# Find all functions $f(z)$ holomorphic on $\mathbb{C}-\{z_0,z_1\}$ such that $f(z)\leq|z-z_0|^a|z-z_1|^b$ for $a,b\in\mathbb{R}$

Question: Find all functions $f(z)$ holomorphic on $\mathbb{C}-\{z_0,z_1\}$ such that $f(z)\leq|z-z_0|^a|z-z_1|^b$ for $a,b\in\mathbb{R}$. you may assume $z_0\neq z_1$.

I found that if $a,b$ are integers then $\frac{f(z)}{(z-z_0)^a(z-z_1)^b}$ is holomorphic on $\mathbb{C}-\{z_0,z_1\}$ and we have: $|\frac{f(z)}{(z-z_0)^a(z-z_1)^b}| \leq 1$ hence it is also holomorphic on all of $\mathbb{C}$ and therefore by Liouvilee theorem it is constant, hence: $f(z)=\lambda(z-z_0)^a(z-z_1)^b$ where $|\lambda| \leq 1$. Altough, I'm not sure how to solve this when $a$ or $b$ are not integers.

Any hint would help, Thanks!

• Where your problem if $a$ & $b$ are not integers? – Empty Mar 13 '15 at 7:09
• the same method wouldn't work, because f(z)/(z-z0)^a(z-z1)^b will need a branch of logarithm to be defined so it will not be defined on all of C\{z0,z1} – user3054303 Mar 13 '15 at 15:36