Derivative wrt Complex Conjugate

[SwordSmith]

I am not sure what the derivative with respect to a complex conjugate is and I have not been able to find it in any books.

I assume I should use the chain rule somehow to figure this out:
$$\frac{\partial z}{\partial z^*}, \quad z=x+iy$$

Maybe you can do like this?
$$\frac{\partial z}{\partial z^*}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial z^*}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial z^*}$$
$$\quad \quad = 1\cdot\frac{\partial x}{\partial (x-iy)} + i\cdot\frac{\partial y}{\partial (x-iy)}$$
$$\quad \quad = 1 + i\cdot\frac{-1}{i} =0$$

I am not sure the above procedure is correct. Can someone in here confirm the result or, if not, help me?

 

[DonAntonio]

I am not sure what the derivative with respect to a complex conjugate is and I have not been able to find it in any books.

I assume I should use the chain rule somehow to figure this out:
$$\frac{\partial z}{\partial z^*}, \quad z=x+iy$$

Maybe you can do like this?
$$\frac{\partial z}{\partial z^*}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial z^*}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial z^*}$$
$$\quad \quad = 1\cdot\frac{\partial x}{\partial (x-iy)} + i\cdot\frac{\partial y}{\partial (x-iy)}$$
$$\quad \quad = 1 + i\cdot\frac{-1}{i} =0$$

I am not sure the above procedure is correct. Can someone in here confirm the result or, if not, help me?

Put $z=x+iy , \overline{z}=x-iy$ , and applying the chain rule:

$\frac {\partial f}{\partial x}=\frac {\partial f}{\partial \overline{z}}\frac {\partial \overline{z}}{\partial x}=\frac{\partial f}{\partial\overline{z}}$

$\frac {\partial f}{\partial y}=\frac {\partial f}{\partial \overline{z}}\frac {\partial \overline{z}}{\partial y}=\frac{\partial f}{\partial\overline{z}}(-i)$

Sum both extreme equalities and get the important and known equation

$\frac{\partial f}{\partial\overline{z}}=\frac{1}{2}\left(\frac {\partial f}{\partial x}+i\frac {\partial f}{\partial y}\right)$ , which equals zero iff the function fulfills the Cauchy-Riemann equations.

Tonio