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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!

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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 $$ \overline \alpha = - \delta, \overline \beta = -\gamma, |\alpha| \ne |\beta| \, . $$

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  • $\begingroup$ 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:) $\endgroup$
    – Oski
    May 31, 2020 at 9:09
  • $\begingroup$ @Oski: $a, b, c, d$ are real numbers, so $a = \operatorname{Im}(C+D) /2 $, $b = \operatorname{Re} (C+D)/2$, etc. $\endgroup$
    – Martin R
    May 31, 2020 at 9:42

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