# Is conjugate of holomorphic function holomorphic?

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 $\overline{\lim_{z\rightarrow a}\dfrac{f(z)-f(a)}{z-a}}$?

• $f(z) = z$ is holomorphic, but $\overline{f(z)} = \overline{z}$ is not. Commented Oct 5, 2013 at 20:07
• Note that if you tried to pull the complex conjugate out of the limit, you would have to conjugate the denominator. Commented Oct 5, 2013 at 20:10
• But $$g(z) = \overline{f(\overline{z})}$$ is holomorphic (on the reflection of the domain of $f$). Commented Oct 5, 2013 at 20:33
• @DanielFischer, why is that?
– Ian
Commented Feb 2, 2014 at 22:34
• @Ian Because one conjugates twice. The composition of a $\mathbb{C}$-linear map with an antilinear map is antilinear, the composition of two antilinear maps is linear. In the same way (and for the same reason), the composition of a holomorphic and an antiholomorphic map is antiholomorphic, and the composition of two antiholomorphic maps is holomorphic. Conjugation is antiholomorphic, so the composition of holomorphic maps with an even number of conjugations, in whatever order, is holomorphic; with an odd number of conjugations, antiholomorphic. Commented Feb 2, 2014 at 22:40

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 $u(x,y)=\operatorname{const}$
• By Cauchy - Riemann we get $v_x=v_y$ and $u_x=u_y$, which is the result or the fact to conclude that $u$ and $v$ are constant? Commented May 6, 2021 at 23:33
• $u_x = v_y$ and $u_y = -v_x$ from f holomorphic Commented Mar 3, 2022 at 20:40
• $u_x = -v_y$ and $u_y = v_x$ from conjugate holomorphic so they are all $0$ Commented Mar 3, 2022 at 20:42
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.