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I want to show given that $f:\mathbb{C} \to \mathbb{C}$ is holomorphic then $g(z)=\overline{f(\overline{z})}$ is also holomorphic. I used the definition of the derivative to write $$g'(z_0)=\lim_{z\to z_0} \frac{\overline{f(\overline{z})}-\overline{f(\overline{z}_0)}}{z-z_0}=\lim_{z\to z_0}\frac{f(\overline{z})-f(\overline{z}_0)}{\overline{z}-\overline{z}_0}.$$ Now I argue that conjugation is continuous so if $z\to z_0$ then $\overline{z}\to \overline{z}_0$, does this mean $g'(z_0)=f'(\overline{z}_0)$?

Since $f$ is differentiable on an open set $U$ then $g$ would be differentiable for $\overline{z}_0\in U$ and also holomorphic.

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  • $\begingroup$ Looks OK to me. If the domain of $f$ is a proper subset $U$ then the domain of $g$ is $\bar U$. $\endgroup$ – Ethan Bolker Feb 4 '18 at 15:56
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It should be:\begin{align}g'(z_0)&=\lim_{z\to z_0}\frac{\overline{f\left(\overline z\right)}-\overline{f\left(\overline{z_0}\right)}}{z-z_0}\\&=\overline{\lim_{z\to z_0}\frac{f\left(\overline z\right)-f\left(\overline{z_0}\right)}{\overline z-\overline{z_0}}}\\&=\overline{f'\left(\overline{z_0}\right)}.\end{align}

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