# Thread: Analytical function

1. ## 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:

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

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

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:

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

And

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

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.

2. ## 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.

3. ## 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

$\displaystyle \chi$ $\displaystyle \sigma$

4. ## Re: Analytical function

Thank you both