Let \(f\) be an analytic function on a region \(\Omega\). Show that \(f\) is a constant if there is a simple closed curve \(\gamma\) in \(\Omega\) such that its image \(f(\gamma)\) is contained in the real axis.
Solution
The statement is false for a general region \(\Omega\). Take \[ \Omega=\{z:1/2<|z|<2\}, \qquad f(z)=z+{1\over z}, \] and let \(\gamma\) be the unit circle. Then \(f\) is nonconstant and holomorphic on \(\Omega\), while for \(z=e^{it}\), \[ f(z)=e^{it}+e^{-it}=2\cos t\in\RR. \] Thus \(f(\gamma)\subset\RR\) although \(f\) is not constant.
The natural corrected statement is true if the bounded component \(U\) of \(\CC\setminus\gamma\) is contained in \(\Omega\). Indeed, then \(v=\Im f\) is harmonic on \(U\), continuous on \(\overline U\), and satisfies \(v=0\) on \(\partial U=\gamma\). The maximum and minimum principles for harmonic functions give \(v\equiv0\) on \(U\). Hence \(f(U)\subset\RR\).
If \(f\) were nonconstant on \(U\), the open mapping theorem would make \(f(U)\) open in \(\CC\), which is impossible for a subset of \(\RR\). Thus \(f\) is constant on the nonempty open set \(U\), and the identity theorem then makes \(f\) constant on the connected region \(\Omega\).