Let \(f\) be analytic on \(\{z\in\mathbb C:|z|>1\}\). Prove that if \(f\) is real-valued on \((1,\infty)\), then it is also real-valued on \((-\infty,-1)\).
Solution
Let \[ \Omega\coloneqq\{z\in\CC:\abs z>1\}. \]
1. The function\[ g(z)\coloneqq\overline{f(\overline z)} \]is holomorphic on \(\Omega\).
If \(z_0\in\Omega\), then \(\overline{z_0}\in\Omega\). Expanding \(f\) in a power series around \(\overline{z_0}\), \[ f(w)=\sum_{n=0}^{\infty}a_n(w-\overline{z_0})^n, \] gives, near \(z_0\), \[ g(z) = \sum_{n=0}^{\infty}\overline{a_n}(z-z_0)^n. \] Thus \(g\) is holomorphic.
2. For every real \(x>1\),\[ g(x)=f(x). \]
Since \(x\) is real and \(f(x)\) is real by hypothesis, \[ g(x) = \overline{f(\overline x)} = \overline{f(x)} = f(x). \]
3. One has\[ g=f \]throughout \(\Omega\).
The domain \(\Omega\) is connected. By steps <1>1–<1>2, the holomorphic functions \(f\) and \(g\) agree on the interval \((1,\infty)\), which has accumulation points in \(\Omega\). The identity theorem gives \(f=g\) on all of \(\Omega\).
4. For every real \(x<-1\), the value \(f(x)\) is real.
Such an \(x\) lies in \(\Omega\) and satisfies \(\overline x=x\). By step 3, \[ f(x) = g(x) = \overline{f(\overline x)} = \overline{f(x)}. \] A complex number equal to its conjugate is real. Hence \[ \boxed{f(x)\in\RR\quad\text{for every }x<-1}. \]
5. Q.E.D.
Step 4 is the required conclusion.