Derivative wrt Complex Conjugate

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SUMMARY

The derivative with respect to a complex conjugate, denoted as \(\frac{\partial z}{\partial z^*}\), is evaluated using the chain rule. The correct approach involves expressing \(z\) in terms of its real and imaginary components, \(z = x + iy\) and \(\overline{z} = x - iy\). The resulting derivative is confirmed to be zero if the function satisfies the Cauchy-Riemann equations, leading to the important equation \(\frac{\partial f}{\partial\overline{z}} = \frac{1}{2}\left(\frac {\partial f}{\partial x} + i\frac {\partial f}{\partial y}\right)\).

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SwordSmith
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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?
 
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SwordSmith said:
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
 

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