# Homework Help: Analytical function

1. Sep 22, 2011

### Telemachus

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

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

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

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

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

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

So to acomplish Cauchy Riemann I should get:

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

And

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

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?

2. Sep 22, 2011

### micromass

You did nothing wrong. The function is indeed not analytic anywhere.