well-defined, since
z1z2 = (x1 + iy1)(x2 + iy2) = x1x2 + i2(y1y2) + i(y1x2 + x1y2) = (x1x2 − y1y2, y1x2 + x1y2)
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1.2 Basic Algebraic Properties
• Complex numbers are associative
z1 + (z2 + z3) = (z1 + z2) + z3
and commutativez1 + z2 = z2 + z1
so we can rearrange and drop parenthesis.
• We have an additive identity 0 = (0, 0) ∈ C, and mult. identity 1 = (1, 0) ∈ C. Checkthese!
• For any z = x+ iy we have an additive inverse −z = −x− iy since
x+ iy + (−x− iy) = 0.
For z = x + iy non-zero we have a multiplicative inverse z−1 = u + iv so that zz−1 = 1defined by
(x+ iy)(u+ iv) = 1
and soux− vy + i(uy + vx) = 1
therefore ux− vy = 1 and uy + vx = 0. Solving the linear equation, we get
z−1 =
(x
x2 + y2,−y
x2 + y2
)• If z1z2 = 0 then z1 = 0 or z2 = 0. To see this, suppose z1 6= 0, then
z2 = z2 · 1 = z2 · z1 · z−11 = 0 · z−11 = 0
• Some interesting things happen with quotients, for example
1
i=
1
i· ii
=i
−1= −i
and4 + i
2− 3i=
(4 + i)(2 + 3i)
(2− 3i)(2 + 3i)=
5 + 14i
13=
5
13+ i
14
13.
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1.3 Vectors and Moduli
We can consider z = x+ iy = (x, y) as a vector on the complex plane.
Then the sum can also be interpreted vectorially, as follows.
Definition: The modulus of a complex number z = x + iy gives the distance from z to theorigin, and is given by
| z |=√x2 + y2
extending the notion of absolute value for the reals, and
| z |2= (Re z)2 + (Im z)2
now gives a relation between real numbers. From here
• | z |=| −z |
• Re z ≤| Re z |≤| z | and Im z ≤| Im z |≤| z |
• Note that z1 < z2 is meaningless, but | z1 |<| z2 | has meaning.
Consider the difference
| z1 − z2 |=| (x1 − x2) + i(y1 − y2) |=√
(x1 − x2)2 + (y1 − y2)2
What does this mean geometrically?
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Examples:
1. Consider| −3 + 2i |=
√9 + 4 =
√13
and| 1 + 4i |=
√1 + 16 =
√17
What does this mean, geometrically?
2. The equation | z |= 1 is the unit circle centered at the origin.
3. The equation | z − 1 + 3i |= 2 is the circle with center z = (1,−3) and radius 2.
4. Exercise: Show that | z1 · z2 |=| z1 | · | z2 |.
5. Exercise: Show that | z2 |=| z |2.
1.4 Triangle Inequality
Definition: The triangle inequality is given by
| z1 + z2 |≤| z1 | + | z2 | .
Illustrated by figure 3. Some consequences:
• We have | z1 + z2 |≥|| z1 | − | z2 ||, to see this, observe
| z1 |=| (z1 + z2) + (−z2) |≤| (z1 + z2) | + | −z2 |
and hence| (z1 + z2) |≥| z1 | − | z2 | if | z1 |≥| z2 |
and| (z1 + z2) |≥ −(| z1 | − | z2 |) if | z1 |<| z2 | .
What does this mean geometrically? Length of one side is greater than difference of othertwo.
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Examples:
1. Suppose z is on the circle | z |= 1. Then
| z − 2 |=| z + (−2) |≤| z | + | 2 |= 1 + 2 = 3
and| z − 2 |=| z + (−2) |≥|| z | − | −2 ||=| 1− 2 |= 1.
This means the distance from z to 2 is between 1 and 3.
2. The triangle inequality can be extended inductively, that is
| z1 + z2 + ...+ zn |≤| z1 | + | z2 | +...+ | zn |
for any n. So for example, for z on the circle | z |= 2, we have
| 3 + z + z2 |≤ 3+ | z | + | z2 |= 3+ | z | + | z |2≤ 9.
3. Consider the degree n polynomial given by
P (z) = a0 + a1z + a2z2 + ...+ anz
n
where n ∈ Z+ and a0, ..., an ∈ C. We will show that for some positive R ∈ R,∣∣∣∣ 1
P (z)
∣∣∣∣ < 2
anRn
whenever | z |> R. Morally: Reciprocal 1/P (z) is bounded from above when z is outsideof the circle | z |= R. Recall,
• | z1 · z2 |=| z1 | · | z2 |• | z2 |=| 2 |2
and write
w =P (z)− anzn
zn=a0zn
+a1zn−1
+ ...+an−1z
.
