Linear Fractional Transformation - proof that it preserves circles

Since linear fractional transformations are compositions of translations, mutliplications by a constant and inversion, I tried proving that an LTF would transform circles into circles by writing the equation of a circle as $z_0 + r\cdot\exp ( i \cdot \theta )$ , $0 \le \theta \le 2\pi$ and then looking at the effect of translations, multiplications by a constant and inversion.

The first two clearly leave me with a circle, as does inversion of a circle centered around the origin, but I can't get this to work for a circle centered at an arbitrary point $z_0$.

On the other hand, if I write down the general equation of a circle: $$A(x^2 + y^2) + Bx +Cy +D = 0$$

...and write $w = u+iv = \frac{1}{z} = \frac{1}{(x+iy)}$ and make the appropriate substitutions, I get the equation of circle in terms of $u$ and $v$. Is it possible to write $$\frac{1}{z_0 + re^{i\theta}}$$ where $0 \le \theta \le 2\pi$, in the form $$w_0 + r'e^{i\phi}$$ with $0 \le \phi \le 2\pi$?

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Please use LaTeX to write mathematics in this site. You can go to the FAQ section and search there. – DonAntonio Dec 14 '12 at 3:24
I realize LaTeX can be intimidating when you need to use it the first couple of times. Thus, I've just suggested an edit on your post that added the necessary LaTeX. If you look over it, you may find some commands you find useful. – apnorton Dec 14 '12 at 3:58
@anorton: I suspect that the 2nd last equation should be $\frac{1}{z_0 + r e^{i \theta}}$. – copper.hat Dec 14 '12 at 7:07
@copper.hat I really have no idea (I haven't had complex analysis). I'll suggest the change... EDIT: Oops. I don't have the rep to suggest that trivial of a change. Never mind. – apnorton Dec 14 '12 at 13:33

It is possible, but it requires some grunt work. Let $w_{z_0}(r, \theta) = \frac{1}{z_0+r e^{i\theta}}$.

We deal with a simple case first: If $z_0 = 0$, then $w_0(r,\theta) = \frac{1}{r}e^{-i\theta}$, which is a circle.

Now reduce the problem to a simpler one. Let $w_{z_0}(r, \theta) = \frac{1}{z_0+r e^{i\theta}}$, and suppose $z_0 = |z_0| e^{i \zeta}$. Then we have $w_{z_0}(r, \theta) = \frac{1}{z_0}\frac{1}{1+\frac{r}{z_0} e^{i\theta}} = \frac{1}{z_0}\frac{1}{1+\frac{r}{|z_0|} e^{i(\theta-\zeta)}} = \frac{1}{z_0} w_1(\frac{r}{|z_0|}, \theta-\zeta)$. So, we will analyze $w_1$ first, and then generalize to arbitrary $z_0 \neq 0$. We have $w_1(r, \theta) = \frac{1}{1+r e^{i\theta}} = \frac{1+r e^{-i\theta}}{1+2 r \cos \theta + r^2} = \frac{1+r \cos \theta - i r \sin \theta}{1+2 r \cos \theta + r^2}$.

Note that if $r=1$, then $w_1(1,-\pi) = \infty$, so we expect $\theta \mapsto w_1(1,\theta)$ to be a line. To see this, look at $w_1(1,\theta) = \frac{1}{2} \frac{1+ \cos \theta - i \sin \theta}{1+\cos \theta } = \frac{1}{2} (1 - i \frac{\sin \theta}{1+\cos \theta })$, so $\theta \mapsto w_1(1,\theta)$ is a line parallel to the imaginary axis.

