TITLE: Is conjugate of holomorphic function holomorphic?
QUESTION [7 upvotes]: If $f(z)$ is holomorphic, does it follow that $g(z)=\overline{f(z)}$ is holomorphic?
I'm looking at $$\lim_{z\rightarrow a}\dfrac{g(z)-g(a)}{z-a} = \lim_{z\rightarrow a}\dfrac{\overline{f(z)-f(a)}}{z-a}$$
Can we pull the limit out to get $\overline{\lim_{z\rightarrow a}\dfrac{f(z)-f(a)}{z-a}}$?

REPLY [11 votes]: If $f(z)=u(x,y)+iv(x,y)$ then $\overline{f(z)}=u(x,y)-iv(x,y)$. The Cauchy-Riemann equations imply $v=\operatorname{const}$. Hence $u(x,y)=\operatorname{const}$

REPLY [8 votes]: No. When you took the conjugation out, it forces you to conjugate $z-a$ in the denominator. 
An easy (and canonical) way to see that the conjugate of a holomorphic function is not holomorphic is to consider $z\mapsto \overline z$. This is easily confirmed by looking at the Cauchy-Riemann equations.