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The problem states: Let $f: \Omega \to \mathbb C$ holomorphic, with $\Omega$ an open set and symmetric with respect to the real line, prove that $g(z)=\overline {f(\overline z)}$ is holomorphic.

I don't know how to solve the exercise, I am trying to prove by definition of differentiability, i.e., by the existence of the limit but I couldn't arrive to anything.

I want to show that $\lim_{\overline z \to \overline z_0} \dfrac{\overline {f(\overline z)}-\overline {f(\overline z_0)}}{\overline z - \overline z_0}$ exists, here I know I must use the symmetric hypothesis but I got stuck, I would appreciate some help.

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  • $\begingroup$ I haven't tried this, but using the fact that if $f = u(x,y) + iv(x,y)$ is holomorphic then $u$ and $v$ are harmonic might be easier. $\endgroup$ – JessicaK Apr 25 '14 at 17:20
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Use the cauchy-riemann differential equations, i.e. that $$ f(z) = f(u + iv) = g(u,v) + ih(u,v) $$ is holomorphic at $x = u + iv$ exactly if $$\begin{eqnarray} \frac{\partial g}{\partial u} &=& \frac{\partial h}{\partial v} \\ \frac{\partial g}{\partial v} &=& -\frac{\partial h}{\partial u} \end{eqnarray}$$ on some neighbourhood of $x$, and if the partial derivatives are continuous there.

For an $f$ as above, you have that $$ \hat f(u + iv) = \overline{f\left(\overline{u + iv}\right)} = g(u,-v) +i(-h(u,-v)) \text{,} $$ now you just have to compute the partial derivatives and use that the cauchy-riemann equations hold for $f$. The symmetrie of $\Omega$ guarantees that you don't leave the domain of $f$ when you conjugate.

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One nice definition of holomorphic is that for each $z_0$ there exists some number $g'(z_0)$ such that $g(z_0+h)=g(z_0)+hg'(z_0)+o(h)$. We have $$\begin{align}g(z_0+h)&=\overline{f(\overline{z_0+h})}\\&=\overline{f(\overline{z_0}+\overline h})\\&=\overline{f(\overline{z_0})+\overline hf'(\overline{z_0})+o(\overline h)}\\&=g(z_0)+h\cdot \underbrace {\overline{f'(\overline{z_0})}}_{=:g'(z_0)}+o(h)\end{align}$$

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You actually need to prove the existence of: $\text{lim}_{z\to z_0}\frac{\overline{f(\overline{z})}-\overline{f(\overline{z_0})}}{z-z_0}$

With your given expression, you are trying to show $\overline{f}$ is holomorphic, which is false.

But $\text{lim}_{z\to z_0}\frac{\overline{f(\overline{z})}-\overline{f(\overline{z_0})}}{z-z_0}= \overline{\text{lim}_{z\to z_0}\frac{f(\overline{z})-f(\overline{z_0})}{\overline{z}-\overline{z_0}}}= \overline{\text{lim}_{\overline{z}\to \overline{z_0}}\frac{f(\overline{z})-f(\overline{z_0})}{\overline{z}-\overline{z_0}}}=\overline{f'(\overline{z_0})}$

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