# MATH3401

### Complex Analysis

Lecture 5

#### Basic idea with new mappings

Example 2: $$z\mapsto \dfrac{1}{z}$$ on $$\C_*$$.

Define $$\xi (z) = \dfrac{z}{|z|^2}$$ on $$\C_*$$; $$\;\;\eta(\xi)=\conj{\xi}$$ on $$\C$$.

$$\eta \circ \xi (z) = \eta \left( \xi \left( z \right) \right) = \conj{\left( \dfrac{z}{|z|^2} \right)} = \dfrac{\conj z }{ |z|^2 }$$

$$\;\;\,\quad\quad = \dfrac{\conj z}{z \conj z} = \dfrac{1}{z}$$. $$\qquad z\in \C_*$$

#### Basic idea with new mappings

$$\xi$$ is called inversion (w.r.t the unit circle),

$$\eta$$ is reflection in the real axis.

#### Basic idea with new mappings

For $$w=\dfrac{1}{z} = \dfrac{\conj z }{|z|^2}$$, we have

$$x+i\,y \mapsto u+i\,v$$ and $$w = \dfrac{x-i\,y}{x^2+y^2}$$, so $u = \dfrac{x}{x^2+y^2} \; (1)\quad \text{ & } \quad v = \dfrac{-y}{x^2+y^2} \; (2)$

We can use this to show that $$1/z$$ maps circles and lines in the
$$z$$-plane to circles and lines in the $$w$$-plane.

#### A note on mapping circles and lines

Note: Circles and lines in the $$z$$-plane can be represented as $A(x^2+y^2)+Bx+Cy +D = 0 \;\;(*)$

with $$A,B, C, D\in \R$$, and $$B^2+C^2>4AD$$.

If $$A = 0$$ then we have a line. If $$A \neq 0$$ then we have a circle.

#### A note on mapping circles and lines

From $$(*)$$ we have $\left( x+\frac{B}{2A} \right)^2 + \left( y+\frac{C}{2A} \right)^2 = \left( \frac{\sqrt{B^2+C^2-4AD}}{2A} \right)^2 .$

Note that for $$w=1/z$$, expressions (1) & (2) imply $D(u^2+v^2)+Bu-Cv+A=0.$

### Terminology reminder

$$*\; f: \Omega \ra \C$$ is 1-1 or injective says $f(z) = f(\xi)\Rightarrow z=\xi.$

$$*\; f: \Omega \ra \Lambda\subseteq \C$$ is onto or surjective says given $$\eta \in \Lambda$$, there exists (at least one) $$z\in \Omega: f(z)=\eta$$.

### Möbius transformation

Definition: Let $$a,b,c,d\in \C$$ with $$ad-bc\neq 0$$. Then $w = T(z) = \frac{az+b}{cz+d}$ is a Möbius (or linear fractional) transformation.

Natural domain of definition:

$$*\; c=0$$: dom$$(w)=\C$$ (note $$c= 0 \Rightarrow d\neq 0$$);

$$*\; c\neq 0$$: dom$$(w)=\C\setminus \{-d/c\}$$.

### Möbius transformation

Goal: Understand $$T$$ geometrically

Case I: $$c=0$$.

Claim: $$T$$ maps $$\C \ra \C$$ 1-1 & onto.

Proof. (i) 1-1: Suppose $$T(z)= T(\xi)\;\; (**).\;$$ We want to show: $$z=\xi$$. From $$(**)$$ we have $\frac{a}{d}z+\frac{b}{d} = \frac{a}{d}\xi +\frac{b}{d} \implies z=\xi.$

(ii) onto: given $$w\in \C$$ we need $$z\in \C$$ such that $$T(z)=w$$.

📝 Check $$z=\frac{d}{a}\left(w-\frac{b}{d}\right)$$ works.

### Möbius transformations

Goal: Understand $$T$$ geometrically

Case II: $$c\neq 0$$.

$$w = \dfrac{az+b}{cz+d} =$$ $$\dfrac{a\left(z+\dfrac{d}{c}\right)- \dfrac{ad}{c}+b}{c\left(z+\dfrac{d}{c}\right)}$$ $$=\dfrac{a}{c}+\left(\dfrac{bc-ad}{c}\right)\cdot \dfrac{1}{cz+d}$$

So in this case, $$T$$ is the composition of the mappings $Z= cz+d; W = \frac{1}{Z}; \;\text{ & } \; w = \frac{a}{c} + \frac{bc-ad}{c}W.$

So in both cases I & II, $$T$$ is a composition of maps previously studied.