$f$ holomorphic in $\mathbb{C}\setminus\{0\}$ and $|f(z)|\le \sqrt{|z|} + \frac{1}{\sqrt{|z|}}$ then $f$ is constant

Question

given $f$ a holomorphic function in $\mathbb{C}\setminus\{0\}$ satisfies $|f(z)|\le \sqrt{|z|} + \frac{1}{\sqrt{|z|}}$ for all $z\in \mathbb{C}\setminus \{0\}$. I wish to show that $f$ is constant, well what I did so far was to take the limit: $0\le \lim_{z\rightarrow0} |z||f(z)| \le \lim_{z\rightarrow 0} |z|^{1.5}+|z|^{0.5} =0$ so $\lim_{z\rightarrow0} |zf(z)|=0$ thus $\lim_{z\rightarrow0} zf(z)=0$ and by Riemann's continuations theorem $0$ is a removable singularity. So $f$ is can be extended holomorphicaly to $\tilde{f}$. I thought maybe its possible to bound $\tilde{f}$ and then use Liouville's theorem. But couldn't find a way to do so.

Answer 1

Let $g(z)=zf(z)$. Then $\bigl\lvert g(z)\bigr\rvert\leqslant\lvert z\rvert^{3/2}+\sqrt{\lvert z\rvert}$ and therefore $\lim_{z\to0}g(z)=0$. So, $g$ has a removable singularity at $0$ and you can extend $g$ to an analytic map $G\colon\mathbb C\longrightarrow\mathbb C$ such that $G(0)=0$. But then $f$ can be extended to an entire function $F$. And you know that $\bigl\lvert F(z)\bigr\rvert\leqslant\sqrt{\lvert z\rvert}+1$ if $\lvert z\rvert\geqslant1$. Now you can use the Cauchy inequalities in order to prove that, if $F(z)=a_0+a_1z+a_2z^2+\cdots$, then $(\forall n\in\mathbb N):a_n=0$. Therefore, $F$ is constant.