- Show that if \(\abs{f} = 0\) on \(\bd \Omega\) then either \(f\) is constant or \(f\) has a zero in \(\Omega\).
Solution
Goal: If \(f\) is holomorphic on a bounded region \(\Omega\), continuous on \(\bar\Omega\), and \(\abs f = 0\) (i.e. \(f = 0\)) on \(\bd\Omega\), show that either \(f\) is constant or \(f\) has a zero in \(\Omega\).
1. If \(f\) has no zeros in \(\Omega\), then \(\abs{f}\) attains its maximum on \(\bd\Omega\).
\(\abs{f}\) is continuous on the compact set \(\bar\Omega\), so it attains its maximum somewhere in \(\bar\Omega\); by the maximum modulus principle (or its version for \(1/f\), valid because \(f\) has no zeros), the maximum is attained on the boundary \(\bd\Omega\).
2. If \(f\) has no zeros in \(\Omega\), then \(f \equiv 0\) on \(\bar\Omega\).
By 1 and the hypothesis \(\abs f = 0\) on \(\bd\Omega\), \(\max_{\bar\Omega}\abs f = 0\), so \(f \equiv 0\) on \(\bar\Omega\).
3. If \(f\) is not constant, then \(f\) has a zero in \(\Omega\).
Contrapositive of 2: if \(f\) has no zero in \(\Omega\), then by 2 \(f \equiv 0\) on \(\bar\Omega\), so \(f\) is constant (the zero function).
4. Q.E.D.
3 shows that a nonconstant \(f\) must vanish somewhere in \(\Omega\), which together with the trivial constant case proves the claim.