Why is $f$ constant in this case? Reference: Conway - Functions of one complex variable p.55 Exercise 17.
Let $\epsilon>0$.
Let $f:B(0,\epsilon)\rightarrow \mathbb{C}$ be an analytic function such that $|f(z)|=|f(w)|$ for each $z,w\in B(0,\epsilon)$.
In this case, how do I prove that $f$ is constant?
I'm completely lost where to start..

if $|f(z)|$ is a nonzero constant $k$ then $f(z)\overline{f(z)} = k^2$, and so $\overline f = k^2/f$ is also a holomorphic function. Then $g_1(z) = \Re(f(z)) = \frac 12(f(z) + \overline {f(z)})$ is also holomorphic (and only takes real values). $g_2(z) = \Re(f(iz))$ is also holomorphic and also takes real values. Thus $g_1'$ and $g_2'$ have to be real. But $g_2'(z) = ig_1'(iz) \in i\Bbb R$, so we must have $g_2' = 0$. Hence $\Re(f(z))$ is constant, so $f$ can take atm ost two different values, and because it is continuous, $f$ is also constant.
