Is $g(z)$ holomorphic if $f$ is holomorphic in $G$?

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Here is this week's POTW, suggested by Chris L T521:

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$\newcommand{\CC}{\mathbb{C}}$ Let the function $f$ be holomorphic in the open set $G\subset\CC$. Prove that the function $g(z)=\overline{f(\overline{z})}$ is holomorphic in the set $G^{\ast}=\{\overline{z}:z\in G\}$.
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This week's problem was answered correctly by Cbarker1. You can read his solution below.

Spoiler

Let $g: G^{\ast} \to \Bbb{C}$ is given by $g(z)=\overline{f(\overline{z})}$. By definition of holomorphic, let $z_0 \in G^{\ast}$ be arbitrary, we have
\begin{align*}
&g'(z_0)=\lim_{{z}\to{z_0}}\frac{g(z)-g(z_0)}{z-z_0}\\
& =\lim_{{z}\to{z_0}}\frac{\overline{f(\overline{z})}-\overline{f(\overline{z_0})}}{z-z_0} & (\text{by the definition of g})\\
& =\lim_{{z}\to{z_0}}\frac{\overline{f(\overline{z})-f(\overline{z_0})}}{z-z_0} \\
&=\overline{f'(\overline{z_0})} &(\text{f is holomorphic}).
\end{align*}Thus, $g$ is holomorphic in the set $G^{\ast}$.