Complex Analysis Chapter 1 Summary

Complex Analysis Chapter 1 Summary Fall 2021

The Complex Plane

Let C = (R², +, ·), where the addition, multiplication by α ∈ R and multiplication for (a, b), (c, d) ∈ C are defined by

(a, b) + (c, d) = (a + c, b + d) α (a, b) = (α a, α b)

(a, b) · (c, d) = (ac − bd, ad + bc)

Write C = R² with the identication (a, b) = a + bi. Then

(a) C is a field. In particular, its an additively commutative group, and multiplicatively, C \ {0}

is a commutative group.

(b) Every nonzero complex number z = a + bi ∈ C has a multiplicative inverse.

Proof In fact, z⁻¹ := a

a²+ b² − b

a²+ b²i works.

For z = a + bi ∈ C, define the complex conjugate ¯z := a − bi = a + b(i). Then for z, w ∈ C, z + w = ¯z + ¯w

z w = ¯z ¯w

For z = a + bi ∈ C, define the norm |z| :=√

a²+ b² =√

z ¯z. Then |¯z| = |z| and

|z| ≥ 0 ∀ z ∈ C and |z| = 0 ⇐⇒ z = 0,

|z w| = |z| |w| ∀ z, w ∈ C,

|z + w| = |z| + |w| ∀ z, w ∈ C

|Re z|, |Im z| ≤ |z| ≤ |Re z| + |Im z| ∀ z = Re z + Im z i = a + bi ∈ C

Viewing z, w ∈ C as vectors, addition is visualized as a parallelogram, and multiplication is best seen in polar coordinates. Polar coordinates can be obtained by r = |z| and θ = Arg(z), the argument of z. So we get the pretty important identity

z = r cos θ + ir sin θ = r(cos θ + i sin θ) := re^iθ. For z₁ = r₁(cos θ₁+ i sin θ₁) and z₂ = r₂(cos θ₂+ i sin θ₂), we have

z1z2 = r1r2(cos(θ1+ θ2) + i sin(θ1+ θ2)) = r1r2e^i(θ1+θ2) Arg(z₁z₂) = Arg(z₁) + Arg(z₂) mod 2π.

Definitions

(a) Let f be a function defined on E ⊆ C. We say that f is continuous at a point p ∈ E if lim

z∈E, z→pf (z) = f (p),

and we say that f is continuous in E if it is continuous at each point in E.

Complex Analysis Chapter 1 Summary

(b) Let {f_k} be a sequence of functions defined on E ⊆ C. We say that {fk} is pointwise convergent on E if lim

k→∞f_k(z) exists for all z ∈ E; the pointwise limit is defined as f (z).

(c) Let {fk} be a sequence of functions defined on E ⊆ C. We say that {f^k} converges uniformly to f in E if for each ε > 0, there exists N ∈ N such that if n ≥ N, then

|f_n(z) − f (z)| ≤ ε ∀ z ∈ E.

Theorem If {f_k(z)} are continuous and converge uniformly on E, then the pointwise limit f (z) is continuous on E.

Theorem (M-Test) Let {f_k(z)} be a sequence of continuous functions defined on D. If there exists a sequence {M_k} of nonnegative numbers such that |f_k(z)| ≤ M_k for all z ∈ D and for each k ∈ N and if

∞

X

k=1

M_k converges, then

∞

X

k=1

f_k(z) converges uniformly to a function f which is continuous in D.

Stereographic Projection

Consider a sphere Σ := {(u, v, w) | u² + v² + (w − 1

2)² = 1

4} in R³. We have a bijection ϕ : Σ \ N → C, where N = (0, 0, 1) is the north pole (point at ∞), given by stereographic projection. Indeed, we define the Riemann sphere ˆC := C ∪ {∞} ≡ Σ.

Let ϕ : Σ \ N → R² be defined by mapping a point (u, v, w) ∈ Σ \ N to the point (x, y) ∈ R² such that N = (0, 0, 1), (u, v, w) and (x, y, 0) are collinear, i.e. there exists t > 0 such that

(x, y, −1) = t (u, v, w − 1)

=⇒ t² = x²+ y²+ 1

u² + v²+ (w − 1)² = x²+ y²+ 1

u²+ v²+ [(w − 1/2) − 1/2]²

= x²+ y²+ 1

u²+ v²+ (w − 1/2)²− (w − 1/2) + 1/4 = x²+ y²+ 1 1/4 − (w − 1/2) + 1/4

= x²+ y²+ 1

1 − w = t(x²+ y²+ 1)

=⇒ t = x²+ y² + 1 or t = 1 1 − w. Thus the bijection ϕ and ϕ⁻¹ are given by

(∗) ϕ(u, v, w) =

 u

1 − w, v 1 − w



; ϕ⁻¹(x, y) =

 x

x²+ y² + 1, y

x²+ y²+ 1, x² + y² x²+ y²+ 1

 . Note that

lim

(u,v,w)→(0,0,1)(x²+ y²) = lim

(u,v,w)→(0,0,1)

u²+ v²

(1 − w)² = lim

(u,v,w)→(0,0,1)

w

1 − w = ∞ and

|z|→∞lim (u, v, w) = lim

|z|→∞

 x

x²+ y²+ 1, y

x²+ y²+ 1, x²+ y² x²+ y²+ 1



= (0, 0, 1).

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Complex Analysis Chapter 1 Summary

A circle on Σ is the intersection of Σ with a plane of the form Au + Bv + Cw = D. According to (∗), if S is such a circle and T is the corresponding set in C, then

(†) (C − D)(x²+ y²) + Ax + By = D for (x, y) ∈ T.

Note that if C 6= D, and if (u, v, w ∈ Σ ∩ {Au + Bv + Cw = D}, since Au + Bv + C(w − ¹₂) = D − ^C₂ = ^2D−C₂

=⇒ A²+ B²+ C²
1/2

u²+ v²+ (w − ¹₂)²
1/2

> ^|2D−C|₂

=⇒ A²+ B²+ C²
_{1/2 1}

2 > ^|2D−C|₂

=⇒ A²+ B²+ C² > (2D − C)²

=⇒ A²+ B²+ 4D(C − D) > 0

(†) is an equation of a circle. If C = D, (†) represents a line. Since C = D if and only if S intersects (0, 0, 1), we have the following proposition.

Proposition Let S be a circle on Σ and let T be its projection on C. Then (a) if S contains (0, 0, 1), T is a line;

(b) if S does not contain (0, 0, 1), T is a circle.

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