# Explicit calculation of the center of a circle, image of a circle by a Möbius transformation

It's a warm up calculation I decided to carry out while reading "PCT,Spin and statistics, and all that" by Streater and Wightmann. However I do not find what they have.

p.79 within the proof of Thm 2-14 p.77 (the calculation has not much to do with the proof, at least at this point. But if you are reading the book, notice that in the figure 2-7 p.79 they consider a function of z and $$|u|\neq 1$$ while just above, it was a function of u on the unit circle...): let's consider the following Möbius transformation

$$T: z \mapsto \frac{u+z}{1+uz}\ ,\quad |u|\neq 1$$ (otherwise the unit circle is mapped to $$\mathbb{R}$$, as can be seen by a calculation analogous to the following)

The unit circle is mapped to another circle, whose center I wish to find. I recall that the inverse of a Möbius transformation (in particular, such maps are invertible...) $$S: z \mapsto \frac{az+b}{cz+d}\ ,\ a,b,c,d \in \mathbb{C} \quad \text{is}\quad S^{-1}: z \mapsto \frac{dz-b}{-cz+a}$$ so in our case (as can also be checked directly) $$T^{-1}: w \mapsto \frac{w-u}{1-uw}$$ Let's now write the condition for $$z$$ to be on the unit circle and see what conditions its image $$w:=T(z)$$ will then satisfy: $$|z|²= 1 \quad \Leftrightarrow \quad |T^{-1}(w)|^2 =1 \quad \Leftrightarrow \quad \left(\frac{w-u}{1-uw}\right) \overline{\left(\frac{w-u}{1-uw}\right)}=1$$ $$\Leftrightarrow \quad |w|^2 - 2\, \mathop{Re}(w\overline{u}) + |u|^2 = 1 - 2\, \mathop{Re}(w u) + |uw|^2$$ $$\Leftrightarrow \quad (1-|u|²)|w|^2 - 2\, \mathop{Re}(w(\overline{u}-u)) + |u|^2 -1 = 0$$ $$\Leftrightarrow \quad |w|^2 - 2\, \mathop{Re}\left(w\ \frac{2\,i \mathop{Im}(u)}{1-|u|²}\right) -1 = 0$$ Identifying with the equation of a circle of center $$c\in \mathbb{C}$$ and radius $$r \in \mathbb{R}_+$$: $$|w-c|^2=r² \quad \Leftrightarrow \quad |w|^2 - 2\, \mathop{Re}(w \overline{c}) +|c|^2 - r² = 0$$ one obtains $$c=- \frac{2\,i \mathop{Im}(u)}{1-|u|²} \quad \text{and}\quad r= \sqrt{1 + |c|²}$$

However, in the book it seems that they find $$c= \frac{4 \left[ u(1+|u|²)- (1+|u|²) \mathop{Re}(u) \right]}{\left[ (1+|u|²)(1+u²) - 4 u \mathop{Re}(u) \right]}$$

So if a benevolent mind double checks the present calculation (or does something of its own), I'll be happy to discuss the result.

I almost agree with your answer, except you made a mistake in the sign: $$\bar u-u=-2i\operatorname{Im}u$$, not $$2i\operatorname{Im}u$$, so $$c=\frac{2i\operatorname{Im}u}{1-\lvert u\rvert^2}.$$

Here is an alternative method.

Since $$u\neq\pm 1$$, the fixed points $$z=T(z)$$ is easily seen to be $$z=\pm 1$$. Hence $$T$$ maps the unit circle to a circle containing $$\pm 1$$, so the centre has to lie on the imaginary axis. Moreover, we compute derivative $$T'(z)=\frac{1-u^2}{(1+uz)^2},\quad T'(\pm1)=\frac{1\mp u}{1\pm u}$$ So the unit circle (with normal direction $$1$$ at $$\pm 1$$) is mapped to a circle whose normal direction at $$\pm 1$$ is given by $$\frac{1\mp u}{1\pm u}$$. Now $$\frac{1\mp u}{1\pm u}=\frac{(1\mp u)(1\pm\bar u)}{(1\pm u)(1\pm\bar u)}= \frac{(1-\lvert u\rvert^2)\mp (u-\bar u)}{\lvert 1\pm u\rvert^2}$$ So both normals intersect the imaginary axis at $$c=\frac{(u-\bar u)}{1-\lvert u\rvert^2}.$$

$$-1/u$$ is mapped to $$\infty$$, therefore its conjugate point wrt the unit circle, which is $$-\overline u$$, is mapped to the center of the image circle.

• I didn't know this "rule", insteresting... – Noix07 Nov 12 '18 at 10:39
• It's not about the point at infinity specifically; Mobius transformations preserve conjugation. – Maxim Nov 12 '18 at 12:14
• Thanks, got it. To make a useful comment, for those like me who did not know about these stuffs, "preserve conjugation" means "if two points are conjugate w.r.t. a circle or a line, then there image are conjugate w.r.t. the image circle/line" – Noix07 Nov 16 '18 at 13:32

As shown in this answer, given the LFT $$\frac{z+u}{uz+1}$$ and the circle with radius $$1$$ and center $$0$$, we find the antipodal points $$0\pm\frac{0+1/u}{|0+1/u|}\cdot1=\pm\frac{|u|}{u}$$ These points get mapped to antipodal points in the image of $$\frac{z+u}{uz+1}$$: $$\frac{u+\frac{|u|}{u}}{1+|u|}\qquad\frac{u-\frac{|u|}{u}}{1-|u|}$$ Therefore, the center is $$\newcommand{\Im}{\operatorname{Im}} \frac12\left(\frac{u+\frac{|u|}{u}}{1+|u|}+\frac{u-\frac{|u|}{u}}{1-|u|}\right) =\frac{2i\Im(u)}{1-|u|^2}$$ and the radius is $$\frac12\left|\frac{u+\frac{|u|}{u}}{1+|u|}-\frac{u-\frac{|u|}{u}}{1-|u|}\right| =\left|\frac{1-u^2}{1-|u|^2}\right|$$