Maximal value of real part of holomorphic function

Question

Let $f:U \rightarrow C$ be a non-constant holomorphic function. $U$ is open, connected and $D(0,1+\epsilon) \subset U$. I'd like to show that there exists $z_0 \in \partial D(0,1)$ such that $Re(f(z))<Re(f(z_0))$ for $|z|<1$.Additionaly for such a point $z_0$ it is true that $Im(z_0f^{'}(z_0))=0$. The first part is easy since $D(0,1) \cup \partial D(0,1)$ is compact, $Re(f(z)):U \rightarrow R$ is continuous and $f$ sends open sets to open sets.I'm struggling to show the second part, though. I'll be grateful for any hints.

Answer 1

It's actually from direct computation: $$ \Im(zf'(z)) = y\frac{\partial u}{\partial x} - x\frac{\partial u}{\partial y}. $$ Evaluated at $z_0$, it's the directional derivative of $u$ in the direction $(y_0,-x_0)$, tangent to the unit circle.