Prove that if \(z\mapsto f(z)\) is analytic, then \(z \mapsto \bar{f(\bar z)}\) is analytic.
Solution
Goal: Prove that if \(z \mapsto f(z)\) is analytic on a domain \(\Omega\), then \(z \mapsto \overline{f(\bar z)}\) is analytic on \(\overline{\Omega} \definedas \theset{\bar z \suchthat z \in \Omega}\).
1. Fix \(z_0 \in \Omega\); near \(z_0\), \(f\) has a convergent power series \(f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n\).
Analytic functions are locally representable by power series (Taylor expansion).
2. Compute \(\overline{f(\bar z)}\) for \(z\) near \(\bar z_0\).
2.1. \(\overline{f(\bar z)} = \sum_{n=0}^{\infty} \bar a_n (z - \bar z_0)^n\).
Substitute \(\bar z\) for \(z\) in 1 and conjugate termwise: \(\overline{a_n(\bar z - z_0)^n} = \bar a_n \overline{(\bar z - z_0)}^n = \bar a_n (z - \bar z_0)^n\).
2.2. The series \(\sum \bar a_n (z - \bar z_0)^n\) converges for \(\abs{z - \bar z_0} < r\), where \(r\) is the radius of convergence of
1.
\(\abs{z - \bar z_0} = \abs{\bar z - z_0}\), so the set of convergence is the mirror image of that of 1.
3. \(g(z) := \overline{f(\bar z)}\) is analytic at \(\bar z_0\).
2.1 gives a power series representation of \(g\) centered at \(\bar z_0\) with positive radius of convergence, which characterizes analyticity.
4. Q.E.D.
\(\bar z_0\) was an arbitrary point of \(\overline{\Omega}\), so \(g\) is analytic throughout \(\overline{\Omega}\).