Now suppose $r\neq 1$. Since we expect $\theta \mapsto w_1(r,\theta)$ to be a circle, we should look for the center first. If we have two points on a circle, we know that the perpendicular bisector of the segment joining the points passes through the center. Now note that $w_1(r,0) = \frac{1}{1-r}$, $w_1(r,\pi) = \frac{1}{1+r}$, hence the real part of the center must line on the line $\text{Re}\, z = \frac{1}{2} (\frac{1}{1-r} + \frac{1}{1+r}) = \frac{1}{1- r^2}$. Similarly, since $w_1(r,\frac{\pi}{2}) = \frac{1-i r}{1+r^2}$, $w_1(r,-\frac{\pi}{2}) = \frac{1+i r}{1+r^2}$, we see that the center must lie on the line $\text{Im}\, z = \frac{1}{2} (\frac{1-i r}{1+r^2} + \frac{1+i r}{1+r^2}) = 0$. So, we expect the center to be $\frac{1}{1- r^2}$. To confirm, look at $|w_1(r,\theta) - \frac{1}{1- r^2}|$:

\begin{eqnarray} |w_1(r,\theta) - \frac{1}{1- r^2}| &=& \frac{1}{|1-r^2|}\left| \frac{(1-r^2)(1+r \cos \theta - i r \sin \theta)}{1+2 r \cos \theta + r^2} - 1 \right| \\ &=& \left| \frac{(1-r^2)(1+r \cos \theta - i r \sin \theta)- (1+2 r \cos \theta + r^2)}{(1-r^2)(1+2 r \cos \theta + r^2)}\right| \\ &=& \frac{|r|}{|1-r^2|} \left| \frac{-(2r+(1+r^2)\cos \theta) - i (1-r^2) \sin \theta}{1+2 r \cos \theta + r^2}\right| \\ &=& \frac{|r|}{|1-r^2|} \frac{\sqrt{(2r+(1+r^2)\cos \theta)^2+((1-r^2) \sin \theta)^2}}{1+2 r \cos \theta + r^2} \\ &=& \frac{|r|}{|1-r^2|} \end{eqnarray} Hence $w_1(r,\theta)$ lies on a circle $\Gamma$ of radius $\frac{|r|}{|1-r^2|}$ centered at $\frac{1}{1- r^2}$.

To see that the map $\theta \mapsto w_1(r,\theta)$ is onto $\Gamma$, we note that the map $\phi: \mathbb{C}_\infty \to \mathbb{C}_\infty$ given by $\xi(z) = \frac{1}{z}$ is a homeomorphism (continuous with a continuous inverse). If we let $D=\{z||z|=1\}$, we note that $D^C$ is disconnected, and also $w_1(r,[0,2 \pi]) = \xi(D) \subset \Gamma$. However, if $\xi(D) \neq \Gamma$, then $\xi(D)^C$ is connected and by continuity of $\xi^{-1}$, $\xi^{-1}(\xi(D)^C) = \xi^{-1}(\xi(D^C)) = D^C$ is connected, which is a contradiction.

If we let $\phi(r,\theta) = i\,\mathbb{arg} (-(2r+(1+r^2)\cos \theta) - i (1-r^2) \sin \theta)$, then we can write $w_1(r,\theta) = \frac{1}{1- r^2}+\frac{r}{|1- r^2|}e^{i \phi(r,\theta)}$.

Now to return to the general problem. As above, we see that $w_0(r,\theta) = \frac{1}{r}e^{-i\theta}$, and for $z_0 = |z_0| e^{i \zeta} \neq 0$, we have $w_{z_0}(r, \theta) = \frac{1}{z_0} w_1(\frac{r}{|z_0|}, \theta-\zeta)$. Hence if $r=|z_0|$, we have the line $w_{z_0}(|z_0|, \theta) = \frac{1}{2 z_0} (1 - i \frac{\sin (\theta-\zeta)}{1+\cos (\theta-\zeta) })$. Finally, if $r \neq |z_0|$, we have $w_{z_0}(r, \theta) = \frac{1}{z_0}\frac{1}{1- (\frac{r}{|z_0|})^2}+\frac{\frac{r}{|z_0|}}{|z_0||1- (\frac{r}{|z_0|})^2|}e^{i (\phi(\frac{r}{|z_0|},\theta)-\zeta)}$.

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