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I need to show that any complex valued function, $f$, which is entire and $\text{Re }f \leq 0$ is constant.

I know that by Liouville’s Theorem that any bounded entire function is constant.

So, I aim to show that $f$, where $\text{Re }f \leq 0$ is bounded.

Any help in showing this would be appreciated, thanks.

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    $\begingroup$ Apply a simple transformation to $f$ that creates a bounded entire function. If your transformation is invertible, the constancy of $f$ follows. $\endgroup$ – Daniel Fischer Jan 16 '14 at 16:50
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Hint: $$\forall z\in\mathbb C,\ \left|e^z\right|=e^{\mathrm{Re}(z)}$$

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Consider the function $$g(z)=\frac1{f(z)-1},$$ noting in particular that it is entire, and that the denominator is bounded away from $0$--that is, there is some $m>0$ such that $\left|f(z)-1\right|\ge m$ for all $z$. This makes it fairly easy to show that $g$ is bounded, so constant (and non-zero), and so $f$ is constant, since $$f(z)=\frac1{g(z)}+1.$$

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