Is conjugate of holomorphic function holomorphic?

Question

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. — tylerc0816, Oct 05 '13 at 20:07
Note that if you tried to pull the complex conjugate out of the limit, you would have to conjugate the denominator. — Cameron Williams, Oct 05 '13 at 20:10
But $$g(z) = \overline{f(\overline{z})}$$ is holomorphic (on the reflection of the domain of $f$). — Daniel Fischer, Oct 05 '13 at 20:33
@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. — Daniel Fischer, Feb 02 '14 at 22:40

score 14 · Answer 1 · answered Oct 05 '13 at 20:09

14

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 $u(x,y)=\operatorname{const}$

answered Oct 05 '13 at 20:09

njguliyev

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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? – George May 06 '21 at 23:33
$u_x = v_y$ and $u_y = -v_x$ from f holomorphic – jg mr chapb Mar 03 '22 at 20:40
$u_x = -v_y$ and $u_y = v_x$ from conjugate holomorphic so they are all $0$ – jg mr chapb Mar 03 '22 at 20:42

score 11 · Accepted Answer · answered Oct 05 '13 at 20:10

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.

Is conjugate of holomorphic function holomorphic?

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