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Let $f(z)=u(x,y)+iv(x,y)$ be an analytic function $f: D \to E$ and $\exists z_0 \in D : f(z_0)=c$ where $z,z_0 \in \mathbb{C}$ and $x,y \in \mathbb{R}$. Then

$$ f(z)=2u\left(\frac{z+\bar{z_0}}{2},\frac{z-\bar{z_0}}{2i}\right)-\bar{c} \tag1$$ $$ f(z)=2iv\left(\frac{z+\bar{z_0}}{2},\frac{z-\bar{z_0}}{2i}\right)+\bar{c} \tag2$$

There is also a question which mentions an equivalent result from Ahlfors' book. But how do I prove either of those identities?

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Minethlos
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    You must know that the expressions on the right hand side of $(1)$ and $(2)$ make sense. If you know that harmonic functions are real-analytic, you can just plug the arguments into the power series of $u$ resp. $v$ and use the identity theorem for real-analytic functions. – Daniel Fischer Nov 13 '14 at 20:45
  • @DanielFischer Uh, and how would I find the coefficients of the power series? – Minethlos Nov 13 '14 at 20:50
  • You can get them from the coefficients of the power series of $f$. It's going to be somewhat tedious, but I don't know how to avoid tedium without resorting to hand-waving here. – Daniel Fischer Nov 13 '14 at 20:54

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