Bounded entire fuctions.

Question

Let $f$ be an entire function and assume f(0) = 1 and $|f(z)| \large \geq\frac{1}{3}\left| {\LARGE e^{z^{3}}}\right|$ for all $z$.

Show $f(z) = {\huge e^{z^{3}}}$ for all $z$.

Can this be shown with the Cauchy integral formula?

Answer 1

I think it's more direct to use Liouville's theorem. (You could use the Cauchy integral formula, but it would then probably be emulating the proof of Liouville's theorem)

Note that $g(z) = \frac{e^{z^3}}{f(z)}$ is entire (in particular because $f(z) \ne 0$ for any $z$ by the hypothesis), and it satisfies $|g(z)| \le 3$. Thus, $g(z)$ is constant, and therefore the conclusion follows immediately.