Complex Analysis. Analytic functions

Question

How could I solve this problem?:

"Supose an open set A $\subset$ $\mathbb C$ , so that $A^*= \lbrace z \in \mathbb C : \bar{z} \in A \rbrace$. If f is an analytic function in A, demonstrate that $F(z)=\overline{f(\bar{z})}$ is an analytic function in $A^*$".

Thanks for your attention!

Answer 1

Given $z \in A^*$,

$$\lim_{w \to z} \frac{F(w) - F(z)}{w - z} = \lim_{w\to z} \frac{\overline{f(\bar{w})} - \overline{f(\bar{z})}}{w - z} = \overline{\lim_{\bar{w}\to \bar{z}} \frac{f(\bar{w}) - f(\bar{z})}{\bar{w} - \bar{z}}} = \overline{f'(\bar{z})}.$$

Therefore, $F$ is complex differentiable with $F'(z) = \overline{f'(\bar{z})}$, for all $z \in A^*$. So $F$ is analytic in $A^*$.

Answer 2

Fix $w\in A$. Then $f$ can be expressed by a convergent power series, that is, $f(z)=\sum_{k=0}^\infty a_k(z-w)^k$ for $z$ in a neighbourhood of $w$. Now, $F(z)=\sum_{k=0}^\infty\overline{a_k}(z-\overline{w})^k$ for $z$ in a neighbourhood of $\overline{w}$, and since the radius of convergence is still the same, $F$ is analytic in a neighbourhood of $\overline{w}$.

Answer 3

Recall $f(z)$ is analytic in $D$ iff $\frac{\partial f(z)}{\partial \bar{z}}\equiv 0$ for $z\in D$. Now for $z\in A^*$, let $w=\bar{z}\in A$. In terms of $w$, $f(w)$ is analytic in $A$, namely $\frac{\partial f(w)}{\partial \bar{w}}\equiv 0$ for $w\in A$. Then $$ \frac{\partial F(z)}{\partial \bar{z}}=\overline{\left(\frac{\partial \overline{F(z)}}{\partial z}\right)}=\overline{\left(\frac{\partial f(\overline{z})}{\partial z}\right)}=\overline{\left(\frac{\partial f(w)}{\partial \bar{w}}\right)}=0. $$ Thus $F(z)$ is analytic in $A^*$.

Answer 4

Interesting question.

Let $\gamma$ be a simple closed curve contained (with its interior) in $A^*$. We claim that $\int_\gamma F(z)\,dz = 0$. If we show this holds for all closed curves in $A^*$, it will prove that $F$ is analytic in $A^*$.

Let $\gamma^*$ be the complex conjugate curve in $A$. We know $f$ is analytic, so $\int_{\gamma^*} f(z)\,dz=0$. Compute: $$ \int_\gamma F(z)\,dz = \int_\gamma\overline{f(\,\overline{z}\,)}\;dz = \int_\gamma \overline{f(\,\overline{z}\,)\,d\overline{z}} \\ =\int_{\gamma^*}\overline{f(z)\,dz} = \overline{\int_{\gamma^*}f(z)\,dz} = \overline{0} = 0 . $$

Maybe $\gamma^*$ has reversed orientation? So maybe we get a minus sign? Anyway, it's still zero.