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Derivative wrt Complex Conjugate

  1. Mar 30, 2012 #1
    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?
  2. jcsd
  3. Mar 30, 2012 #2

    Put [itex]z=x+iy , \overline{z}=x-iy[/itex] , and applying the chain rule:

    [itex]\frac {\partial f}{\partial x}=\frac {\partial f}{\partial \overline{z}}\frac {\partial \overline{z}}{\partial x}=\frac{\partial f}{\partial\overline{z}}[/itex]

    [itex]\frac {\partial f}{\partial y}=\frac {\partial f}{\partial \overline{z}}\frac {\partial \overline{z}}{\partial y}=\frac{\partial f}{\partial\overline{z}}(-i)[/itex]

    Sum both extreme equalities and get the important and known equation

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

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