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Analytic Function Derivative

  1. Jun 6, 2005 #1

    sat

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    Take [itex]x=r\cos\theta[/itex] and [itex]y=r\sin\theta[/itex]

    If [itex]f(z)=u(r,\theta) + iv(r,\theta)[/itex], is analytic with u and v real, show that the derivative is given by
    [tex]f'(z) = \left( \cos\theta \frac{\partial u}{\partial r}- \sin\theta\frac{1}{r}\frac{\partial u}{\partial\theta}\right) + i\left( \cos\theta\frac{\partial v}{\partial r} - \sin\theta\frac{1}{r}\frac{\partial v}{\partial\theta} \right)[/tex]

    Since f is analytic, I use the result
    [tex]f'(z)=\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}[/tex]
    Though this seems to give
    [tex]f'(z) = \left( (1/\cos\theta) \frac{\partial u}{\partial r}- (1/\sin\theta)\frac{1}{r}\frac{\partial u}{\partial\theta}\right) + i\left( (1/\cos\theta)\frac{\partial v}{\partial r} - (1/\sin\theta)\frac{1}{r}\frac{\partial v}{\partial\theta} \right)[/tex]

    Can anyone see why this isn't correct?
     
  2. jcsd
  3. Jun 6, 2005 #2

    dextercioby

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    How did you apply the chain rule...?I think there's a problem with that.

    Daniel.
     
  4. Jun 6, 2005 #3

    sat

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    I just did:

    du/dx = (du/dr)(dr/dx) + (du/dtheta)(dtheta/dx)

    though I think that's probably not quite right when I think about the dependence of the variables on eachother...

    Though the "units" look OK at first sight, I'm not sure that I can do that.
     
  5. Jun 6, 2005 #4

    dextercioby

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    Here's where you went wrong

    [tex] \frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial\theta}\frac{\partial\theta}{\partial x} [/tex] (1)

    [tex] \frac{\partial r}{\partial x}=\frac{\partial}{\partial x}\sqrt{x^{2}+y^{2}}=\frac{x}{\sqrt{x^{2}+y^{2}}}=\frac{r\cos\theta}{r}=\cos\theta [/tex] (2)

    Can u do the other derivative...?

    Daniel.
     
  6. Jun 6, 2005 #5

    sat

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    Thanks. I'll look at that.
     
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