proof of Möbius circle transformation theorem

Case 1: $f(z)=az+b$.

Case 1a: The points on $|z-C|=R$ can be written as $z=C+R{e}^{i\theta }$. They are mapped to the points $w=aC+b+aR{e}^{i\theta }$ which all lie on the circle $|w-(aC+b)|=|a|R$.

Case 1b: The line $\text{Re}(e^{i\theta }z)=k$ are mapped to the line $\text{Re}\left(\frac{e^{i\theta }w}{a}\right)=k+\text{Re}\left(\frac{b}{a}\right)$.

Case 2: $f(z)=\frac{1}{z}$.

Case 2a: Consider a circle passing through the origin. This can be written as $|z-C|=|C|$. This circle is mapped to the line $\text{Re}(Cw)=\frac{1}{2}$ which does not pass through the origin. To show this, write $z=C+|C|e^{i\theta }$. $w=\frac{1}{z}=\frac{1}{C+|C|e^{i\theta }}$.

$$\text{Re}(Cw)=\frac{1}{2}(Cw+\overline{Cw})=\frac{1}{2}\left(\frac{C}{C+|C|e^{i\theta }}+\frac{\overline{C}}{\overline{C}+|C|e^{-i\theta }}\right)$$

$$=\frac{1}{2}\left(\frac{C}{C+|C|e^{i\theta }}+\frac{\overline{C}}{\overline{C}+|C|e^{-i\theta }}\frac{e^{i\theta }}{e^{i\theta }}\frac{C/|C|}{C/|C|}\right)=\frac{1}{2}\left(\frac{C}{C+|C|e^{i\theta }}+\frac{|C|e^{i\theta }}{|C|e^{i\theta }+C}\right)=\frac{1}{2}$$

Case 2b: Consider the line which does not pass through the origin. This can be written as $\text{Re}(az)=1$ for $a\ne 0$. Then $az+\overline{az}=2$ which is mapped to $\frac{a}{w}+\frac{\overline{a}}{\overline{w}}=2$. This is simplified as $a\overline{w}+\overline{a}w=2w\overline{w}$ which becomes $(w-a/2)(\overline{w}-\overline{a}/2)=a\overline{a}/4$ or $\left|w-\frac{a}{2}\right|=\frac{|a|}{2}$ which is a circle passing through the origin.

Case 2c: Consider a circle which does not pass through the origin. This can be written as $|z-C|=R$ with $|C|\ne R$. This circle is mapped to the circle

$$\left|w-\frac{\overline{C}}{{|C|}^2-R^2}\right|=\frac{R}{\left|{|C|}^2-R^2\right|}$$

which is another circle not passing through the origin. To show this, we will demonstrate that

$$\frac{\overline{C}}{{|C|}^2-R^2}+\frac{C-z}{R}\frac{\overline{z}}{z}\frac{R}{{|C|}^2-R^2}=\frac{1}{z}$$

Note:$\left|\frac{C-z}{R}\frac{\overline{z}}{z}\right|=1$.

$$\frac{\overline{C}}{{|C|}^2-R^2}+\frac{C-z}{R}\frac{\overline{z}}{z}\frac{R}{{|C|}^2-R^2}=\frac{z\overline{C}-z\overline{z}+\overline{z}C}{z({|C|}^2-R^2)}$$

$$=\frac{C\overline{C}-(z-C)(\overline{z}-\overline{C})}{z({|C|}^2-R^2)}=\frac{{|C|}^2-R^2}{z({|C|}^2-R^2)}=\frac{1}{z}$$

Case 2d: Consider a line passing through the origin. This can be written as $\text{Re}(e^{i\theta }z)=0$. This is mapped to the set $\text{Re}\left(\frac{e^{i\theta }}{w}\right)=0$, which can be rewritten as $\text{Re}(e^{i\theta }\overline{w})=0$ or $\text{Re}(w{e}^{-i\theta })=0$ which is another line passing through the origin.

Case 3: An arbitrary Mobius transformation can be written as $f(z)=\frac{az+b}{cz+d}$. If $c=0$, this falls into Case 1, so we will assume that $c\ne 0$. Let

$$f_1(z)=cz+d\mathit{\hspace{1em}\hspace{1em}}{f}_2(z)=\frac{1}{z}\mathit{\hspace{1em}\hspace{1em}}{f}_3(z)=\frac{bc-ad}{c}z+\frac{a}{c}$$

Then $f=f_3\circ f_2\circ f_1$. By Case 1, $f_1$ and $f_3$ map circles to circles and by Case 2, $f_2$ maps circles to circles.

Title	proof of Möbius circle transformation theorem
Canonical name	ProofOfMobiusCircleTransformationTheorem
Date of creation	2013-03-22 13:38:00
Last modified on	2013-03-22 13:38:00
Owner	brianbirgen (2180)
Last modified by	brianbirgen (2180)
Numerical id	5
Author	brianbirgen (2180)
Entry type	Proof
Classification	msc 30E20