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Let S be the "slit plane" $S = \mathbb{C} - \{t \in \mathbb{R} : t \leq 0\}$ and deduce that if $f$ is an entire function whose image is contained in $S$, then $f$ is constant.

Clearly $S$ is simply-connected and does not contain zero, so I attempted to go through the analytic branch of $\log(f(z))$ but was not successful. The closest I've gotten seemed rather roundabout:

Let $w: S \to \{z: Im(z) > 0\}$, for instance $w(z) = e^{i\pi /2} \sqrt z = i\sqrt z$. Then the image of $w(f(z))= (w \circ f)(z)$ is contained in the upper half plane. Since $f$ is entire and $f(z) \neq 0$ for any $z$, $\sqrt{f(z)} = e^{\frac{1}{2}\log(f(z))}$ is an entire function, thus $w \circ f = i\sqrt f$ is entire. So $e^{i w \circ f}$ is an entire function whose image lies in the unit disk. In other words, $e^{i w \circ f}$ is an entire bounded function and thus is constant by Liouville's theorem. But now $$e^{i w \circ f}=c \iff i w \circ f = \log c \iff w \circ f = -i\log c \iff w(f(z)) = -i \log c$$ and so we conclude that $f$ must be constant.

Does this solution seem valid?

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    $\begingroup$ I think you can just take $g(z) = e^{i\log(f(z))}$ and apply Liouville's theorem to conclude that $\textrm{Im}(\log(f(z)))$ is constant, which implies $f$ is constant. $\endgroup$ – Ethan Alwaise Oct 15 '16 at 21:07
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Let $g(z) = {1 \over \log f(z) + i (\pi +1) }$, then $|g(z)| \le 1$ and $g$ is entire, hence constant.

Since $\log f(z) = {1 \over g(0)} -i(\pi+1)$, we see that $f$ is constant too.

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