# All holomorphic functions such that $|f(z)| \leq |z|^k$ for some non-negative integer $k$?

I'm trying to reduce a well-known result by finding all holomorphic functions $f(z)$ (on the whole of $\mathbb{C}$) such that $|f(z)| \leq |z|^k$ for some non-negative integer $k$ and for all $z\in \mathbb{C}$. I know that $f(z)$ is a polynomial of degree at most $k$. So I've tried to write a general form of such a polynomial and take particular values of $z$ to restrict the possibilities for the coefficient, but I don't find anything concluent...

Also, is it possible to extend even more by finding holomorphic functions such that $|f(z)|\leq|p(z)|$ for some polynomial $p(z)$?

Thank you all very much for your help!

$f$ is a polynomial of degree $k$. It is also clear that $f$ has a zero of order $k$ at $z=0$. This implies that $f(z)=c\,z^k$ with $|c|\le1$.
For the general case $|f(z)|\le|p(z)|$ consider the zeroes of $p$.
• The result you give for the first part is what I expected... But I'm affraid I'm not sure I see why $z=0$ is a zero of order $k$ of $f$... Most probably I'm missing a stupid point, what is it? – Noom Nov 10 '14 at 10:22
• If $k=0$ there is nothing to prove ($f$ is constant.) If $k>0$ then $f(0)=0$. Now consider $f(z)/z$. It is a polynolial of degree $k-1$ and $|f(z)/z|\le|z|^{k-1}$. – Julián Aguirre Nov 10 '14 at 10:44
• It is certainly a possibility. Or you can study $f(z)/z^k$. It is bounded, so it has an avoidable singularity at $z=0$. It is the a bounded entire function. – Julián Aguirre Nov 10 '14 at 11:39