- #1
Telemachus
- 820
- 30
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}=<br /> \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?
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}=<br /> \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?