This is an old question from Ph.D Qualifying Exam of Complex Analysis.

Without using Picard's theorem directly, prove that if $f$ is an entire function such that $\text{Im}f(z)\le (\text{Re}f(z))^2$ for all $z\in \mathbb{C}$ then $f$ is constant.

My attempt: I'd like to apply Liouville's theorem, but I can't find the entire function to apply the theorem. First I tried for $e^{f^2}$, but its real part becomes $e^{(Ref)^2-(Imf)^2}\ge e^{Imf-(Imf)^2}$ so I failed to find the bound.

How should I do?

Thanks in advance!


2 Answers 2


The image of $f$ does not intersect a disc centred at $i$. Therefore $z\mapsto 1/(f(z)-i)$ is entire and bounded. Use Liouville.

  • $\begingroup$ Clever, generalizable trick. Do you know of any other good techniques for these kinds of problems? $\endgroup$
    – user217285
    Commented Jul 19, 2018 at 6:07

One can prove more general result using Lord Shark the Unknown's argument as follows:

If $f$ is a nonconstant entire function, then $f(\mathbb{C})$ is dense in $\mathbb{C}$.

Proof: Suppose $f(\mathbb{C})$ is not dense in $\mathbb{C}$. Then there exist $z_0\in \mathbb{C}$ and $\delta>0$ such that $D_\delta(z_0)\cap f(\mathbb{C})=\emptyset$. Now consider the function $g(z)=\dfrac{1}{f(z)-z_0}$, then $|g(z)|\le \dfrac{1}{\delta}$ for all $z\in \mathbb{C}$ because $|f(z)-z_0|\ge \delta$. Therefore, $g$ is constant by Liouville's theorem, and it follows that $f$ is constant.


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