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I don't understand the final step of an argument I read.

Suppose $f$ is holomorphic in a neighborhood containing the closed unit disk, nonconstant, and $|f(z)|=1$ when $|z|=1$. There is some point $z_0$ in the unit disk such that $f(z_0)=0$.

By the maximum modulus principle, it follows that $|f(z)|<1$ in the open unit disk. Since the closed disk is compact, $f$ obtains a minimum on the closed disk, necessarily on the interior in this situation.

But why does that imply that $f(z_0)=0$ for some $z_0$? I'm aware of the minimum modulus principle, that the modulus of a holomorphic, nonconstant, nonzero function on a domain does not obtain a minimum in the domain. But I'm not sure if that applies here.

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If $f$ has modulus 1 on the entire unit disk, then the function has to be constant. However, what you write in the body of the question just requires $|f(z)=1|$ on the unit circle. –  Henning Makholm Apr 2 '12 at 21:46
    
Dear @HenningMakholm, thanks, I accidentally used the wrong word. I've now changed it to circle. –  Hana Bailey Apr 2 '12 at 21:48
    
"I'm aware of the minimum modulus principle, that the modulus of a holomorphic, nonconstant, nonzero function on a domain does not obtain a minimum in the domain. But I'm not sure if that applies here." Yes, it applies here. Why wouldn't it? If $f$ had no zeros in the unit disk then that theorem would be contradicted. –  Jonas Meyer Apr 3 '12 at 0:52

1 Answer 1

up vote 11 down vote accepted

If not, consider $g(z)=\frac 1{f(z)}$ on the closure of the unit disc. We have $|g(z)|=1$ if $|z|=1$ and $|g(z)|>1$ if $|z|<1$. Since $g$ is holomorphic on the unit disk, the maximum modulus principle yields a contradiction.

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Thanks. So the key point is that if $f(z)\neq 0$ for any $z$ in the disk, then $g$ is well-defined and this argument works? So we must have $f(z)=0$ for some $z$ somewhere in the interior. –  Hana Bailey Apr 2 '12 at 21:51
    
Yes, that's indeed the key point. –  Davide Giraudo Apr 2 '12 at 21:53

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