1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook