Let \(f(z)\) be analytic in a domain, and prove that \(f\) is constant if it satisfies any of the following conditions:
- \(|f(z)|\) is constant.
- \(\Re(f(z))\) is constant.
- \(\arg(f(z))\) is constant.
- \(\overline{f(z)}\) is analytic.
Solution
Write \(f=u+iv\) on the connected domain.
1. If \(|f|\) is constant, then either \(f\equiv0\), or \(f\) is nonzero and its image lies in a circle. A nonconstant holomorphic map is open, but a circle has empty interior. Hence \(f\) is constant.
2. If \(\Re f=u\) is constant, then \(u_x=u_y=0\). By the Cauchy–Riemann equations,\[ v_y=u_x=0,\qquad v_x=-u_y=0, \]so \(v\) is constant and therefore \(f\) is constant.
3. If \(\arg f\equiv\theta_0\), then \(f\) never vanishes and\[ e^{-i\theta_0}f \]is real-valued. It is holomorphic, so by part (2) it is constant. Hence \(f\) is constant.
4. If \(\overline f=u-iv\) is also holomorphic, then applying the Cauchy–Riemann equations to both \(u+iv\) and \(u-iv\) gives\[ u_x=v_y=-v_y, \qquad u_y=-v_x=v_x. \]Thus \(u_x=u_y=v_x=v_y=0\), so \(f\) is constant.