Suppose \(f\) is entire and maps a circle \(|z-a|=r\) into \(\mathbb{R}\). Prove that \(f\) is constant.
Solution
Define \[ g(z)=\overline{f\!\left(a+\frac{r^2}{\overline{z-a}}\right)} \] for \(z\ne a\). Because \(f\) is entire, \(g\) is holomorphic on \(\mathbb C\setminus\{a\}\). If \(|z-a|=r\), then \[ a+\frac{r^2}{\overline{z-a}}=z, \] and since \(f(z)\in\mathbb R\) on the circle, \[ g(z)=\overline{f(z)}=f(z). \] Thus the holomorphic functions \(f\) and \(g\) agree on the circle, hence by the identity theorem they agree on the connected punctured plane.
Therefore, for every \(z\ne a\), \[ f(z)=\overline{f\!\left(a+\frac{r^2}{\overline{z-a}}\right)}. \] As \(z\to\infty\), the point inside \(f\) on the right tends to \(a\), so the right-hand side tends to \(\overline{f(a)}\). Hence \(f(z)\) has a finite limit at infinity. An entire function with a finite limit at infinity is bounded, so by Liouville’s theorem \(f\) is constant.