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The following is a problem from Conway chapter 6 section 2: Suppose f is analytic in some region containing $\bar{B}(0;1)$ and $ |f(z)| = 1$ where $|z| = 1$. Find a formula for $f$. (Hint: First consider the case where f has no zeros in B(0;1).)

I have figured that $f$ is analytic over $B(0;1)$, so by Maximum Modulus principle it attains its maximum on the boundary of $B(0;1)$, and so again by the maximum modulus principle $f$ must be constant in $B(0;1)$. So $f(z)=c$ where $|c|=1$ for all $z\in$ $\bar{B}(0;1)$.

Any hints how to continue from here? Also I'm not entirely sure the argument I gave so far is totally correct.

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  • $\begingroup$ Could you finish this problem? Because I can´t even with the hints. $\endgroup$ – Peter Languilla Nov 20 '19 at 13:25
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By the Maximum Modulus Principle we see that $|f(z)|\le 1$ for $z\in B(0,1)$. To show that $f$ is constant a little more work is needed. If $f$ does not vanish in $B(0,1)$ we can consider the function $1/f$.

If $f$ vanishes at $B(0,1)$, may be non constant. Example: $f(z)=z$. Can there be any other example?

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Hint: $f(z)$ can only be a finite Blaschke product, up to some unimodular constant. You should justify this.

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