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Problem Statement

If $F(z)$ is analytic in a domain $D$, is $\overline{F(\overline{z})}$ also analytic in $D$?

My Try

Suppose that $F(z)$ is analytic in a domain $D$. It follows that $F$ is differentiable at each $z \in D$.

I know that the partial derivative $$\frac{\partial F(z)}{\partial \overline{z}} = 0,$$ since $F$ is differentiable at each point $z \in D$. What I am having trouble with is how to get an expression for $F(\overline{z})$.

Question

Of course, I cannot conclude at once that $\overline{z} \in D$ also holds. Or can I? Is this even the correct approach to the problem? If not, how do I solve this conundrum? Hints will be most welcome!

Archimedes Plutonium
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1 Answers1

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What you wrote is just the Cauchy-Riemann equations (using the Wirtinger derivative $\frac{\partial}{\partial \overline{z}}$). I am not sure about how could you do the problem with this approach but I think it might not be "the correct one" since you can easily prove $\overline{f(\overline{z})}$ is analytic from the definition of complex derivative: Let $z_0\in\mathbb{D}$, and let $g:\mathbb{D}\to\mathbb{D}$ be the function defined by $g(z)=\overline{z}$. You need to check that the limit $$ \lim_{z\to z_0}\frac{\overline{f(\overline{z})}-\overline{f(\overline{z_0})}}{z-z_0} $$ exists. Noticing the expression inside the limit equals $$ g\left(\frac{f(\overline{z})-f(\overline{z_0})}{\overline{z}-\overline{z_0}}\right) $$ and using that $g$ is a continuous function you should be able to conclude that the limit above is $g(f'(\overline{z}))=\overline{f'(\overline{z_0})}$. As this limit exists for every $z_0$ you can conclude $\overline{f(\overline{z})}$ is holomorphic in $\mathbb{D}$, and thus analytic.

Jonatan B. Bastos
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