# An analytic function on the open unit disk with a zero of order k

I have the following question: Let $f$ be an analytic function from the open unit disk to itself. Assume that $f$ has a zero of order $k$ at zero where $k\geq 1$. Prove that $|f(z)|\leq |z|^{k}$ for all $z$ in the open unit disk.

I define $h(z)=\frac{f(z)}{z^k}\,\,$if $z\neq 0$ and $\lim_{z\rightarrow0} \frac{f(z)}{z^k}$ if $z=0$. Now $h$ is analytic on the open unit disk by the Morera theorem. I think I can use maximum modulus principle to conclude statement but $f$ is not defined and continuous on the boundary of the open unit disk (for $z$ that $|z|=1$).

• – Daniel Fischer Aug 4 '15 at 16:52
• Exactly the same problem arises in the proof of the Schwarz Lemma (which is just the case $k=1$). Look up that proof; the same solution applies here. – David C. Ullrich Aug 4 '15 at 17:04
• Now, I see, thanks. – delueze Aug 4 '15 at 17:18