Root of $f(z)$ inside $|z|<1$

Question

Let $c\in\mathbb{R}$. A non-constant function $f(z)$ is holomorphic in $|z|<2$. Suppose $|f(z)|=c$ for all $|z|=1$. Show that $f(z)$ must have a root in $|z|<1$.

I'm thinking about the maximum principle, which says $f(z)$ cannot attain a maximum inside $|z|<1$. But that still doesn't yield a root. Also, Rouche's theorem might be applicable if there's another function $g(z)$ to be used.

score 4 · Accepted Answer · answered Nov 18 '13 at 20:24

4

$f$ is non-constant. So you must have $\lvert f(z)\rvert < c$ for all $z$ in the unit disk. If $f$ had no zero in the unit disk, what would the maximum principle have to say about $$g(z) = \frac{1}{f(z)}\; ?$$

answered Nov 18 '13 at 20:24

Daniel Fischer

206,697

1

Since $g$ is also holomorphic, the principle would say $|g(z)|<1/c$ inside the unit disk. But then we would have $1=|f(z)g(z)|<1$, impossible. – JJ Beck Nov 18 '13 at 20:30
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Exactly, it would say "impossible". – Daniel Fischer Nov 18 '13 at 20:31

Root of $f(z)$ inside $|z|<1$

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