2

Suppose that $f(z)$ is analytic and satisfies $|f(z)| \le 1$ for $|z| < 1$. Show that if $f(z)$ has a zero of order $m$ at $z_0$, then $|z_0|^m \ge |f(0)|$ Hint: Let $\psi(z) = \frac{z-z_0}{1-\bar{z_0}z}$ and then show that $|f(z)| \leq |\psi(z)|^m$

I tried by representing $|f(z)| = |z-z_0|^m g(z)$ because $f$ is analytic with a zero at $z_0$. Comparing to $\psi(z)$ it would suffice to show that $|g(z)||1-\bar{z_0}z| \leq 1$ but I cannot find a proper bound for $g(z)$ or how to extend this into the final conclusion.

At most I have found that $|g(0)| \leq \frac{1}{|z_0^m|}$ from $f$ being analytic, yet I cannot get further. The solution may have something to do with the Schwarz lemma, but again I could not figure out the relationship. Any help would be greatly appreciated!

Kenny Wong
  • 32,192
Aidan McNabb
  • 152

1 Answers1

0

Noticing that $f(z)$ can be written in the form $f(z) = (z - z_0)^m g(z)$ for some holomorphic $g(z)$ is a good start.

What this observation implies is that the function $$ h(z) := \frac{f(z)}{\psi(z)^m} = g(z) \times (1 - \bar z_0 z)^m$$ is holomorphic on the unit disk. [The $m$th order zero of $f(z)$ has cancelled out the $m$th order pole of $1/\psi(z)^m$, so the singularity at $z_0$ is removable.]

I think the hint is prompting you to apply the maximum principle to the function $h(z)$ to deduce that $|h(z)| \leq 1$ for all $z$ in the unit disk.

I hope this is enough for you to get going with. If you're still stuck after giving this considerable thought, then by all means leave a comment.

Kenny Wong
  • 32,192