Let \(f:\mathbb C\to\mathbb C\) be analytic and let \(U\subset\mathbb C\) be a bounded region. Suppose \(|f(z)|\) is constant on \(\partial U\). Show that either \(f\) is constant or \(f\) has a zero in \(U\).
Solution
1. Let\[ |f(z)|=c\qquad(z\in\partial U). \]If \(c=0\), then \(f\) is identically zero.
If \(c=0\), then \(|f|=0\) on \(\partial U\). Since \(f\) is entire, it is continuous on the compact set \(\overline U\). If \(|f|\) were positive somewhere in \(U\), its maximum on \(\overline U\) would be positive and would occur at an interior point. The maximum modulus principle would force \(f\) to be constant, contradicting its zero boundary values unless that constant were zero. Hence \(|f|=0\) throughout \(U\), and therefore \(f\equiv0\) on \(U\) and, by the identity theorem, on all of \(\mathbb C\).
2. Assume \(c>0\) and suppose that \(f\) has no zero in \(U\). Then\[ |f(z)|\le c \qquad(z\in U). \]
The closure \(\overline U\) is compact because \(U\) is bounded. The entire function \(f\) is continuous on \(\overline U\), so \(|f|\) attains its maximum there.
If some point of \(U\) had modulus strictly larger than \(c\), then the maximum of \(|f|\) on \(\overline U\) would occur at an interior point of \(U\), because \(|f|=c\) on \(\partial U\). The maximum modulus principle would then force \(f\) to be constant. In that event there is nothing further to prove. Therefore, in the only remaining case, \[ |f(z)|\le c \] throughout \(U\).
3. Under the same hypotheses,\[ |f(z)|\ge c \qquad(z\in U). \]
Because \(f\) has no zero in \(U\) and \(c>0\) on \(\partial U\), it has no zero on \(\overline U\). Hence \[ g(z)=\frac1{f(z)} \] is holomorphic on \(U\) and continuous on \(\overline U\). On the boundary, \[ |g(z)|=\frac1c. \] Applying the same maximum-modulus argument as in 2 to \(g\) gives \[ |g(z)|\le\frac1c \] throughout \(U\). Equivalently, \[ |f(z)|\ge c. \]
4. Therefore, if \(f\) has no zero in \(U\), then \(f\) is constant.
Combining 2 and 3 gives \[ |f(z)|=c \] for every \(z\in U\).
If \(f\) were nonconstant, the open mapping theorem would imply that \(f(U)\) is open in \(\mathbb C\). But \[ f(U)\subset\{w\in\mathbb C:|w|=c\}, \] and a circle has empty interior. This is impossible. Thus \(f\) is constant.
Consequently, every nonconstant \(f\) satisfying the boundary-modulus hypothesis must have a zero in \(U\).