I'm learning about complex analysis and need some help with this problem :

Given $f(z)$ entire function and $\left| f(z) \right| \le 1 + \left| z \right|^3$ for all $z \in \mathbb{C}$ show that $f$ is a polynomial. What is the degree of the polynomial?

Here's my attempt so far :

By Cauchy's Integral Formula for Derivatives we have: $$f^{(n)}(z) = \frac{n!}{2 \pi i} \int_{C} \frac{f(\zeta)}{(\zeta - z)^{n+1}} \, d\zeta $$

Let $C_R = \{z \in \mathbb{C} : \left| z \right| = R \}$.

For $\left| z \right| < R$ we have: $$\left| f^{(n)}(z) \right| = \frac{n!}{2 \pi} \left| \int_{\left| \zeta \right| = R} \frac{f(\zeta)}{(\zeta - z)^{n+1}} \, d\zeta \right| = \frac{n!}{2 \pi} \left| \int_{0}^{2 \pi}\frac{\left|f(\zeta_t) \right|}{\left|\zeta_t - z\right|^{n+1}} \, \left| \zeta'_t \right| d\zeta \right| \le$$

$$\le \frac{n!}{2 \pi} \int_{0}^{2 \pi}\frac{1 + \left| \zeta(t) \right|^3}{(\left|\zeta(t) \right| - \left| z \right|)^{n+1}} \, \left| \zeta'(t) \right| dt = \frac{n!}{2\pi} \frac{1+R^3}{(R - \left| z \right|)^{n+1}} 2\pi R =$$

$$= n! \frac{1+R^3}{(R - \left| z \right|)^{n+1}} R$$

We want $\left| f^{(n)}(z) \right| \rightarrow 0$ when $R \rightarrow \infty$.

This is where I'm stuck. Is my work correct so far and how do I continue from here to find the degree of the polynomial?


The limit tends to zero if $n > 3$. Thus, $|f^{n}(z)| = 0$ for all $n > 3$ and $z \in \mathbb{C}$ which implies that $f$ is a polynomial of degree at most three.

  • 1
    $\begingroup$ I think I got it : $n! \frac{1+R^3}{(R - \left| z \right|)^{n+1}} R = n! \frac{R+R^4}{(R - \left| z \right|)^{n+1}}$. We want the degree of the denominator to be superior to the degree of the numerator for the limit to tend to zero. Thus, $n + 1 > 4 \iff n > 3$. Is that correct? $\endgroup$ – Von Kar Nov 29 '15 at 15:00
  • $\begingroup$ Yes, that is correct. $\endgroup$ – levap Nov 29 '15 at 15:01

No reason to think you're stuck. For fixed $z$ your final expression is $\sim R^{3-n}$.

  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – Daniel Robert-Nicoud Nov 29 '15 at 14:44
  • $\begingroup$ @DanielRobert-Nicoud Well, "No reason to think you're stuck" implicitly answers the question "Is my work correct so far?" positively, and the observation "$\sim R^{3-n}$" is the next step to "continue from here" ... $\endgroup$ – Hagen von Eitzen Nov 29 '15 at 14:51
  • $\begingroup$ I know. It's just that I think that your answer would be more suited as a comment, as it doesn't contain any details. $\endgroup$ – Daniel Robert-Nicoud Nov 29 '15 at 16:20

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