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Math Help - Analytical function

  1. #1
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    Analytical function

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

    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:

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

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

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

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

    So to acomplish Cauchy Riemann I should get:

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

    And

    \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
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    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
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    Re: Analytical function

    If z=x+i y then is...

    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

    \chi \sigma
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  4. #4
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    Re: Analytical function

    Thank you both
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