The question: Let $L \subset\mathbb{C}$ be the line $L = \{ z = x+iy : x = y $}. Assume that $f : \mathbb{C} \rightarrow \mathbb{C}$ is an entire function such that for any $z \in L$ we have $f(z) \in L$. Assume that $f(1) = 0$. Prove that $f(i) = 0$.
The attempt: I think I have some sort of an idea how to work this out. I think I have to use the Riemann Mapping Theorem on the Line to the disc. Is that the right way or do I have to Schwartz Lemma?
Please only give me hints. Try not to solve the problem completely. Thank you guys!!!
Let $g(z)=(1-i)f\bigl((1+i)z\bigr)$. If $x\in\mathbb R$, then $(1+i)x\in L$, and therefore $f\bigl((1+i)x\bigr)\in L$. But then, since $(1-i)(1+i)=2\in\mathbb R$, $g(x)\in\mathbb R$. So, $g$ is an entire function which maps $\mathbb R$ into $\mathbb R$. But then $(\forall z\in\mathbb C):\overline{g(z)}=g\left(\overline z\right)$. In particular, $g(z)=0\implies g\left(\overline z\right)=0$. But $f(1)=0\iff g\left(\frac{1-i}2\right)=0$. Therefore, $g\left(\frac{1+i}2\right)=0$, which means that $f(i)=0$.