# MATH3401

### Complex Analysis

Lecture 6

#### Implicit form of Möbius transformations (MT)

The Möbius transformation $$T(z) = w = \dfrac{az+b}{cz+d}\; (*)$$, with $$ad-bc \neq 0$$, can be written as $A z w + B z + C w + D = 0$ with $$A=c, B=-a, C=d$$ and $$D=b$$; called the implicit form.

#### Implicit form of MT

Case I: $$c=0$$. $$T$$ is a bijection (1-1 and onto): $$\C\ra\C$$.

Case II: $$c\neq 0$$. $$(*)$$ $$\Rightarrow z=\dfrac{-dw+b}{cw-a}$$. Thus $T^{-1}(w)= \dfrac{-dw+b}{cw-a}.$

Then in Case II, $$T$$ is 1-1 and onto $$\C \setminus \left\{-d/c\right\} \ra \C \setminus \left\{a/c\right\}$$.

Question: Can we extend $$T$$ to a function $$\C \ra \C$$ in Case II? In particular, so extension is 1-1 and onto.

Ans: Yes! Take $$T\left(-d/c\right)=a/c$$. Not great: discontinuous.

### Important concept

Extend $$\C$$ to the extended complex plane, denoted by $$\conj{\C}$$,
"by adding a point at infinity", which we call $$\infty$$.

Define $$T\left(-d/c\right)=\infty$$ and $$T\left(\infty\right)=a/c$$.

This extends $$T$$ to a map $$\conj{\C} \ra \conj{\C}$$, which is 1-1 & onto.

Remark: $$\conj{\C}$$ is a topological space and the given extension is continuous.

#### Riemann sphere

$$\varphi(z)$$ is on the Riemann sphere.

#### Final remarks on MT

Remark 1: Given 3 distinct points in $$\conj{\C}$$, $$z_1,z_2,z_3$$ and 3 distinct points in $$\conj{\C}$$, $$w_1,w_2,w_3$$, $$\exists !$$ (there exists a unique) Möbius transformation $$T$$ such that $$T(z_j)=w_j$$.

Here $$T=w(z)$$ is given by $\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)} = \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}\;(**)$

#### Final remarks on MT

Note 1: In practice, may be easier to solve directly for $$a,b,c,d$$ rather than use $$(**)$$.

Note 2: How does this work with $$\infty$$?

$$T(\infty)=\dfrac{a}{c}$$ and $$T\left(-\dfrac{d}{c}\right)=\infty$$, rigorously later.

$$T(\infty)=w \iff \ds \lim_{|z|\ra \infty}T(z) = w$$.

$$T(\xi)=\infty \iff \ds \lim_{z\ra \xi }\dfrac{1}{T(z)} = 0$$.