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Thread: Analytical function

  1. #1
    Senior Member
    May 2010

    Analytical function

    Hi there. I have to study the analyticity for the complex function:

    $\displaystyle f(z)=\displaystyle\frac{y-ix}{x^2+y^2}$

    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:

    $\displaystyle e^{i\theta}=\cos\theta+i\sin\theta$
    $\displaystyle -ie^{i\theta}=-i \cos\theta+\sin\theta$

    $\displaystyle f(z)=\frac{y-ix}{x^2+y^2}=\frac{-\rho ie^{i\theta}}{\rho^2}=\frac{ \sin \theta -i \cos\theta}{\rho}$

    The Cauchy-Riemann conditions in polar coordinates are

    $\displaystyle \displaystyle\frac{\partial u}{\partial \rho}=\displaystyle\frac{1}{\rho}\frac{\partial v}{\partial \theta}$
    $\displaystyle \displaystyle\frac{\partial v}{\partial \rho}=\displaystyle\frac{-1}{\rho}\frac{\partial u}{\partial \theta}$

    And for my function I got:
    $\displaystyle \displaystyle\frac{\partial u}{\partial \rho}=\frac{-1}{\rho^2}\sin\theta$
    $\displaystyle \displaystyle\frac{\partial v}{\partial \theta}\displaystyle\frac{1}{\rho}\sin\theta$

    $\displaystyle \displaystyle\frac{\partial v}{\partial \rho}=\displaystyle\frac{1}{\rho^2}\cos\theta$
    $\displaystyle \displaystyle\frac{\partial u}{\partial \theta}=\displaystyle\frac{1}{\rho}\cos\theta$

    So to acomplish Cauchy Riemann I should get:

    $\displaystyle \displaystyle\frac{-1}{\rho^2}\sin\theta=\frac{\partial v}{\partial \theta}=\frac{1}{\rho^2}\sin\theta \rightarrow -\sin\theta=\sin\theta$


    $\displaystyle \displaystyle\frac{1}{\rho^2}\cos\theta=\frac{-1}{\rho^2}\cos\theta\rightarrow \cos\theta=-\cos\theta$

    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?

    PD: I didn't know where to post complex analysis question, so I'm not sure this is the propper section, if its not, please move it. Thanks.
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  2. #2
    MHF Contributor

    Apr 2005

    Re: Analytical function

    Your assumption is wrong. This is NOT "a division between polynomials" because in functions of complex variables a "polynomial" is specifically a polynomial in the variable z, not in the real and imaginary parts of z.
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  3. #3
    MHF Contributor chisigma's Avatar
    Mar 2009
    near Piacenza (Italy)

    Re: Analytical function

    If $\displaystyle z=x+i y$ then is...

    $\displaystyle f(z)= \frac{y-i x}{x^{2}+y^{2}}= \frac{1}{i\ \overline{z}}$ (1)

    ... so that f(z) isn't analytic in any point of the complex plane...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  4. #4
    Senior Member
    May 2010

    Re: Analytical function

    Thank you both
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