- #1
SwordSmith
- 8
- 0
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:
[tex] \frac{\partial z}{\partial z^*}, \quad z=x+iy [/tex]
Maybe you can do like this?
[tex] \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^*} [/tex]
[tex] \quad \quad = 1\cdot\frac{\partial x}{\partial (x-iy)} + i\cdot\frac{\partial y}{\partial (x-iy)} [/tex]
[tex] \quad \quad = 1 + i\cdot\frac{-1}{i} =0[/tex]
I am not sure the above procedure is correct. Can someone in here confirm the result or, if not, help me?
I assume I should use the chain rule somehow to figure this out:
[tex] \frac{\partial z}{\partial z^*}, \quad z=x+iy [/tex]
Maybe you can do like this?
[tex] \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^*} [/tex]
[tex] \quad \quad = 1\cdot\frac{\partial x}{\partial (x-iy)} + i\cdot\frac{\partial y}{\partial (x-iy)} [/tex]
[tex] \quad \quad = 1 + i\cdot\frac{-1}{i} =0[/tex]
I am not sure the above procedure is correct. Can someone in here confirm the result or, if not, help me?