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For integers $j=1,\dots,n$, let $\alpha_j\in\mathbb{C}$ with $|\alpha_j|<1$ and let $\theta\in\mathbb{R}$, we define $$b(z)=e^{i\theta}\prod_{j=1}^n\frac{z-\alpha_j}{1-\overline{\alpha_j}z}.$$

a) Show that there is $\delta>1$ such that $b$ is analytic on $D(0,\delta)$. Conclude that $b$ is $C$-analytic on $\mathbb{D}=\{z\in\mathbb{C}:|z|<1\}.$

b) Show that if $z\in\mathbb{D}$ then we have for the modulus of each factor: $|(z-\alpha_j)/(1-\overline{\alpha_j}z)|<1$.

c) Prove that $|b(z)|=1$ if and only if $|z|=1$.

I am trying to do part (a), and I am having trouble grasping the expression $b(z)$ itself. If I am not mistaken, $b(z)$ represents a complex number of radius 1 (the $e^{i\theta}$ part) multiplied by the quotient of two polynomials. So this function should be analytic away from any poles. I'm stuck on how to show this right now. Haven't even gotten around to parts (b) or (c) yet. Any help understanding the function $b$ would be appreciated along with some possible hints. I don't necesaarily need a full answer.

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For part a): Without loss of generality, assume none of the $\alpha_j$ are zero. Choosing $\delta = \min\{1/|\alpha_1|,\ldots, 1/|\alpha_n|\}$ will make $\delta > 1$ and $b$ analytic on $B(0,\delta)$. To see that $b$ is analytic on $B(0,\delta)$, note that for all $z \in B(0,\delta)$, and $j = 1,2,\ldots, n$, $$|1 - \bar{\alpha_j}z| \ge 1 - |\alpha_j||z| > 1 - 1 = 0.$$ Therefore, the denominators of the rational factors of $b(z)$ do not vanish. It follows that $b$ is analytic on $B(0,\delta)$. Since $\Bbb D \subset B(0,\delta)$, $b$ is analytic on $\Bbb D$.

In parts b) and c), compare $|z - \alpha_j|^2$ with $|1 - \bar{\alpha_j}z|^2$ (it'll help to expand these terms).

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  • $\begingroup$ Thank you, I had the same idea for a) while I was waiting for an answer here. But I don't understand why we can assume without loss of generality that all of the $\alpha_j$ are not zero. $\endgroup$ Mar 22 '15 at 19:31
  • $\begingroup$ If an $\alpha_j$ was zero, then the rational function $(z - \alpha_j)/(1 - \bar{\alpha_j}z)$ is equal to $z$. $\endgroup$
    – kobe
    Mar 22 '15 at 19:33
  • $\begingroup$ Also, $f(z) = z$ is analytic (in fact, entire). $\endgroup$
    – kobe
    Mar 22 '15 at 19:35

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