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Let $f(z)$ be analytic and nonzero in a region R. Show that $|f(z)|$ has a minimum value in R that occurs on the boundary.

I think you should use the Maximum-Modulus Theorem for the function $1/f(z)$

The Maximum-Modulus Theorem

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ALternatively, $\log|f(z)|$ is harmonic, if you know about harmonic functions. –  GEdgar Apr 29 '12 at 2:56
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Do you know anything about $R$? –  Brett Frankel Apr 29 '12 at 2:59
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A region is an open set, so if $f$ nonzero in $R$, and $|f|$ has a minimum value in $R$, then by Strong Maximum-Modulus Theorem for $1/f$, $f$ must be constant. –  Yuchen Liu Apr 29 '12 at 3:02
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1 Answer 1

Since it is not clear what is meant by a "region" in the problem (and careless formulation leads to easy counterexamples), I will rephrase.

Claim. Suppose $f$ is holomorphic and nonzero in open connected bounded set $U$ and is continuous on $\overline{U}$. Then there exists $\zeta\in\partial U$ such that $$|f(\zeta)|=\min_{\overline U}|f| \tag1$$

Proof The existence of $\zeta \in \overline{U}$ with the property (1) follows by compactness of $\overline{U}$. Suppose $\zeta\in U$. Then $f(\zeta)\ne 0$ by assumption. Since the right-hand side of (1) is nonzero, the function $g(z)=1/f(z)$ is holomorphic in $U$. In terms of $g$ (1) becomes $$|g(\zeta)|=\max_{\overline U}|g| \tag2$$ which, by the strong maximum principle, implies $g$ is constant. Hence $f$ is constant, and for a constant function the claim is obviously true. $\Box$

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