Can you help on complex analysis problem

Question

The question: Let $D = {z : |z| < 1}$, and let $f : D → D$ have a zero of order $n$ at zero. Show that $|f(z)| ≤ |z|^{n}$ on $D$.

My attempt: I am not sure what theorem's are applied to this problem I think we use Rouche's Theorem but there may be some problems with that theorem. Do I consider the function $g(z) = f(z) - z$; or could I use the function $h(z) = \frac{f(z)}{z}$ and try to use the maximum modulus principle. I think I am not heading in the right direction. Can you please give me some hints on how to approach this problem? Thanks a lot guys!

Answer 1

Since $f$ has at $z=0$ a zero of order $n$, then $$ g(z)=\frac{f(z)}{z^n} $$ is analytic in the unit disk.

If $0<r<1$ and $M_r=\max_{|z|\le r}|g(z)|$, then $$ M_r=\max_{|z|\le r}|g(z)|=\max_{|z|= r}\frac{|f(z)|}{|z|^n}\le \frac{1}{r^n} $$ Maximum principle implies that $$ \sup_{|z|<1}|g(z)|=\lim_{r\to 1}M_r\le \lim_{r\to 1}\frac{1}{r^n}= 1. $$ Hence $$ |f(z)|=|g(z)||z|^n\le |z|^n, $$ for all $|z|<1$.

Note. If $|f(z)|=|z|^n$, for some $|z|<1$, then there exists an $a\in\mathbb C$, with $|a|=1$, such that $f(z)=az^n$, for all $|z|<1$, due to maximum principle.

Answer 2

1. Since the image of $f$ is $\mathbb D$, $|f|\le 1$.
2. Consider $h(z)=f(z)z^{-n}$ which is clearly analytic on $\mathbb D$.
3. By maximum modulus principle, $|h(z)|$ is maximized on $\partial\mathbb D$.
4. Thus, $$\max |h(z)|=\max_{|z|=1}|f(z)||z|^{-n}=\max_{|z|=1}|f(z)|\le 1$$
5. $|h|\le 1\implies|f|\le|z|^n$ on $\mathbb D$.