ThenP (z) = (an + w)zn
Then the equation above becomes
wzn = a0 + a1z + ...+ an−1zn−1
and hence| w || zn |≤| a0 | + | a1z | +...+ | an−1zn−1 |
implying
| w |≤ | a0 || z |n
+| a1 || z |n−1
+ ...+| an−1 || z |
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Now pick R large enough so that
| ai || z |n−i
<| an |2n
when | z |> R. Then,
w < n · | an |2n
=| an |
2whenever | z |< R.
Consequently,
| an + w |≥|| an | − | w ||>| an |
2
and hence
| P (z) |=| an + w | · | zn |> | an |2·Rn.
1.5 Complex Conjugates
Definition: The complex conjugate of a number z = x+ iy is given by
z = x− iy
Some observations:
• | z |=| z |
• z = z.
• z1 + z2 = z1 + z2, since
z1 + z2 = x1 + iy1 + x2 + iy2 = xi + x2 − i(y1 + y2) = x1 − iy1 + x2 − iy2 = z1 + z2.
• z1 − z2 = z1 − z2
• z1 · z2 = z1 · z2
•(z1z2
)= z1
z2
• z + z = 2Re z.
• z − z = 2iIm z.
• zz = (x+ iy)(x− iy) = x2 + y2 =| z |2.
Examples:
1. −1+3i2−i = (−1+3i)(2+i)
(2−i)(2+i) = −5+5i|2−i|2 = −5+5i
5 = −1 + i
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2. We can use conjugates to establish modulus properties. We know zz =| z |2. Therefore,we obtain
| z1 · z2 |2 = (z1 · z2)(z1 · z2)= z1 · z2 · z1 · z2= z1 · z1 · z2 · z2= | z1 |2 · | x2 |2
= (| z1 | · | z2 |)2
from which we conclude that | z1 · z2 |=| z1 | · | z2 |, since modulus is always positive.
1.6 Exponential Form
Let r > 0 and θ be polar coordinates for z = (x, y), then
z = r(cos θ + i sin θ)
then| z |= r
√cos2 θ + sin2 θ = r
Definition: The value θ = arg(z) is an argument for z, and it is the angle z = (x, y) makeswith the real axis. Infinite arguments exist for z, namely,
θ + 2π, θ + 4π, ..., θ + 2kπ, ...
but we call Θ = Arg(z) with −π < Θ ≤ π the principal argument of z.
Examples:
1. The complex number −1− i has
Arg(−1− i) = −3π
4
but
arg(−1− i) =5π
4+ 2kπ
for n ∈ Z.
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Definition: The symbol eiθ is defined by Euler’s formula as
eiθ = cos θ + i sin θ
and hencez = reiθ = r(cosθ + i sin θ)
Examples:
1. Writing −1− i in exponential form we have
−1− i =√
2ei−3π4
or−1− i =
√2ei(
−3π4
+2πk)
2. The set of numbers z = eiθ are the set of numbers on the unit circle. Accordingly,
eiπ = −1 and eiπ/2 = i and ei4π = 1.
Also,z = Reiθ where 0 ≤ θ ≤ 2π
is a parametrization of the unit circle, moving counter-clockwise, and
z = z0 +Reiθ where 0 ≤ θ ≤ 2π
moves the circle to be centered at z0.
Some basic properties of exponential forms:
• eiθeiθ′ = ei(θ+θ′)
• reiθ · r′eiθ′ = (rr′) · ei(θ + θ′)
• reiθ
r′eiθ′= r
r′ ei(θ−θ′)
• If z = reiθ then
z−1 =1
reiθ=
1
re−iθ
andzn = rneinθ.
Examples:
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1. From the properties above we see that
arg(z1z2) = arg(z1) + arg(z2).
but this doesn’t necessarily hold when arg is replaced with Arg. For example, z1 = −1and z2 = i, then
Arg(−1) = π and Arg(i) =π
2so
Arg(−1) + Arg(i) = π +π
2=
3π
2
but
Arg(−1 · i) = Arg(−i) =−π2
2. Find the principal argument ofi
−1− i.
We already know that i = 1eiπ2 and −1− i =
√2e−i
3π4 and so
i
−1− i=
eiπ2
√2e−i
3π4
=1√2ei(π/2+3π/4)
Therefore
arg
(i
−1− i
)=π
2+
3π
4=
5π
4
and hence
Arg
(i
−1− i
)=
5π
4− 2π =
−3π
4
3. Let’s write the expression (−1 + i)7 in rectangular form. First, for z = (−1 + i) we have
| z |=√
1 + i =√
2
and
Arg(z) =3π
4
so −1 + i =√
2ei3π4 , and
(−1 + i)7 =(√
2eii3π4
)7= 8√
2ei21π4 = 8
√2ei5πe
iπ4 = −8
√2e
iπ4
But we can easily see that
(1 + i) =√
2eiπ4
and so(−1 + i)7 = −8(1 + i)
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4. When r = 1, for any value n ∈ Z, we have(eiθ)n
= einθ
but this means(cos θ + i sin θ)n = cosnθ + i sinnθ
this is known as De Moivre’s formula.
