$g(z)$ analytic implies $\overline{g(\overline{z})}$ analytic?

Question

Let $D = \{x \in \mathbb{C} : \text{Re}(z) > 0\}$, and $g: D \to \mathbb{C}$ an analytic function. Does it follow that$$f(z) = \overline{g(\overline{z})}$$is also analytic?

Answer 1

It suffices to show that $f$ satisfies the Cauchy-Riemann equations. Let $g(z) = u(x, y) + iv(x, y)$, where $u$ and $v$ take real values. Then$$f(z) = u(x, -y) - iv(x, -y).$$We have$${\partial\over{\partial x}}u(x, -y) = {{\partial u}\over{\partial x}}(x, -y) = {{\partial v}\over{\partial y}}(x, -y) = {\partial\over{\partial y}}(-v(x, -y)).$$Similarly,$${\partial\over{\partial y}}u(x, -y) = -{{\partial u}\over{\partial y}}(x, -y) = {{\partial v}\over{\partial x}}(x, -y) = {\partial\over{\partial x}}v(x, -y).$$The Cauchy-Riemann equations are satisfied, so $f$ is analytic.

Answer 2

Fix $z_0\in D=\bar D$. Then $$f(z_0+h)-f(z_0)=\overline{g(\bar z_0+\bar h)}-\overline{g(\bar z_0)}=\overline{A\bar h+o(h)}=\bar A h+o(h)\qquad(h\to0)\ ,$$ where $A:=g'(\bar z_0)$. This proves that $f$ is complex differentiable at $z_0$ with $f'(z_0)=\bar A=\overline{g'(\bar z_0)}$.

Answer 3

By definition of differentiability, and because complex conjugation is a continuous involutive automorphism of $\mathbf{C}$, and because $\zeta \to 0$ if and only if $\overline{\zeta} \to 0$, $$f'(z) = \lim_{\zeta \to 0} \frac{f(z+\zeta)-f(z)}{\zeta} = \lim_{\zeta \to 0} \frac{\overline{g(\overline{z+\zeta})} - \overline{g(\overline{z})}}{\zeta} = \lim_{\zeta \to 0} \frac{\overline{g(\overline{z}+\overline{\zeta}) - g(\overline{z})}}{\overline{\overline{\zeta}}} = \overline{\lim_{\zeta \to 0} \frac{g(\overline{z}+\overline{\zeta}) - g(\overline{z})}{\overline{\zeta}}} = \overline{g'(\overline{z})},$$ so $f$ is differentiable, hence analytic.

Answer 4

$D$ doesn't need to be a half-plane, it is sufficient that $D$ is open and that $z \in D$ implies $\overline{z} \in D$. This is a natural condition, as the function $\overline{f(\overline{z})}$ wouldn't be well-defined otherwise.

It is easy to prove this statement by using power series. Let $z_0 \in D$ be an arbitrary complex number. Choose a real number $R > 0$ so that both closed balls $\overline{B_R}(z_0)$ and $\overline{B_r}(\overline{z_0})$ are subsets of $D$. Since $f$ is analytic, it can be written as a power series around $\overline{z_0}$, i.e. for all $z$ with $|z - \overline{z_0}| < R$ the equality

$$f(z) = \sum \limits_{n = 0}^\infty a_n (z - \overline{z_0})^n$$

holds. This implies

$$\overline{f(\overline{z})} = \overline{\sum \limits_{n = 0}^\infty a_n (\overline{z} - \overline{z_0})^n} = \sum \limits_{n = 0}^\infty \overline{a_n} (z - z_0)^n$$

for all $z$ with $|\overline{z} - \overline{z_0}| = |z - z_0| < R$, which means that $\overline{f(\overline{z})}$ is analytic at $z_0$. The claim now follows since $z_0 \in D$ was arbitrary.