11. Conformal Maps

11.1 The Extended Complex Plane $C_{\infty }$

11.1.1 Stereographic Projection

Stereographic Projection

Let

$$\mathbb{S}=\{(x,y,z)\in {\mathbb{R}}^3:x^2+y^2+z^2=1\}$$

be the unit sphere of radius $1$ centred at the origin in ${\mathbb{R}}^3$

Let $N=(0,0,1)$ of $\mathbb{S}$ (the north pole)

Define bijective map $S:\mathbb{C}\to \mathbb{S}\setminus \{N\}$ as follow
Join $z$ to $N$ by a straight line and let $S(z)$ be the point where the line meets sphere $\mathbb{S}$
then $S$ is called the stereographic projection

Viewing the Complex Plane inside ${\mathbb{R}}^3$

By viewing the complex plane $\mathbb{C}$ as the copy of ${\mathbb{R}}^2$ inside ${\mathbb{R}}^3$ by plane

$$\{(x,y,0)\in \mathbb{R}:x,y\in \mathbb{R}\}$$

then $z=x+iy$ corresponds to point $(x,y,0)$

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Explicit Formula for the Stenographic Projection lemma

$$S(z)=\left(\frac{2x}{x^2+y^2+1},\frac{2y}{x^2+y^2+1},\frac{x^2+y^2-1}{x^2+y^2+1}\right)$$

Proof

General Point on the line joining $z$ and $N$ is

$$t(0,0,1)+(1-t)(x,y,0)$$

There is a unique value of $t$ with $t\ne 1$ such that the point lies on sphere, that is

$$t=\frac{(x^2+y^2-1)}{(x^2+y^2+1)}$$

This can also be rewritten as

$$S(z)=\frac{1}{1+|z{|}^2}(2\mathrm{Re}(z)+2\mathrm{Im}(z),|z{|}^2-1)$$

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Distance Metric of the Stenographic Projection

Let $z,w\in \mathbb{C}$

$$d(z,w):=||S(z)-S(w)||=\frac{2|z-w|}{\sqrt{1+|z{|}^2}\sqrt{1+|w{|}^2}}$$

Proof

Since $||S(z)||=||S(w)||=1$ then

$$||S(z)-S(w)|{|}^2=2-2⟨S(z),S(w)⟩$$

where $⟨\cdot ,\cdot ⟩$ refers to the usual inner product in ${\mathbb{R}}^3$

So using the formulae and computation then

$$⟨S(z),S(w)⟩=1-\frac{2|z-w{|}^2}{(1+|z{|}^2)(1+|w{|}^2)}$$

Thus

$$||S(z)-S(w)|{|}^2=\frac{4|z-w{|}^2}{(1+|z|{)}^2(1+|w{|}^2)}$$

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11.1.2 Adding in $\infty$

Extension of the Distance Metric to Infinity lemma

As $|z|\to \infty$ then $S(z)\to N$ hence

$${\mathbb{C}}_{\infty }=\mathbb{C}\cup \{\infty \}$$

then for $z\in \mathbb{C}$

$$d(z,\infty )=\frac{2}{\sqrt{1+|z{|}^2}}$$

Proof

By definition we have

$$d(z,\infty )=||S(z)-S(\infty )||=||S(z)-N||$$

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Translation Map

Let $a\in \mathbb{C}$ then

Define $f:C_{\infty }\to C_{\infty }$ by

$$f(z)=z+a\text{ for }z\in \mathbb{C}\text{ and }f(\infty )=\infty$$

then $f$ is a continuous bijection

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Dilations

Let $b\in \mathbb{C}$ with $b\ne 0$ then

Define $f:C_{\infty }\to C_{\infty }$ by

$$f(z)=bz\text{ for }z\in \mathbb{C}\text{ and }f(\infty )=\infty$$

then $f$ is a continuous bijection

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Inversion

Define $f:{\mathbb{C}}_{\infty }\to {\mathbb{C}}_{\infty }$ by

$$f(z)=\frac{1}{z}\text{ for }z\in \mathbb{C}\setminus \{0\}\text{ and }f(0)=0\text{ and }f(\infty )=\infty$$

