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Differentiable functions in complex analysis

  1. Nov 7, 2013 #1
    Hello all, I have the following problem from Complex Analysis that I would like for someone to check my understanding on:
    1. The problem statement, all variables and given/known data
    The problem is to find the derivative if it exists of
    [itex] f(z) = \frac{e^{i\theta}}{r^2} = r^{-2}\cos \theta + i r^{-2}\sin \theta [/itex]
    where I have already changed the complex function f(z) into polar form.



    2. Relevant equations
    The Cauchy-Riemann conditions in polar coordinates are
    [itex] \frac{\partial{u}}{\partial{r}} = \frac{1}{r}\frac{\partial{v}}{\partial{\theta}} [/itex]
    and
    [itex] \frac{\partial{u}}{\partial{\theta}} = -r\frac{\partial{v}}{\partial{r}} [/itex]



    3. The attempt at a solution
    For the first Cauchy-Riemann condition above, I get
    [itex] \frac{\partial{u}}{\partial{r}} = \frac{-2}{r^3}\cos \theta [/itex] and
    [itex] \frac{1}{r}\frac{\partial{v}}{\partial{\theta}} = \frac{1}{r^3}\cos \theta [/itex]
    For the second Cauchy-Riemann condition above I get
    [itex] \frac{\partial{u}}{\partial{\theta}} = \frac{-1}{r^{2}}\sin \theta [/itex] and
    [itex] -r\frac{\partial{v}}{\partial{r}} = \frac{2}{r^2}\sin \theta [/itex]

    To check my understanding of this, the Cauchy-Reimann conditions are not satisfied "symbolically", but there could be certain values of [itex] r [/itex] or [itex] \theta [/itex] where the Cauchy-Reimann conditions are satisfied, right? However, in this problem, I am not seeing any combination of [itex] r [/itex] or [itex] \theta [/itex] where the Cauchy-Riemann conditions could be satisfied, and so this function is not differentiable anywhere. Is this the right understanding of complex functions being differentiable or not?
     
  2. jcsd
  3. Nov 7, 2013 #2
    Usually we differentiate with respect to z. If the function has a derivative with respect to z it is analytic throughout some neighborhood of z.

    If you break the function back down either into x,y or r ##\theta##, getting it all back together is a major hassle.

    So what was the original function of z?
     
  4. Nov 7, 2013 #3
    Well, it wasn't in terms of z. It was [itex] \frac{x+iy}{x^2 + y^2} [/itex] which i thought would work better if I changed it to polar form.
     
  5. Nov 7, 2013 #4

    Dick

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    Yes, it is easier in polar. But you don't have the right polar form. x+iy isn't the same as ##e^{i \theta}##.
     
  6. Nov 7, 2013 #5
    Ahh, I'm seeing that I should have written [itex] x+iy = re^{i\theta} [/itex], is this what you are hinting at?

    Re-working through it all I get down to similar expressions for the C-R conditions above with 2 changing to 1, and the powers of r decreasing by 1. My conclusion above is still the same.
     
    Last edited: Nov 7, 2013
  7. Nov 7, 2013 #6

    Dick

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    Yes, it's not differentiable everywhere, but there are some values of r and θ where it is differentiable. Can you find them? And could you spell out what you got for the new CR equations?
     
  8. Nov 7, 2013 #7
    New CR equations are

    1st set:
    [itex]\frac{\partial{u}}{\partial{r}} = -\frac{1}{r^2}\cos \theta [/itex] and
    [itex]\frac{1}{r}\frac{\partial{v}}{\partial{\theta}} = \frac{1}{r^2}\cos \theta [/itex]

    2nd set:
    [itex]\frac{\partial{u}}{\partial{\theta}} = -\frac{1}{r}\sin \theta [/itex] and
    [itex]-\frac{r\partial{v}}{\partial{r}} = \frac{1}{r}\sin \theta [/itex]

    This one I'm having some trouble with:
    If I try [itex] \theta = 0 [/itex]
    then the 2nd set of CR equations is satisfied, but the 1st set is not. And vice versa for [itex]\theta = \pi / 2 [/itex].
    If I try angles like [itex] \frac{\pi}{4}, \frac{3\pi}{4}, \ldots [/itex] I get one negative and one positive number for each set of CR equations.
    Any r I choose except r=0, I get one negative and one positive number for each set of CR equations.

    Are my CR equations right?
     
  9. Nov 7, 2013 #8

    Dick

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    Ooops. Actually, I made a sign error. I think you are right.
     
    Last edited: Nov 7, 2013
  10. Nov 7, 2013 #9

    D H

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    That is correct given the corrected function.

    Even easier is to use the fact that x2+y2=zz*, which means f(z)=z/(zz*)=1/z*. What does that tell you about differentiability?
     
  11. Nov 7, 2013 #10
    Thank you everyone.

    I wish I would have thought of this in the first place; thanks for the extra tip.
     
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