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Suppose $f(z)$ is analytic for $|z|\le 1$ and $f(0) = a_0 \ne 0$. If $M = \max_{|z|=1} |f(z)|$, then show $f(z)\ne 0$ for all $z$ with $|z| < \frac{|a_0|}{|a_0|+M} =:r$.

I know we can write $f(z) = a_0 + z^kg(z)$, some $k\ge 1$ and $g$ analytic and $g(0)\ne 0$. From here, I've tried various techniques, like contradiction by assuming $f$ has a root in the disk $\{|z| < r\}$, or trying to use Rouche's Theorem on the disk by examining $|f(z)-a_0|$, but I haven't really gotten anywhere. Any hints would be greatly appreciated.

Curious
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2 Answers2

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Let $g(z) = f(z)-a_0.$ Then $g(0)=0$ and $|g|\le M + |a_0|.$ The Schwarz lemma then shows

$$\frac{|g(z)|}{M+|a_0|}\le |z|$$

for $|z|<1.$ Thus if $|z|< |a_0|/(M+|a_0|),$ then $|g(z)| < |a_0|.$ For such $z$ we then have $|f(z)| = |a_0 + g(z)| \ge |a_0|-|g(z)| > 0.$ (Note that we only need $f$ holomorphic in the open unit disc, with $M = \sup_{|z|<1}|f(z)|.$)

Daniele Tampieri
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zhw.
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The map $g(z):=f(z)/M$ is a analytic map from the unit disc in itself.

Then by the Schwarz-Pick Theorem, for $|z|< 1$, $$\left|\frac{g(z)-g(0)}{1-\overline{g(z)}g(0)}\right|\leq |z|.$$ If $f(z)=0$ and $|z|<r<1$ then $g(z)=0$ and we obtain $$\frac{|a_0|}{M}=|g(0)|\leq |z| <r= \frac{|a_0|}{|a_0|+M}$$ which is a contradiction because $f(0) = a_0 \ne 0$.

Robert Z
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