Let $A\subseteq\mathbb{C}$ be a region in the complex plane.
Let $f:A\rightarrow\mathbb{C}$ be an analytic function on $A$ and $f\left(A\right)\subseteq\left\{z\in\mathbb{C}:|z|=3 \right\}$.
Show that f is a constant function.
I'm having a bad while with this problem. The first thing I tried was to get $f'(z)=0, \forall z\in A$ so the function would be a constant, but I failed to get it.
The second idea I have is to try to get $f'(z)\neq0$ so I can aply the Inverse Function Theorem. I'm not getting into this problem.
Any ideas? Thanks in advanced.
If you don't know the open mapping theorem, write $f=u+iv$ where $u$ and $v$ are real-valued. Then by assumpton, $u^2+v^2=9$. Differentiate and use Cauchy-Riemann's equations to show that all four derivatives $u'_x$, $u'_y$, $v'_x$ and $v'_y$ must be identically $0$.