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Spring 2016 - Problem 11

Rouché's theorem

Assume that $f\p{z}$ is holomorphic on $\abs{z} < 2$ . Show that

\max_{\abs{z}=1}\,\abs{f\p{z} - \frac{1}{z}} \geq 1.

Solution.

Suppose otherwise and that $\max_{\abs{z}=1} \abs{f\p{z} - \frac{1}{z}} < 1$ . If $g\p{z} = zf\p{z} - 1$ , then we have

\max_{\abs{z}=1}\,\abs{g\p{z}} = \max_{\abs{z}=1}\, \abs{z}\abs{f\p{z} - \frac{1}{z}} = \max_{\abs{z}=1}\, \abs{f\p{z} - \frac{1}{z}} < 1,

so by Rouché's theorem, $1$ and $1 + g$ have the same number of zeroes, i.e., $1 + g$ does not vanish in $\D$ . But this means that

1 + g\p{z} = zf\p{z}

does not vanish in $\D$ , which is impossible as this implies that $f$ is unbounded near $z = 0$ . Thus, the maximum must have been larger than $1$ to begin with.