How to find this Laurent series?

[Noone1982]

I understand perfectly well how to do Taylor series, but I am foggy on these Laurent series. Say, we have something like,

$$f\left( z \right)\; =\; \frac{1}{z^{2}\cdot \sin \left( z \right)}$$

I think I need to use the taylor series expressions for sin(z) but otherwise, I am not sure what to do about that z^2. If I use that in a taylor series with z=0, then I get a singularity.

Since its 1/sin(z), do I just inverse the taylor series for sin(z) Yes, you can see I am very unknowledgeable about this. I turn here because the explanations I found in my book are totally lacking.

 

[AiRAVATA]

It's clear that z=0 is a pole for your function, that's why the Taylor series is not well defined there. The Laurent series will tell you how this singularity behaves (how it grows to infinity).

A Laurent series has the form

$$\sum_{n=1}^\infty \frac{b_n}{z^n}+\sum_{n=0}^\infty a_nz^n,$$

so you are right, in order to calculate the series, you just have write down the series for $\csc z$ and multiply it by $z^{-2}$.

 

[tacman]

To find the Laurent series for a function, we need to first understand the concept of singularities. A singularity is a point in the complex plane where the function is not defined or becomes infinite. In the given function f(z), the point z=0 is a singularity because the denominator becomes 0, making the function undefined.

To find the Laurent series, we need to expand the function in powers of (z-a), where a is the singularity. In this case, a=0. So, we can write the function as:

f(z) = 1/(z^2 * sin(z)) = 1/z^2 * 1/sin(z)

Now, we can expand 1/sin(z) using the Taylor series for sin(z) around z=0. This gives us:

1/sin(z) = 1/z - z/3! + z^3/5! - z^5/7! + ...

Substituting this in the original function, we get:

f(z) = 1/z^2 * (1/z - z/3! + z^3/5! - z^5/7! + ...)

= 1/z^3 - z/3!z^2 + z^3/5!z^2 - z^5/7!z^2 + ...

= 1/z^3 - 1/6z + z/120 - z^3/5040 + ...

This is the Laurent series for the given function f(z). We can see that there is a pole (singularity) at z=0, which is reflected in the negative powers of z in the series. The coefficients of the series can be found by using the formula:

c_n = 1/(2πi) * ∮f(z)(z-a)^n dz

where n is the power of (z-a) and the integral is taken around a small circle centered at the singularity a.

In summary, to find the Laurent series for a function, we need to first identify the singularities and then expand the function in powers of (z-a), where a is the singularity. We can use the Taylor series for any sub-functions that appear in the original function. With practice, you will become more comfortable with finding the Laurent series for different types of functions.