27 - Conformal Transformations

Conformal Transformations

Map that preserves angles

Tangent Line

If $\gamma :[-1,1]\to \mathbb{C}$ is a $C^1$ path which has ${\gamma }^{\prime }(t)\ne 0$ for all $t$ then

$$\{\gamma (t)+s{\gamma }^{\prime }(t):s\in \mathbb{R}\}$$

is a tangent line to $\gamma$ at $\gamma (t)$, and the vector ${\gamma }^{\prime }(t)$ is a tangent vector at $\gamma (t)\in \mathbb{C}$

Conformal Maps

Let $U$ be an open subset of $\mathbb{C}$

Suppose that $T:U\to \mathbb{C}$ ( or $\mathbb{S}$) is continuously differentiable in the real sense
so that its partial derivatives exist and are continuous

if ${\gamma }_1,{\gamma }_2:[-1,1]\to U$ are two paths with $z_0={\gamma }_1(0)={\gamma }_2(0)$ then

$${\gamma }_1^{\prime }(0)\text{ and }{\gamma }_2^{\prime }(0)\text{ are two tangent vectors at }{z}_0$$

Consider the angle between them (difference in arguments)

From the assumption of $T$ then compositions $T\circ {\gamma }_1$ and $T\circ {\gamma }_2$ are $C^1$ paths through $T(z_0)$
Hence we obtain a pair of tangent vectors at $T(z_0)$

$T$ is conformal at $z_0$ if for every pair of $C^1$ paths ${\gamma }_1,{\gamma }_2$ through $z_0$ then
Angle between tangent vectors at $z_0$ is equal to the angle between tangent vectors at $T(z_0)$
given by the $C^1$ paths $T\circ {\gamma }_1$ and $T\circ g_2$

Conformal

$T$ is conformal on $U$ if it is conformal at every $z\in U$

Relation between Holomorphic and Conformal

Let $f:U\to \mathbb{C}$ be a holomorphic map
Let $z_0\in U$ be such that $f^{\prime }(z_0)\ne 0$ then

$$f\text{ is conformal at }{z}_0$$

Especially if $f:U\to \mathbb{C}$ has non-vanishing derivative on all of $U$ then it is conformal on all of $U$

Proof

Need to show that $f$ preserves angles at $z_0$

Let ${\gamma }_1,{\gamma }_2$ be $C^1$-paths with ${\gamma }_1(0)={\gamma }_2(0)=z_0$ then
Obtain paths ${\eta }_1,{\eta }_2$ through $f(z_0)$ where

$${\eta }_1(t)=f({\gamma }_1(t))\text{ and }{\eta }_2(t)=f({\gamma }_2(t))$$

For $i=1,2$ then

$${\eta }_i^{\prime }(0)=\underset{h\to 0}{\lim }\frac{f({\gamma }_1(h))-f({\gamma }_i(0))}{h}=\underset{h\to 0}{\lim }\frac{f({\gamma }_i(h))-f(z_0)}{{\gamma }_i(h)-z_0}\cdot \frac{{\gamma }_i(h)-z_0}{h}$$

For small $h$ then ${\gamma }_i(h)\ne z_0$ as ${\gamma }_i(0)\ne 0$
Also

$$\underset{h\to 0}{\lim }\frac{f({\gamma }_i(h))-f(z_0)}{{\gamma }_i(h)-z_0}=f^{\prime }(z_0)$$

So by setting $f^{\prime }(z_0)=p{e}^{i\theta }$ then

$${\eta }_i^{\prime }(0)=f^{\prime }(z_0){\gamma }_i(0)=p{e}^{i\theta }{\gamma }_i^{\prime }(0),i=1,2$$

Hence if ${\varphi }_1,{\varphi }_2$ are the arguments of ${\gamma }_1^{\prime }(0)$ and ${\gamma }_2^{\prime }(0)$, then
Arguments of ${\eta }_1^{\prime }(0)$ and ${\eta }_2^{\prime }(0)$ are ${\varphi }_1+\theta$ and ${\varphi }_2+\theta$ respectively

Hence the difference between two pairs of arguments and angles between curves at $z_0$ and $f(z_0)$ are the same

Stereographic Projection Maps being Conformal lemma

Stereographic Projection map $S:\mathbb{C}\to \mathbb{S}$ is conformal

Möbius Transformations are Conformal lemma

Möbius Transformations are conformal

Proof

For a Möbius Map we have $f(z)=\frac{az+b}{cz+d}$ and

$$f^{\prime }(z)=\frac{ad-bc}{(cz+d{)}^2}\ne 0$$

for all $z\ne -\frac{d}{c}$

Hence $f$ is conformal at each $z\in \mathbb{C}\setminus \left\{-\frac{d}{c}\right\}$
by Relation between holomorphic and conformal

Uniqueness of Möbius Transformation

If $z_1,z_2,z_3$ and $w_1,w_2,w_3$ are triples of pairwise distinct complex numbers then
There exists unique Möbius Transformation $f$ such that

$$f(z_1)=w_1\text{ for each }i=1,2,3$$

Proof