proof of Möbius circle transformation theorem

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

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

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

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 ${\mbox{Re}(Cw)}=\frac{1}{2}$ which does not pass through the origin. To show this, write $z=C+|C|e^{i\theta}$. $w=\frac{1}{z}=\frac{1}{C+|C|e^{i\theta}}$.

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

Case 2b: Consider the line which does not pass through the origin. This can be written as $\mbox{Re}(az)=1$ for $a\neq 0$. 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 $|C|\neq R$. 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 $\mbox{Re}(e^{i\theta}z)=0$. This is mapped to the set $\mbox{Re}\left(\frac{e^{i\theta}}{w}\right)=0$, which can be rewritten as $\mbox{Re}(e^{i\theta}\overline{w})=0$ or $\mbox{Re}(we^{-i\theta})=0$ 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 $c\neq 0$. Let

 $f_{1}(z)=cz+d\qquad f_{2}(z)=\frac{1}{z}\qquad f_{3}(z)=\frac{bc-ad}{c}z+\frac% {a}{c}$

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 ProofOfMobiusCircleTransformationTheorem 2013-03-22 13:38:00 2013-03-22 13:38:00 brianbirgen (2180) brianbirgen (2180) 5 brianbirgen (2180) Proof msc 30E20