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Homework Help: Analytical function

  1. Sep 22, 2011 #1
    Hi there. I have to study the analyticity for the complex function:


    The exercise suggest me to use polar coordinates. So, I do this kind of exercise using a theorem that says that if the function acomplishes the Cauchy-Riemann conditions, and the partial derivatives are continuous in the vecinity of a point, then its analytical in that region.

    At first glance I would say its analytical over the entire complex plane except at the point zero, because its a division between polynomials, and the denominator is zero at zero. But I tried to demonstrate it as the exercise suggests me using polar coordinates.

    So this is what I did:

    [tex]-ie^{i\theta}=-i \cos\theta+\sin\theta[/tex]

    \frac{-\rho ie^{i\theta}}{\rho^2}=\frac{\sin\theta-i\cos\theta}{\rho}[/tex]

    The Cauchy-Riemann conditions in polar coordinates are

    [tex]\frac{\partial u}{\partial \rho}=\frac{1}{\rho}\frac{\partial v}{\partial \theta}[/tex]
    [tex]\frac{\partial v}{\partial \rho}=\frac{-1}{\rho}\frac{\partial u}{\partial \theta}[/tex]

    And for my function I got:
    [tex]\frac{\partial u}{\partial \rho}=\frac{-1}{\rho^2}\sin\theta[/tex]
    [tex]\frac{\partial v}{\partial \theta}=\frac{1}{\rho}\sin\theta[/tex]

    [tex]\frac{\partial v}{\partial \rho}=\frac{1}{\rho^2}\cos\theta[/tex]
    [tex]\frac{\partial u}{\partial \theta}=\frac{1}{\rho}\cos\theta[/tex]

    So to acomplish Cauchy Riemann I should get:

    [tex]\frac{-1}{\rho^2}\sin\theta=\frac{\partial v}{\partial \theta}=\frac{1}{\rho^2}\sin\theta \rightarrow -\sin\theta=\sin\theta[/tex]


    [tex]\frac{1}{\rho^2}\cos\theta=\frac{-1}{\rho^2}\cos\theta\rightarrow \cos\theta=-\cos\theta[/tex]

    Then it isn't analytical over the entire complex plane, which contradicts the assumption that I've made at first, so what did I do wrong?
  2. jcsd
  3. Sep 22, 2011 #2
    You did nothing wrong. The function is indeed not analytic anywhere.
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