# Möbius transformation that maps the unit circle to itself

I need to find necessary and sufficient conditions on the coefficients of a Möbius transform $$T(z)=\frac{\tilde a z+\tilde b}{\tilde c z+ \tilde d}$$ so that it maps the unit circle $$\{z: |z|=1\}$$ into itself.

I initially thought that, since any Möbius transformation can be written as a finite composition of simple transformations (translations ($$z+a$$), rotations ($$e^{i\theta}z$$), dilations ($$az$$) and inversions ($$\frac1z$$)) and since we do not want to dilate or move the unit circle, we then can write the required transformation as $$T(z)=e^{i\alpha}z$$ or $$T(z)=\frac{e^{i\alpha}}{z}$$ for some $$\alpha\in (-\pi,\pi]$$. However, this does not look like the result I'm supposed to get. Where is my mistake?

I then read the exercise hint, which says I should first write a transformation $$R$$ that maps the unit circle to $$\mathbb{R}_\infty$$, and use transformations $$S(z)$$ that map $$\mathbb{R}_\infty$$ to $$\mathbb{R}_\infty$$, which I believe are of the form $$S(z)=\frac{az+b}{cz+d}$$ where $$a,b,c,d\in\mathbb{R}$$.

I then chose $$R(z)=\frac{z+1}{z-1}i$$ and tried composing $$T=R^{-1}\circ S\circ R$$ to find the answer. However, I'm getting an ugly expression that does not seem to be correct either: $$T(z)=\frac{(A+Bi)z-\overline{(A-Bi)}}{(A-Bi)z-\overline{(A+Bi)}}$$, where $$A=b+ai$$ and $$B=d+ci$$.

Could you help me see how to use the hint? I know there are other solutions for this problem on this site, but they solve it in different ways. Thank you!

Your calculations are correct so far. $$S(z)=\frac{az+b}{cz+d}$$ with $$a,b,c,d\in\mathbb{R}$$, $$ad-bc \ne 0$$, is the general Möbius transformation mapping $$\Bbb R \cup \{ \infty \}$$ onto itself. (It maps the upper halfplane onto itself iff $$ad-bc > 0$$.)
Your expression for $$T$$ becomes simpler if we set $$C = A+iB = (b-c) + i(a+d) \, ,\\ D = \overline{A-iB} = (b+c) +i (a-d) \, .$$ Then $$|C|^2 - |D|^2 = 4(ad-bc) \ne 0 \implies |D| \ne |C|$$ and $$T(z) = \frac{Cz-D}{\overline D z - \overline C} \, .$$ This is the general form of a Möbius transformation mapping the unit circle onto itself. (It maps the unit disk onto itself iff $$|D| < |C|$$).
So the conditions on $$T(z)=\frac{\alpha z+\beta}{\gamma z+ \delta}$$ are that $$\overline \alpha = - \delta, \overline \beta = -\gamma, |\alpha| \ne |\beta| \, .$$
• Thank you so much:) I have a bit of trouble seeing how the parameters $C,D$ can be reduced to being any complex numbers with the only condition $|C|\neq |D|$. I tried setting $C=C_1+iC_2$ and $D=D_1+iD_2$ to find the necessary values for $a,b,c,d$ and I found the solutions $a=\frac{C_2+D_2}{2}$, $b=\frac{C_1+D_1}{2}$, $c=\frac{D_1-C_1}{2}$, $d=\frac{C_2-D_2}{2}$, so it seems like $C$ and $D$ could be set to any complex values. The condition $ad-bc\neq 0$ forbids that, of course, but I wouldn't have noticed it and I'm not sure how to see why there aren't more conditions. Thank you:)
• @Oski: $a, b, c, d$ are real numbers, so $a = \operatorname{Im}(C+D) /2$, $b = \operatorname{Re} (C+D)/2$, etc. May 31, 2020 at 9:42