Let \(f\) be analytic on \(\mathbb C\setminus\{0\}\) and suppose \[ f(1/z)=f(z). \] Assume also that \(f(z)\) is real whenever \(|z|=1\). Prove that \(f(z)\) is real for every nonzero real \(z\).
Solution
1. The function\[ g(z)=\overline{f(\overline z)} \]is analytic on \(\CC\setminus\{0\}\).
Fix \(z_0\neq0\). Since \(f\) is analytic near \(\overline{z_0}\), it has a convergent power-series expansion \[ f(w)=\sum_{n=0}^{\infty}a_n(w-\overline{z_0})^n \] there. Hence near \(z_0\), \[ \begin{aligned} g(z) &= \overline{ \sum_{n=0}^{\infty} a_n(\overline z-\overline{z_0})^n }\\ &= \sum_{n=0}^{\infty} \overline{a_n}(z-z_0)^n. \end{aligned} \] Thus \(g\) is analytic near every nonzero \(z_0\).
2. For every \(z\) on the unit circle,\[ g(z)=f(z). \]
If \(\abs z=1\), then \[ \overline z=\frac1z. \] The inversion hypothesis gives \[ f(\overline z) = f(1/z) = f(z). \] Since \(f(z)\) is real on the unit circle, \[ g(z) = \overline{f(\overline z)} = \overline{f(z)} = f(z). \]
3. One has\[ g(z)=f(z) \]for every \(z\neq0\).
The functions \(f\) and \(g\) are analytic on the connected domain \[ \CC\setminus\{0\}. \] By step 2, they agree on the unit circle, which has accumulation points in this domain. The identity theorem therefore gives \[ f\equiv g \] on \(\CC\setminus\{0\}\).
4. If \(x\in\RR\setminus\{0\}\), then \(f(x)\) is real.
Since \(\overline x=x\), step 3 gives \[ f(x) = g(x) = \overline{f(\overline x)} = \overline{f(x)}. \] A complex number equals its conjugate exactly when it is real. Hence \(f(x)\in\RR\).
5. Q.E.D.
Step 4 is the required conclusion.