If $T(z)=\dfrac{az+b}{cz+d}$, find necessary and sufficient conditions that $T(\Gamma)=\Gamma$ where $\Gamma=\{z\in \mathbb{C}: |z|=1\}.$
My solution: Let $T$ be a Möbius map such that $T(\Gamma)=\Gamma$. Since $T:\mathbb{C}_{\infty}\to \mathbb{C}_{\infty}$ then $\exists a\in \mathbb{C}_{\infty}$ such that $T(a)=0$. Consider two cases: 1) $a\in \mathbb{C}$ and 2) $a=\infty$.
1) If $a\in \mathbb{C}$ such that $T(a)=0$. Then the point $a^{*}=\dfrac{1}{\overline{a}}$ is symmetric with $a$. Then $T(a^{*})=\infty$. Also it's obvious that exists $b\in \Gamma$ such that $T(b)=1$. Then $$T(z)=\left(z,b,a,\dfrac{1}{\overline{a}}\right)=\lambda \dfrac{z-a}{\overline{a}z-1}$$ for some $\lambda \in \mathbb{C}$. Since $|T(1)|=1$ then $|\lambda|=1$ and $\lambda=e^{i\theta}$ where $\theta \in[0,2\pi)$. Thus $$T(z)=e^{i\theta}\dfrac{z-a}{\overline{a}z-1}, \quad \theta\in[0,2\pi), \quad a\in \mathbb{C}. \qquad \qquad \qquad (*)$$
2) If $a=\infty$ such that $T(a)=T(\infty)=0$. Then since $0$ and $\infty$ are symmetric hence $T(0)=\infty$. Also $T(b)=1$ for $b\in \Gamma$. Then $T(z)=(z,b,\infty,0)=\dfrac{b}{z}$.
Thus, $$T(z)=\dfrac{e^{i\theta}}{z}, \quad \theta\in[0,2\pi). \qquad \qquad \qquad \qquad \qquad (**)$$
Eventually, the set of Möbius maps such that $T(\Gamma)=\Gamma$ consists of $(*)$ and $(**)$.
Is my solution correct? Would be very grateful if anyone can check my solution, please!