show holomorphic function is 0

Question

I'm trying to show that a holomorphic function $f$ on $\mathbb{C}$ satisfying $$|f(z)|\leq\sqrt{|z|}$$ for all $z\in \mathbb{C}$ is $0$. Is it possible to use Poisson's formula here? Otherwise, how can I go about it? Thanks for the help!

Answer 1

If you write $\Gamma_{r,z}$ for the circle of radius $r$ around $z$, then by Cauchy's formula \begin{align} |f(z)|&=\frac1{2\pi}\,\left|\int_{\Gamma_{r,z}}\frac{f(\gamma)}{\gamma-z}\,d\gamma\right| \leq\frac1{2\pi}\,\int_{\Gamma_{r,z}}\frac{|f(\gamma)|}{r}\,d\gamma \leq\frac1{2\pi r}\int_{\Gamma_{r,z}}\sqrt{|\gamma|}\,d\gamma\\ \ \\ &\leq\frac{2\pi\sqrt{|z|+ r}}{2\pi r} =\frac{\sqrt{1+|z|/r}}{\sqrt r}\xrightarrow[r\to\infty]{}0. \end{align}