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.
Best Answer
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 $$ c=\frac{(u-\bar u)}{1-\lvert u\rvert^2}. $$