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Let $\Omega$ be a simply connected domain that is properly contained in $\mathbb C$, and $u(x,y)$ is harmonic on the unit disk $\mathbb D $, then there is a funtion $f(z)$, that is one-one and analytic on $\Omega $, so that $u$ o $f$ is harmonic on $\Omega$.

All I can think is there exists a harmonic conjugate $v$ in simply connected domain $\Omega$ such that $f=u+iv$ is analytic there. I do not see how that same function will be injective and the composition will be harmonic. Or the function $f$ in the question is not really the one I mentioned above? Help!

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Related: math.stackexchange.com/questions/243193/… –  user53153 Dec 28 '12 at 5:12

1 Answer 1

Since $\Omega$ is simply connected, open, and not the whole of $\mathbb{C}$, we can use the Riemann Mapping Theorem, which guarantees that there exists a biholomorphic map $f:\Omega \to \mathbb{D}$. You can then use the fact that a composition of an analytic and a harmonic function function is harmonic, as indicated in Pavel's comment, and conclude that $u \circ f$ is indeed a harmonic function.

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