Odd Function on a pointsymmetric domain 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!! 
 A: 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.
A: 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 $a_n(-z)^n=-a_nz^n$. 
