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Let $f$ be non-constant analytic in a neighborhood of the closed unit disk such that $|f|=1$ on the unit circle. Show that $f$ has a zero inside the unit disk. It has been suggested to argue this by contradiction and using the maximum principle. However, is it possible to argue this question using properties of conformal maps since doesn't $f$ map the unit circle to itself? Regardless, I don't understand what information the Maximum Principle would give me to argue by contradiction. Why couldn't $0<|f|<1$ inside the disk?

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    $\begingroup$ I give you a hint: if $f$ has no zero inside the disk, consider 1/f and use the maximum modulus principle $\endgroup$
    – miku
    May 1, 2014 at 15:24

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If $f$ does not have a zero inside the unit disk, then $g=1/f$ is also analytic in a neighborhood of the closed unit disk with $|g|=1$ on the unit circle. Using the maximum principle implies that $|g(0)|\leq 1$, or equivalently $|f(0)|\geq1$, so $f$ attains its maximum on the closed unit disk at a point inside the disk. By the maximum principle this implies that $f$ is constant which is absurd. This contradiction proves that $f$ must have a zero inside the unit disk.

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  • $\begingroup$ yes, I clarified my answer. $\endgroup$ May 1, 2014 at 16:11

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