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Doubt in Partial derivative of complex variables

  1. May 14, 2015 #1
    Today, I had a class on Complex analysis and my professor wrote this on the board :

    The Laplacian satisfies this equation :


    So, how did he arrive at that equation?
  2. jcsd
  3. May 14, 2015 #2


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    ## \triangle=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}##. Now you should just use chain rule to write derivatives w.r.t. x and y, in terms of derivative w.r.t. ## z ## and ## \bar z ##.
  4. May 14, 2015 #3
    x = (z + z¯)/2
    y = (z - z¯)/2i

    How do I solve this further?

    EDIT:Sorry, I don't know how to write the latex code to represent the "Bar" above "z"
  5. May 14, 2015 #4
    I guess, I'll have to use Wirtinger derivatives.
  6. May 14, 2015 #5
    By the definition $$\frac{\partial}{\partial z}=\frac12\left(\frac{\partial}{\partial x} -i \frac{\partial}{\partial y} \right), \qquad \frac{\partial}{\partial\overline z}=\frac12\left(\frac{\partial}{\partial x} +i \frac{\partial}{\partial y} \right).$$ So if you multiply these two differential operators and use the fact that $$\frac{\partial}{\partial x}\frac{\partial}{\partial y} = \frac{\partial}{\partial y}\frac{\partial}{\partial x}$$ (equality of mixed partial derivatives), you get exactly the Laplacian.
  7. May 14, 2015 #6
    I meant ##1/4## of the Laplacian, i.e. $$\frac14\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \right).$$
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