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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.