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11.1.3 Möbius Maps

Möbius Maps

Each element $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\mathrm{GL}}_2(\mathbb{C})$ gives Möbius Map ${\Psi }_g:{\mathbb{C}}_{\infty }\to C_{\infty }$ given by

$${\Psi }_g(z):=\frac{az+b}{cz+d}$$

Need to be careful about $\infty$ so

• If $c\ne 0$ then define ${\Psi }_g\left(-\frac{d}{c}\right)=\infty$ and ${\Psi }_g(\infty )=\frac{a}{c}$

• If $c=0$ then define ${\Psi }_g(\infty )=\infty$

General Linear Group ${\mathrm{GL}}_2(\mathbb{C})$

General Linear Group ${\mathrm{GL}}_2(\mathbb{C})$ consists of $2\times 2$ matrices

$$g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\text{ with }a,b,c,d\in \mathbb{C}$$

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Scalar-Uniqueness of the Möbius Maps

If $g,g^{\prime }\in {\mathrm{GL}}_2(\mathbb{C})$ give the same Möbius Map then

$$g=\lambda g^{\prime }\text{ for some }\lambda \ne 0$$

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Composition of Möbius Maps

For $g_1,g_2\in {\mathrm{GL}}_2(\mathbb{C})$ then

$${\Psi }_{g_1{g}_2}={\Psi }_{g_1}\circ {\Psi }_{g_2}$$

so we have ${\mathrm{GL}}_2(\mathbb{C})$ acts on ${\mathbb{C}}_{\infty }$ via Möbius Maps

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11.1.4 Decomposing Möbius Maps

Translation Möbius Maps

Translation $z\mapsto z+a$ is Möbius Map ${\Psi }_{T(a)}$ where

$$T(a)=\left(\begin{array}{cc}1& a\\ 0& 1\end{array}\right)$$

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Dilation Möbius Maps

Dilation $z\mapsto bz$ is Möbius Map ${\Psi }_{D(b)}$ where

$$D(b)=\left(\begin{array}{cc}b& 0\\ 0& 1\end{array}\right)$$

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Inversion Möbius Maps

Inversion $z\mapsto \frac{1}{z}$ is Möbius Map ${\Psi }_J$ where

$$J=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$$

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Decomposition of a Möbius Map

Every Möbius Maps can be written as a composition of translation, dilations and inversions

Proof

Let ${\Psi }_g,g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ be the Möbius Map

Suppose first that $c\ne 0$ then without worrying about $\infty$ then

$$z\stackrel{{\Psi }_{D(c)}}{\to }cz\mapsto \stackrel{{\Psi }_{T(d)}}{\to }\mapsto cz+d\stackrel{{\Psi }_J}{\to }\frac{1}{cz+d}\stackrel{{\Psi }_{D\left(\frac{bc-ad}{c}\right)}}{\to }\frac{b-\frac{ad}{c}}{cz+d}\stackrel{{\Psi }_{T\left(\frac{a}{c}\right)}}{\to }\frac{az+b}{cz+d}$$

which is a chain of compositions

Hence we get

$${\Psi }_g={\Psi }_{T\left(\frac{a}{c}\right)}\circ {\Psi }_{D\left(\frac{bc-ad}{c}\right)}\circ {\Psi }_J\circ {\Psi }_{T(d)}\circ {\Psi }_{D(c)}$$

And also

$$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=T\left(\frac{a}{c}\right)\cdot D\left(\frac{bc-ad}{c}\right)\cdot J\cdot T(d)\cdot D(c)$$

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11.1.5 Basic Geometry of Möbius Maps

Circline

Either

1. Circle in $\mathbb{C}$ (considered as a subset of $C_{\infty }$
2. Line in $\mathbb{C}$ (considered as a subset of ${\mathbb{C}}_{\infty }$) together with point $\{\infty \}$

Note that lines in $\mathbb{C}$ are given by equations of the form $|z-a|=|z-b|$ for $a$ and $b$ distinct in $\mathbb{C}$

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Invariance of Circlines under Möbius Map

Möbius Map take circlines to circlines

Proof - Mini

By Decomposition of a möbius map,
It is enough to check this for translations, dilations and inversion
Hence is a straightforward case-by-case analysis

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11.2 Conformal Transformations

Conformal Transformations

Map that preserves angles

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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 $\gamma (t)\in \mathbb{C}$

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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 $T\circ g_2$

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Conformal

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

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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

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Stereographic Projection Maps being Conformal lemma

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

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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

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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

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