Suppose that $f: \mathbb{C} \rightarrow \mathbb{C}$ is entire and bounded on the set $\{z \in \mathbb{C}; Re(z) \leq 0\}$. Is $f$ a constant function.

I know by Picards theorem that a non-constant entire function assumes all but one value in the complex plane. Can this result be tweaked and applied here? Any hints?

  • $\begingroup$ here's the full question, asked much earlier. $\endgroup$ – Jesse P Francis Jun 22 '15 at 11:28

Take $f(z) = e^z$. Then, for any $z$ such that $Re(z) \leq 0$, we have that $f(z)$ is contained in the (closed) unit circle, yet $f$ is not constant.

  • 1
    $\begingroup$ ..with radius $1$.. $\endgroup$ – Berci Jun 22 '15 at 11:27
  • $\begingroup$ @Berci While, strictly speaking, I wasn't wrong, I'll give you that one. $\endgroup$ – Arthur Jun 22 '15 at 11:29

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