# Prove that an analytic function on complex plane to its proper open and simply connected set is constant

Suppose that $V$ is an open, simply connected, proper subset of $\mathbb{C}$. Suppose that $f\colon\mathbb{C}\rightarrow V$ is holomorphic. Prove that $f$ is constant function. Give counter example to the case in which $V$ is not simply connected.

It seems to be application of Riemann Mapping theorem but I am stuck ....

This problem is from old comprehensive problem, now I am preparing for this exam, Could you please help me to find the solution for this?

• If you already thought of the Riemann mapping theorem, what problem are you still facing? – Daniel Fischer Jul 15 '13 at 23:22
• By Picard, you have that there is more than one point that $f$ omits in the codomain $\mathbb{C}$ and so it must be constant (for the first part). – Cameron Williams Jul 15 '13 at 23:23
• For the counter-example: $f:\mathbb{C}\rightarrow\mathbb{C}-\{0\}$ where $f(z) = e^z$. – Cameron Williams Jul 15 '13 at 23:31

If you want to use the Riemann mapping theorem instead of Picard, let $g:V\to D$ be a (biholomorphic) map from $V$ to the unit disc. Then $g\circ f$ is holomorphic on $\mathbb C$ and bounded. Liouville finishes things off.
Since $V$ is open and simply connected, $\mathbb{C}\backslash V$ contains more than one point. By the little Picard theorem, since the range of $f$ omits more than one point in $\mathbb{C}$, it must be constant.
As for an example of a nonconstant holomorphic function on a non-simply connected set, let $f:\mathbb{C} \to \mathbb{C}-\{0\}$ such that $f(z) = e^z$.
• @ Cameron thank you for your solution but how do we guarantee that $C$-$V$ contains more than one point? expecting help – Ganga Jul 15 '13 at 23:45
• @Ganga if it only has one point, then $V$ has a hole which is not simply connected. – Cameron Williams Jul 15 '13 at 23:48