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Let $\Omega\neq\mathbb{C}$ be a simply connected domain containing a point $c$. Let $\phi:\Omega\rightarrow D$ be a conformal mapping such that $\phi(c)=0$. The function $g_{c}(z)=\log|\phi(z)|$ is called the Green's function of $\Omega$ corresponding to $c$. Show that $g_{a}(b)=g_{b}(a)$ for any $a,b\in\Omega$.

I'm not sure about this. I took $\phi_{a}:\Omega\rightarrow D$ with $\phi_{a}(a)=0$ and $\phi_{b}:\Omega\rightarrow D$ with $\phi_{b}(b)=0$ so that $g_{a}(z)=\log|\phi_{a}(z)|$ and $g_{b}(z)=\log|\phi_{b}(z)|$. Then it seems I need some map from $\Omega$ to $\Omega$ that perhaps interchanges $a$ and $b$ but I can't justify the existence of such a function and don't know how to continue.

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up vote 2 down vote accepted

How to exchange two points of a domain by a conformal automorphism of the domain?

Consider the simplest nontrivial case: exchange $0$ and $b\in D$ by a conformal automorphism of $D$. You should know a Möbius map that sends $b$ to $0$, namely $\frac{z-b}{1-\bar b z}$. However, it sends $0$ into $-b$. This is easily fixed: take $-\frac{z-b}{1-\bar b z}$ instead.

You can now swap any pair $a,b\in D$ by first moving $a$ to $0$ by a Möbius map $f$, then exchanging $0$ and $f(b)$, then applying $f^{-1}$.

And if two points lie in some simply connected domain $\Omega\ne \mathbb C$, you can map the domain onto $D$ by some conformal map $f$, exchange $f(a)$ and $f(b)$, and then apply $f^{-1}$.

That said, there is a more streamlined way to do this problem:

  • compute $g$ for $\Omega=D$. In this case $\phi_a$ is known explicitly, and you can see directly that $g$ is symmetric.
  • show that if $f:\Omega \to D$ is a conformal map, then for any $a,b\in \Omega $ $$g_a^{\Omega }(b) = g_{f(a)}^{D}(f(b)) $$ where I used subscripts to distinguish Green's functions of different domains. This is a consequence of the fact that composition of conformal maps is conformal.
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