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Suppose $R(z)$ is a rational function such that $|R(z)|=1$ for $|z|=1$. Let ${\alpha_j}^N_{j=1}$ be zeros and poles of $R(z)$ of order $m_j$ in the unit disc $|z|\lt 1$. Here $m_j\gt 0$ is $\alpha_j$ is a zero and $m_j\lt 0$ is it is a pole. Define $B(z)=[(z-\alpha_1)/(1-z\alpha'1)]^{m_1}...[(z-\alpha_N)/(1-z\alpha'_N)]^{m_N}$ (Where $\alpha'$ stands for the complex conjugate of $\alpha$.) Then how to show that $R(z)/B(z)$ is a bounded entire function? And how to show that $R(z)=\lambda B(z)$ for some $\lambda$ belongs to complex number with $|\lambda|=1$.

I've proved that $\alpha$ is a zero or a pole of order m iff $1/\alpha'$ is a pole or zero of order m respectively, but don't know how to proceed from there. This problem seems pretty difficult. Any help?

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  • $\begingroup$ 1. $R(z)/B(z)$ has no zeros in the closed unit disk, 2. What do you know about $\lvert R(z)/B(z)\rvert$ on the unit circle? $\endgroup$
    – Daniel Fischer
    Oct 26, 2016 at 22:04

1 Answer 1

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Hint: Let $\mathbb D$ denote the open unit disc. Suppose $f$ is continuous on $\overline {\mathbb D}$ and holomorphic on $\mathbb D.$ If $f(z)$ is nonzero in $\overline {\mathbb D}$ and $|f|=1$ on $\partial D,$ then $f$ is constant. This follows from the maximum modulus theorem.

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  • $\begingroup$ I didn't do it.. $\endgroup$
    – J.doe
    Oct 27, 2016 at 4:00
  • $\begingroup$ I've up voted it. I didn't quite follow. Could you please explain more? $\endgroup$
    – J.doe
    Oct 27, 2016 at 4:03
  • $\begingroup$ OK, never mind about the down vote; someone I think was dive-bombing a lot of posts of mine, downvoting them, for reasons I don't understand. Anyway, the idea is that if you divide and multiply $R$ by these Blaschke factors, you end up with an $f$ as in my answer. $\endgroup$
    – zhw.
    Oct 27, 2016 at 4:26

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