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This problem is totally out of my ability. Not even sure what it is talking about. Somebody please help me to solve this...

A finite Blaschke product of degree $n-1$ is a function of the form

$ \ B_{n-1} (z) = e^{i \varphi} \prod_{k=1}^{n-1} \frac{z - \alpha_k}{1 - \bar{\alpha}_k z} ,~~ | \alpha_k | < 1 \ $

(a) Explain why $B_{n-1}$ is analytic inside and on the unit circle $C = \{ z : | z | = 1 \}$.

(b) Show that $| B_{n-1} (z) | = 1$ at all points $z$ on the unit circle. [Hint: Show that each factor in the product has absolute value $1$ if $z = e^{i \theta}$.]

(c) Suppose $g(z)$ is a function that is analytic inside and on the unit circle $C$ and matches $B_{n-1}$ at $n$ points inside $C$:

$ \ g( \lambda_j ) = B_{n-1} ( \lambda_j ) ,~~j = 1, \ldots , n ,~~~| \lambda_j | < 1. \ $

Use Rouche's theorem to show that $| g(z) | \geq | B_{n-1} (z) |$ at some point $z$ on the unit circle.

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You say you're not sure what it is talking about, but could you please be more specific about which parts you're stuck on? Do you know what it means for a function of a complex variable to be analytic, and that polynomials and rational functions are examples? Do you know how to find the absolute value of a complex number? (a) and (b) shouldn't be very difficult. – Jonas Meyer Dec 5 '12 at 0:21
I know what it means for a function of a complex variable to be analytic. But I can't seem to tell anything about this function since it is written in such way (Product). – Angus Leo Dec 5 '12 at 0:30
Angus: A product of two analytic functions is analytic. As a consequence, a finite product of analytic functions is analytic. Can you see why each factor in the product is analytic? – Jonas Meyer Dec 5 '12 at 0:34
Because there is no way that 1-akz = 0 given that |z|<=1? – Angus Leo Dec 5 '12 at 0:50
up vote 1 down vote accepted


(a) $B_{n-1}$ is analytic on the closed unit disc since its singularities are $\bar{\alpha_k}^{-1}$, and they lie outside the closed unit disc, so we can choose and open ball $B$ centered at the origin containing $\bar{C}$ which does not containg any $\bar{\alpha_k}^{-1}$, and clearly $\frac{1}{1-\bar{\alpha_k}}$ is holomorphic in $B$, so $B_{n-1}$ is holomorphic on $\bar{C}$

(b)To see $|B_{n-1}(z)|=1$ at all points $z$ on te unit circle it sufficies to show that $|\frac{z-\alpha_k}{1-\bar{\alpha_k}z}|=1$ at all points on the unit circle,this is easy; just find $|\frac{z-\alpha_k}{1-\bar{\alpha_k}z}|^2$ for $|z|=1$.

(c)If, otherwise, we had $|g(z)|<{|B_{n-1}(z)|}$ for all $z$ on the unit disc, by Rouché's theorem $B_{n-1}(z)$ and $g(z)-B_{n-1}(z)$ would have the same roots in the unit disc, but $B_{n-1}(z)$ has only $n-1$ roots on the unit disc, contradicting your assumption on $g$.

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