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Let $f: \mathbb{C} \rightarrow \mathbb{C^*}$ be a holomorphic function, such that, for every $z \in \mathbb{C}$ it holds that : $|f(z)| \leq 2|z|^{\frac{1}{2}} + 3|z|^{\frac{-1}{3}}$

I have to show that f is constant.

I have thought about using the fact that $|a_{n}| \leq \frac{1}{2\pi}\frac{M(r)}{r^{n+1}2}2\pi r$, being

$M(r)= \max\limits_{{\partial B_{r}(z_{0})}}$ $|f|$ $= 2|z|^{\frac{1}{2}} + 3|z|^{\frac{-1}{3}}$

but I don´t really know how to use it or if that would help.

Thank you in advance

Javier Martínez Rubio
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  • $|z| \ge 1$, we have $|f(z)| \le 2\sqrt |z| +3 \le 2|z|+3$ so $f(z)/z$ is bounded on $|z| \ge 1$ hence $|f(z)| \le C|z|+D, C,D>0$ and now show that implies $f$ linear and then constant using that $f(z)/z \to 0, z \to \infty$ – Conrad Jun 09 '20 at 19:07
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    Similar question here: https://math.stackexchange.com/q/1746015/42969 – Martin R Jun 09 '20 at 19:08
  • What is $\mathbb C^*?,$ – zhw. Jun 09 '20 at 21:48
  • @zhw.: $\Bbb C^$ is sometimes used as a notation for $\Bbb C \setminus { 0 }$. I assume however that that should be the domain* of $f$, i.e. $f: \Bbb C^* \to \Bbb C$. – Martin R Jun 10 '20 at 05:24

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