Let \(G\subset\CC\) be a bounded region, and suppose \(f\) is continuous on \(\overline G\) and holomorphic on \(G\). Assume that there is a constant \(c\ge0\) such that \[ |f(z)|=c \qquad(z\in\partial G). \] Show that either \(f\) is constant or \(f\) has a zero in \(G\).
Solution
If \(c=0\), the maximum modulus principle gives \[ |f(z)|\le0 \qquad(z\in G), \] so \(f\equiv0\) and is constant.
Assume \(c>0\) and suppose that \(f\) has no zero in \(G\). Then \(1/f\) is holomorphic on \(G\) and continuous on \(\overline G\). Applying the maximum modulus principle to \(f\) gives \[ |f(z)|\le c \qquad(z\in G), \] while applying it to \(1/f\) gives \[ {1\over|f(z)|}\le {1\over c}, \] hence \(|f(z)|\ge c\). Therefore \(|f|\equiv c\) on \(G\).
A nonconstant holomorphic function has open image, whereas the circle \(\{w:|w|=c\}\) is not open. Thus the open mapping theorem forces \(f\) to be constant. Consequently, if \(f\) is not constant, it must have a zero in \(G\).