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Homework Help: Complex Derivative

  1. Sep 1, 2006 #1
    The question is as follows.

    Let w = 1/z. Check that for [itex]z \neq 0[/itex] the Cauchy-Riemann equations are satisfied and verify that[itex]dw/dz = -1/z^2[/itex]
    -------------

    So I let z = x + iy

    Then, [tex]\frac{1}{z} = \frac{1}{x + iy} = \frac{1}{x + iy}\frac{x - iy}{x - iy} = \frac{x - iy}{x^2 + y^2}[/tex]

    let w = u + iv
    so [tex]u = \frac{x}{x^2 + y^2}[/tex] and [tex]v = \frac{-y}{x^2 + y^2}[/tex]

    [tex]\frac{\partial u}{\partial x} = \frac{-x^2 + y^2}{(x^2 + y^2)^2} = \frac{\partial v}{\partial y}[/tex]

    [tex]\frac{ - \partial u}{\partial y} = \frac{2xy}{(x^2 + y^2)^2 } = \frac{\partial v}{\partial x}[/tex]

    I really don't want to write everything out, but the Cauchy-Riemann equations are satisfied, and thus we can find [tex]dw/dz = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}[/tex]

    My problem is right here when I am trying to manipulate this [tex] \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} [/tex] to get [tex] \frac{1}{z^2} = \frac{1}{(x + iy)^2} = \frac{1}{x^2 + i2xy - y^2}[/tex]

    Is there some trick I am missing or did I take the partial derivatives wrong? I just seem to be stuck here with no ideas. Thanks
     
    Last edited: Sep 1, 2006
  2. jcsd
  3. Sep 2, 2006 #2

    quasar987

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    You're gonna say "duh!". The first thing you found is that

    [tex]\frac{1}{z} = \frac{x - iy}{x^2 + y^2}[/tex]

    remember? Square that, and you get exactly

    [tex] -\frac{\partial u}{\partial x} - i\frac{\partial v}{\partial x} [/tex]
     
  4. Sep 2, 2006 #3
    :rofl: Duh!

    Thank you!!!
     
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