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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 $|f(z)|<M$ 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!

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    $\begingroup$ to be honest, I think Alexandre's solution is more than detailed enough. $\endgroup$ Dec 7, 2019 at 22:30
  • $\begingroup$ @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. $\endgroup$
    – ponchan
    Dec 7, 2019 at 22:44

1 Answer 1

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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 $f=Me^{i\theta}\prod g_k$.

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  • $\begingroup$ How would you go about finding the formula for $f$? $\endgroup$
    – ponchan
    Dec 7, 2019 at 22:01
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    $\begingroup$ to be honest, I think Alexandre's solution is more than detailed enough. $\endgroup$ Dec 7, 2019 at 22:30
  • $\begingroup$ Thanks, I understand that, but where does the $e^{i\theta}$ come from? $\endgroup$
    – ponchan
    Dec 7, 2019 at 22:49
  • $\begingroup$ Is it because we get $|f(z)|=M|g_1\cdots g_n|$? $\endgroup$
    – ponchan
    Dec 7, 2019 at 22:58
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    $\begingroup$ @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 $\theta$. $\endgroup$ Dec 8, 2019 at 1:08

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