Analyzing Analyticity of Complex Function

In summary, the complex function f(z) = (y-ix)/(x^2+y^2) is being studied, and the use of polar coordinates is suggested for the exercise. The Cauchy-Riemann conditions are used to show that the function is not analytic over the entire complex plane except at the point zero, due to a division between polynomials where the denominator is zero. Despite initial assumptions, the function is not analytic anywhere.
  • #1
Telemachus
835
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Hi there. I have to study the analyticity for the complex function:

[tex]f(z)=\frac{y-ix}{x^2+y^2}[/tex]

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]e^{i\theta}=\cos\theta+i\sin\theta[/tex]
Then
[tex]-ie^{i\theta}=-i \cos\theta+\sin\theta[/tex]

And
[tex]f(z)=\frac{y-ix}{x^2+y^2}=
\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]

And

[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?
 
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  • #2
You did nothing wrong. The function is indeed not analytic anywhere.
 

1. What is analyticity of a complex function?

Analyticity of a complex function refers to the property of a function being differentiable at every point within a given domain. This means that the function has a well-defined slope or derivative at every point in its domain.

2. How is analyticity of a complex function determined?

The analyticity of a complex function is determined by checking if the function satisfies the Cauchy-Riemann equations, which are a set of necessary and sufficient conditions for complex differentiability. These equations relate the partial derivatives of the real and imaginary components of the function.

3. What is the significance of analyticity in complex analysis?

Analyticity is a fundamental concept in complex analysis and it allows us to use powerful tools and techniques to study complex functions. It also enables us to express these functions in terms of simpler functions, such as power series, making it easier to analyze and manipulate them.

4. Can a function be analytic at some points and non-analytic at others?

Yes, it is possible for a function to be analytic at some points and non-analytic at others. For example, the absolute value function is analytic everywhere except at the point where it has a sharp turn, which is not differentiable.

5. How is the analyticity of a complex function related to its continuity?

Analyticity is a stronger condition than continuity. A function can be continuous at a point but not differentiable, whereas a function cannot be analytic at a point if it is not continuous at that point. In other words, analyticity implies continuity but not vice versa.

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