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Let $\phi:U\rightarrow \mathbb{C}$, where $U\subseteq\mathbb{C}$ is open and connected, and $\phi$ is analytic in $U$.
Assume that $\bar{\phi}:U\rightarrow \mathbb{C}$ is analytic in $U$ where $\bar{\phi}(x)=\overline{\phi(x)}$.

How do I show that $\phi$ is constant?

What I know:
I was thinking about using the difference quotient $\frac{\phi(a+h)-\phi(a)}{h}$, but I'm not sure what to do with it exactly.

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Since $\overline{\phi}$ is analytic, one has $$0 = \frac{\partial \overline{\phi}}{\partial \overline{z}} = \overline{\left(\frac{\partial{\phi}}{\partial{z}}\right)}.$$ Hence $\frac{\partial{\phi}}{\partial{z}}=0$ and $\frac{\partial{\phi}}{\partial{\overline{z}}}=0$ by hypothesis. This implies $\frac{\partial{\phi}}{\partial{x}}=0$ and $\frac{\partial{\phi}}{\partial{y}}=0$. Hence $\phi$ is constant.

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  • $\begingroup$ Why do we know that $0=\frac{\partial\overline{\phi}}{\partial\overline{z}}=\overline{\left(\frac{ \partial\phi}{\partial z}\right)}$? $\endgroup$ – Spider-Pig Mar 12 '16 at 20:03
  • $\begingroup$ The first equality is by hypothesis ($\overline{\phi}$ is analytic). The second is a true equality for all function. You can obtain it simply by expending the expression. (Write $\phi = u+iv$) $\endgroup$ – C. Dubussy Mar 12 '16 at 20:09
  • $\begingroup$ I thought that being analytic means that $\overline{\phi}$ is complex differentiable at all points in $U$ $\endgroup$ – Spider-Pig Mar 13 '16 at 9:04
  • $\begingroup$ Yeah but the annulation of $\frac{\partial}{\partial \overline{z}}$ is equivalent. en.wikipedia.org/wiki/Holomorphic_function $\endgroup$ – C. Dubussy Mar 13 '16 at 9:27

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