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Helllo,

I have the following problem:

Let $ G \subseteq \mathbb C $ be a domain with $ 0 \in G $. Let $ f: G \to \mathbb {C} $ be a holomorphic function with $ f (G \cap \mathbb R) \subseteq \mathbb R $ and $ f (G \cap \mathbb Ri) \subseteq \mathbb Ri $ , Show that $ f $ must then be an odd function. That means for all $ z \in G \cap (-G) $: $ f (-z) = -f (z) $.

It seems like I should divide the domain into different parts, but I have no clear idea how to get that problem solved. Did anyone have an idea to start? Has it something to do with Schwarz reflection principle?

Thanks for every answer!!

Leviathan
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2 Answers2

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Hints

I would use the fact that $f$ is equal to its Taylor series on $G$ to write that $$f(z)=\sum_{n=0}^{\infty} a_n z^n$$

Then using the fact that $f(z) \in \mathbb R$ for $z \in \mathbb R$ you should be able to prove that the $a_n$ are real.

Using the fact that $f(z) \in i\mathbb R$ for $z \in i\mathbb R$ you can prove that for $n$ even $a_n$ is zero.

fonfonx
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The function is holomorphic and can therefore be decomposed in a power series; \begin{align*} f(z) = \sum_{n=0}^{\infty} a_n z^{n} \end{align*} Furthermore, it is known that $f(i\mathbb{R}) \in i \mathbb{R}$ and $f(\mathbb{R}) \in \mathbb{R}$ . That is why \begin{align} f(r) &= \sum_{n= 0}^{\infty} a_n r^{n}, \quad r\in \mathbb{R} \\ f(ir) &= \sum_{n=0}^{\infty} a_n (ir)^{n} , \quad r\in \mathbb{R} \label{ir}\, . \end{align} This equation can be written as \begin{align*} f(ir) = \sum_{n=0}^{\infty} i^{n} a_n r^{n} = \sum_{n=0}^{\infty} (-1)^{n} a_{2n} r^{2n} + i \sum_{n=0}^{\infty} (-1)^{n} a_{2n+1} r^{2n+1} \end{align*} By assumption, $f(i\mathbb{R}) \in i\mathbb{R}$, that is, $a_{2n} = 0$ consequently we have an odd function because only the odd powers are present: \begin{align*} f(z) = \sum_{n=0}^{\infty}a_{2n+1} z^{2n+1} \end{align*} Which is equivalent for saying $z\in G \cap (-G)$ because $a_n(-z)^n=-a_nz^n$.

Leviathan
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