5. Using the above with n = 2, we reclaim known identities
cos2 θ − sin2 θ + 2i cos θ sin θ = (cos θ + i sin θ)2 = cos 2θ + i sin 2θ.
Now equating real and imaginary parts, we have
cos 2θ = cos2 θ − sin2 θ and sin 2θ = 2 cos θ sin θ
1.7 Roots of Complex Numbers
Two non-zero complex numbers can be arrived at in infinitely many ways, for example
z = reiθ and z′ = r′eiθ′
are equal precisely whenr = r′ and θ = θ′ + 2πk
for k ∈ Z. We also know that zn = rneinθ, so we should know something about nth roots.Definition: An nth root of z0 ∈ C is a complex number z = reiθ such that
zn = z0
that isr0e
iθ0 = rneinθ,
and hencer0 = rn and nθ = θ0 + 2πk.
• Since r0 is real, this just means r = n√r0.
• For k ∈ Z, we know that
θ =θ0n
+2πk
n
So the nth roots of z0 are given by the infinite family
z = n√r0exp
[i
(θ0n
+2πk
n
)]all lying on the circle | z |= n
√r0.
So n of them should be more special than the others, namely, those with
ck = n√r0exp
[i
(θ0n
+2πk
n
)]and k ∈ 0, 1, 2, ..., n− 1.
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• When z0 is a positive real then n√r0 means the familiar unique positive root.
• When θ0 = Arg(z0) then we call c0 the principal root.
• When z0 is a positive real, its principal root is n√r0.
Examples:
1. Let’s find all values of (−16)14 , that is, the 4th roots of −16. Since
−16 = 16ei(π+2kπ)
so the roots are just
ck = 2ei(π4+ kπ
2 )
where k = 0, 1, 2, 3. So we have
ck = 2ei(π4 )ei(
kπ2 ) = 2
(cos
π
4+ i sin
π
4
)ei(
kπ2 ) = 2
(1√2
+ i1√2
)ei(
kπ2 ) =
√2(1 + i)ei(
kπ2 )
and therefore
c0 =√
2(1 + i) · 1 =√
2(1 + i)
c1 =√
2(1 + i) · i =√
2(−1 + i)
c2 =√
2(1 + i) · −1 =√
2(−1− i)c3 =
√2(1 + i) · −i =
√2(1− i)
2. Let’s examine the nth roots of 1, called the nth roots of unity. We start with
1 = 1 · ei(0+2πk)
where k ∈ Z. Then,
ck =n√
1 · ei(0+2πkn ) = ei(
2πkn )
where k ∈ {0, 1, 2, ..., n − 1}. When n = 2 we get c0 = 1 and c1 = −1. For larger n, weget roots lying on the regular n-gon inscribed in the circle | z |= 1.
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1.8 Regions in the Complex Plane
Definition: We define an ε neighborhood around a given point z0 by
| z − z0 |< ε
we can also define the deleted neighborhood, that is the neighborhood minus the point z0 itself,by
0 <| z − z0 |< ε
For a given set S contained in C,
• A point z0 is called interior if there exists a neighborhood containing only z0 and pointsof S.
• A point z0 is called exterior if there exists a neighborhood containing z0 and no points inS.
• If z0 is neither of these, then it is a boundary point.
• A set is open if it contains none of its boundary points.
• A set is closed if it contains all of its boundary points.
• A set is a closure of S if it contains S plus its boundary points. For example, | z |≤ 1 isthe closure of | z |< 1.
• Some sets like 0 <| z |≤ 1 is neither open nor closed.
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• An open set S is connected if every pair of points in S can be joined by a polygonal line.
• A domain is an non-empty open connected set.
• A domain with some, none, or all of its boundary points is called a region.
• A set S is bounded if every point of S lies in some circle | z |< R.
Examples:
1. Consider the set Im(1z
)> 1 in relation to these properties, for z 6= 0 we have
1
z=
z
zz=
z
| z |2=
x− iyx2 + y2
so we want
Im
(1
z
)=
−yx2 + y2
> 1
and thereforex2 + y2 + y < 0.
Completing the square, we get
x2 + y2 + y +1
4<
1
4
which means
(x− 0)2 +
(y +
1
2
)2
<
(1
2
)2
.
This is the region whose boundary is the circle centered at x = 0 and y = −1/2 withradius 1/2.
• A point z0 is an accumulation point, or limit point, of a set S if each deleted neighborhoodof z0 contains at least one point of S.
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