Compositum of two conformal Mappings

Question

Let $g_b: \mathbb{H} \to \mathbb{B}$, $g_b(z) := \frac{z-b}{z-\overline{b}}$ ($b \in \mathbb{H}$) and $f_a : \mathbb{B} \to \mathbb{B}$, $f_a(z) := \frac{z-a}{\overline{a}z-1}$ ($a \in \mathbb{B} := B_1(0)$) be holomorphic functions. I want to show that $f_a \circ g_b$ can be written as $g_c$.

My Idea: \begin{align*} (f_a \circ g_b)(z) &= \frac{\frac{z-b}{z-\overline{b}}-a}{\overline{a}\frac{z-b}{z-\overline{b}}-1} = \frac{z-b - a(z-\overline{b})}{\overline{a} (z-b) - (z-\overline{b})} = \frac{1-a}{\overline{a}-1} \frac{z + \frac{a\overline{b} - b}{1-a}}{z - \frac{\overline{a} b - \overline{b}}{\overline{a}-1}} \\ &= \frac{a-1}{\overline{a}-1} \frac{z+ \frac{a\overline{b} - b}{a-1}}{z - \frac{\overline{a} b - \overline{b}}{\overline{a}-1}} \end{align*} I don't know how to get rid of the the prefactor. I only can define $c:= -i \frac{a\overline{b} - b}{a-1}$ but it's not the same like $g_b$. Please help.

Answer 1

This isn't true, because $g_b$ sends $\infty$ to $1$ and $f_a$ does not send $1$ to $1$; so $f_a \circ g_b$ doesn't send $\infty$ to $1$. You're going to be stuck with a prefactor.