Let \(f\) be holomorphic on an open set containing the closed unit disk and suppose \(f\) is real-valued on the unit circle. Prove that \(f\) is constant.
Find a nonconstant function holomorphic on the complex plane except for a single point of the unit circle and real-valued at every other point of the unit circle.
Constancy on the whole open set requires the open set to be connected. On \[ U=\{\abs z<2\}\union\{\abs{z-4}<1/2\}, \] the function that is \(0\) on the first component and \(1\) on the second is holomorphic on \(U\), is real-valued on the unit circle, and is not constant on \(U\).
Solution
1. Let \(U_0\) be the connected component of the given open set that containsthe closed unit disk. Then \(f\) is constant on \(U_0\).
Write \[ f=u+iv \] on the open unit disk \(\DD\). Since \(f\) is holomorphic, the function \(v=\operatorname{Im}f\) is harmonic on \(\DD\). It is continuous on \(\overline{\DD}\) and vanishes on \(\partial\DD\), because \(f\) is real-valued on the unit circle.
The maximum principle for harmonic functions applied to \(v\) gives \[ v\leq0 \] on \(\DD\), while applying it to \(-v\) gives \[ v\geq0. \] Thus \(v\equiv0\) on \(\DD\), so \[ f(\DD)\subseteq\RR. \] If \(f|_{\DD}\) were nonconstant, the open mapping theorem would make \(f(\DD)\) open in \(\CC\), which is impossible for a subset of \(\RR\). Therefore \(f\) is constant on \(\DD\).
Since \(\DD\subseteq U_0\) and \(U_0\) is connected, the identity theorem implies that \(f\) is constant on all of \(U_0\).
2. In Part (2), one may take\[ \boxed{ F(z)=i\frac{1+z}{1-z} }. \]
The function \(F\) is rational, its denominator vanishes only at \(z=1\), and \[ F'(z)=\frac{2i}{(1-z)^2}, \] so \(F\) is holomorphic on \(\CC\sm\{1\}\) and nonconstant.
If \(\abs{z}=1\) and \(z\neq1\), then \(\bar z=1/z\), and hence \[ \begin{aligned} \overline{F(z)} &= -i\frac{1+\bar z}{1-\bar z}\\ &= -i\frac{1+1/z}{1-1/z}\\ &= -i\frac{z+1}{z-1}\\ &= i\frac{1+z}{1-z}\\ &= F(z). \end{aligned} \] Thus \(F(z)\) is real for every point of the unit circle other than its single singularity at \(z=1\).
3. Q.E.D.
Step 1 proves Part (1) on the connected component containing the closed unit disk, and step 2 supplies the required example for Part (2).