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Find all entire functions such that:

$$\lvert f'(z)\rvert^2 + \lvert f(z)\rvert^2 < 81(\lvert z \rvert^2 + 1), \enspace z \in \mathbb{C},$$

and $f(0)=0$.

I'm struggling with this type of question, I know that I have to bound $\lvert f \rvert$ and then apply Liouville theorem, to find a constant and finally apply the condition $f(0)=0$.

My attempt so far:

$$\lvert f(z) \rvert^2 < 81(\lvert z \rvert^2 + 1)$$ $$ \frac{\lvert f(z) \rvert^2 }{\lvert z \rvert^2 + 1} < 81$$ $$ \frac{\lvert f(z) \rvert^2 }{\lvert z \rvert^2 + 1 + 2 \lvert z \rvert} < 81$$ $$ \frac{\lvert f(z) \rvert^2 }{(\lvert z \rvert + 1)^2} < 81$$ $$ \frac{\lvert f(z) \rvert }{(\lvert z \rvert + 1)} < 9$$

However, I don't know how to go any further. I'd like to say that:

$$ \frac{\lvert f(z) \rvert}{\lvert z +1 \rvert} < 9$$

but it's wrong by the triangle inequality. Is my thinking process correct?

RFTexas
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    Ok, so you have the estimate $|f(z)|<9(|z|+1)$. Now, go over the proof of Louiville's theorem (particularly, Cauchy's inequalities). What does it imply in this case? – peek-a-boo May 09 '21 at 09:05

1 Answers1

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By Riemann's theorem on removable singularities, $f(z)/z$ is entire.

Alternatively, Entire function bounded by a polynomial is a polynomial.

Joshua P. Swanson
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