# Modulus of Analytic Function is Bounded Above

I have the following problem that I am stuck on:

Let $$f$$ be analytic in $$D=\{z\in\mathbb{C}\::\:|z|<1\}$$ and suppose that $$|f(z)| for all $$z\in D$$.

(a) If $$f(z_{k})=0$$ for $$1\leq k\leq n$$, show that $$|f(z)|\leq M\prod_{k=1}^{n}\frac{|z-z_{k}|}{|1-\bar{z}_{k}z|}$$ for $$|z|<1$$.

(b) If $$f(z_{k})=0$$ for $$1\leq k\leq n$$, each $$z_{k}\neq 0$$, and $$f(0)=Me^{i\alpha}(z_{1}z_{2}\cdots z_{n})$$, find a formula for $$f$$.

I honestly have no idea of where to begin with this problem. I think that Schwarz's Lemma may be needed somewhere? Thanks in advance for any help!

• to be honest, I think Alexandre's solution is more than detailed enough. Dec 7, 2019 at 22:30
• @mathworker21 I understand how his solution implies part (a). Yet part (b) I find confusing, and I'm not sure how his solution implies it. Dec 7, 2019 at 22:44

Function $$g_k(z)=\frac{z-z_k}{1-z\overline{z_k}}$$ has the following properties: a) it has a simple zero at $$z_k$$, and no other zeros, b) it is analytic in $$|z|\leq 1$$, and c) $$|g_k(z)|=1$$ for $$|z|=1$$. Consider the ratio $$G(z)=\frac{f(z)}{\prod_{k=1}^n g_k(z)}$$ It is analytic (the poles at $$z_k$$ are removable) and satisfies $$|G(z)|\leq M$$ for $$|z|=1$$. By the Maximum Principle $$|G(z)|\leq M$$ in the unit disk, and this is your inequality.
For b), Maximum Principle says that this inequality is everywhere strict, or everywhere equality. If for example $$|G(0)|=M$$ then $$f=Me^{i\theta}\prod g_k$$.
• How would you go about finding the formula for $f$? Dec 7, 2019 at 22:01
• Thanks, I understand that, but where does the $e^{i\theta}$ come from? Dec 7, 2019 at 22:49
• Is it because we get $|f(z)|=M|g_1\cdots g_n|$? Dec 7, 2019 at 22:58
• @ponchan $G$ is constant. Say $G(z) = z_0$ for all $z$. We know $|G(0)| = M$. Therefore $z_0 = Me^{i\theta}$ for some $\theta$. Dec 8, 2019 at 1:08