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Let $f$ be an entire function whose square $f^2$ is a polynomial. Then is $f$ a polynomial as well?

I think due to the Great Picard Theorem, since $f$ cannot assume any complex value infinitely many times, it is forced to be a polynomial. Is my argument correct?

Also if $f$ is an entire function such that $f(f(z))$ is a polynomial, the same reasoning with the Great Picard Theorem shows that $f$ is again forced to be a polynomial.

Would anyone please give me a comment on my argument?

Keith
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    Both arguments sound good. Make sure that you can solidly argue the essential point (if $f$ assumes a value infinitely often then so do $f^2$ and $f\circ f$). – Erick Wong Aug 28 '19 at 18:24
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    Casorati-Weierstraß is sufficient, see https://math.stackexchange.com/questions/991928/proving-fz-and-gz-are-polynomials. – Martin R Aug 28 '19 at 18:25
  • @MartinR I see. I personally like the Picard Theorem so I used it. Thank you for other solutions. – Keith Aug 28 '19 at 18:28
  • @ErickWong Of course I did so. Thank you for your comment. – Keith Aug 28 '19 at 18:31

1 Answers1

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Picard's Theorem is an overkill for this.

If $g$ is entire and non-constant then $g$ is a polynomial iff $|g(z)| \to \infty$ as $|z| \to \infty$.

[ This is proved by first getting rid of the finite number of zeros of $g$ (by dividing by a polynomial and the considering $\frac 1 g$].

The case of $f^{2}$ follows trivially from this.

For $f(f(z))$ suppose $|z_n| \to \infty$. Then $|f(f(z_n))| \to \infty$ unless it is constant (in which case it is easy to show that $f$ is itself a constant). If $f(z_n)$ has a bounded subsequence then $f(f(z_n))$ would be bounded along the subsequence, a contradiction. Thus $|f(z)| \to \infty$ as $|z| \to \infty$.

Kavi Rama Murthy
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