The question: Let $\ Ω$ be a simply connected domain Let $\phi_{1}$ and $\phi_{2}$ be conformal self maps on $\ Ω$. Let $P, Q$ be distinct points in $\ Ω$ If $\phi_{1} (P) = \phi_{2}(P)$ and $\phi_{1} (Q) = \phi_{2}(Q)$, show that $\phi_{1} = \phi_{2}$.
My attempt: I want to try to show this in either one or two ways. The first way was to consider the function $h(z) = \phi_{1} -\phi_{2}$ and try to show that $h(z) = 0$ on $\ Ω$ but I am having trouble working on that approach. The second idea I had was to consider the conformal map to the unit disc to itself. It was a map like this:
$$ Φ_{a}(z) = ω\frac{z -a}{1-\bar{a}z}. $$
Ιnstead of a, I was going to replace them with points $\phi_{1} (P), \phi_{1}(Q)$ and try to find a relation between the two functions. I am not sure I am on the right track here. I want to use function composition but I am not sure how to do it. Can you guys give me some hints on how to work out this problem? Thanks guys!
In the case $\Omega = \Bbb C$ you can use that the only conformal self-maps of $\Bbb C$ are linear functions, and a linear function is uniquely determined by the values at two distinct points.
Otherwise, transporting the problem to the unit disk is the right approach.
$\phi = \phi_2^{-1} \circ \phi_1$ is a conformal self-map of $\Omega$ with two distinct fixed points $P$ and $Q$, and the goal is to show that $\phi$ is the identity function.
The Riemann mapping theorem gives a conformal map $h$ from $\Omega$ to the unit disk with $h(P) = 0$ and $h'(P) > 0$. Then $f = h \circ \phi \circ h^{-1}$ is a conformal self-map of the unit disk with $f(0) = 0$, $f'(0) > 0$, and the additional fixed point $h(Q)$.
Use the Schwarz lemma to conclude that $f$ (and therefore $\phi$) is the identity function.