# Bounded Complex Function

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.

• Apply a simple transformation to $f$ that creates a bounded entire function. If your transformation is invertible, the constancy of $f$ follows. – Daniel Fischer Jan 16 '14 at 16:50

Hint: $$\forall z\in\mathbb C,\ \left|e^z\right|=e^{\mathrm{Re}(z)}$$
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 $$f(z)=\frac1{g(z)}+1.$$