Complex derivative of complex conjugate is zero

Question

I'm reading a proof for the statement: let $G\subset\mathbb{C}$ be an open set containing all inner regions of piecewise horizontal or vertical closed linear curves in $G$, and let $u$ be harmonic on $G$. Then there exists a harmonic conjugate $v$ of $u$ on $G$.

The proof goes on as follows.

Since $u$ is harmonic, it satisfies $\frac{\partial^2u}{\partial\overline{z}\partial z}=0$ on $G$, thus $g=\frac{\partial u}{\partial z}$ satisfies C-R. Define the holomorphic function $F(z)=\int_a^z g(w)\,\mathrm{d}w$, where we integrate over a piecewise horizontal or vertical linear curve (it follows from Cauchy that it doesn't matter which one). Then $\frac{\partial F}{\partial z}=\frac{\partial u}{\partial z}=g(z)$. Then \begin{align*}\frac{\partial }{\partial z}(F+\overline{F})&=\frac{\partial u}{\partial z};\\\frac{\partial }{\partial \overline{z}}(F+\overline{F})&=\frac{\partial u}{\partial \overline{z}}.\end{align*} The last equation follows because $u$ is a real valued function. Thus $F+\overline{F}-u=C\in\mathbb{R}$, thus $u=2\mathrm{Re}(F)-C$, and we could take $v=2\mathrm{Im}(F)$.

I can follow this proof, except for the step $\frac{\partial }{\partial z}(F+\overline{F})=\frac{\partial u}{\partial z}$. They mention that this is true because $\frac{\partial\overline{F}}{\partial z}=0$. I don't understand how we could just take the derivative of a complex conjugate of a function; also in general this step does not seem to make sense. Similar problems arise at the second line of the aligned part. I'm looking for some explanation for this step. Any help or reference is much appreciated!

Answer 1

Write \begin{align*} F(z) & = F(x,y) = u(x,y) + iv(x,y) \\ \overline{F}(z) & = \overline{F}(x,y) = u(x,y) - iv(x,y). \end{align*} Then: \begin{align*} \frac{\partial F}{\partial z} & = \frac{1}{2}\left( \frac{\partial F}{\partial x} - i \frac{\partial F}{\partial y} \right) \\ & = \frac{1}{2}\left( u_x + iv_x - i(u_y + iv_y) \right) \\ & = \frac{1}{2}\left( (u_x + v_y) + i(v_x - u_y) \right) \tag{1} \end{align*} while on the other hand \begin{align*} \frac{\partial \overline{F}}{\partial \overline{z}} & = \frac{1}{2}\left( \frac{\partial \overline{F}}{\partial x} + i \frac{\partial \overline{F}}{\partial y} \right) \\ & = \frac{1}{2}\left( u_x - iv_x + i(u_y - iv_y) \right) \\ & = \frac{1}{2}\left( (u_x + v_y) + i(- v_x + u_y) \right) \\ & = \frac{1}{2}\left( (u_x + v_y) - i(v_x - u_y) \right) \tag{2} \end{align*} Expressions (1) and (2) are complex conjugates of each other. We deduce that: $$\overline{\frac{\partial F}{\partial z}} = \frac{\partial \overline{F}}{\partial \overline{z}}$$

Answer 2

They are using the Wirtinger derivative: $$\dfrac{\partial}{\partial z}=\dfrac{1}{2}\bigg(\dfrac{\partial}{\partial x}-i\dfrac{\partial}{\partial y}\bigg)\\ \dfrac{\partial}{\partial \bar z}=\dfrac{1}{2}\bigg(\dfrac{\partial}{\partial x}+i\dfrac{\partial}{\partial y}\bigg) $$

It is then easily calculated $$\dfrac{\partial}{\partial z}\bar F(z)=\dfrac{\partial}{\partial \bar z}F(z) =0.$$

EDIT: The calculation. Write $F(x,y)=u(x,y)+iv(x,y)$ so $\bar F(x,y)=u(x,y)-iv(x,y)$. Then $$\begin{align} \dfrac{\partial}{\partial z}\bar F & = \dfrac{1}{2}\bigg(\dfrac{\partial}{\partial x}-i\dfrac{\partial}{\partial y}\bigg)(u-iv)\\ & = \dfrac{1}{2}(u_x-v_y)-\dfrac{i}{2}(u_y+v_x) \\ & = \overline{\dfrac{1}{2}\bigg(\dfrac{\partial}{\partial x}+i\dfrac{\partial}{\partial y}\bigg)(u+iv)}\\ & = \overline{\dfrac{\partial}{\partial\bar z}F}\\ & = \bar 0 =0. \end{align} $$So it would be better to say $$\dfrac{\partial}{\partial \bar z}F(z)=0=\bar 0 =\dfrac{\partial}{\partial z}\bar F(z).$$

Answer 3

"where we integrate over a piecewise horizontal or vertical linear curve (it follows from Cauchy that it doesn't matter which one)": It certainly does matter which one, unless you know something you're not telling us. If $g(z)=1/z$ then the integral of $g$ over the piecewise linear curve joining $-1-i$, $-1+i$ and then $1+i$ is not the same as the integral over the curve from $-1-i$ to $1-i$ and then to $1+i$.