# Laurent Series Expansion of (z^2-1)^(-2)

• jjangub
In summary, the Laurent Series Expansion of (z^2-1)^(-2) is a mathematical representation that approximates the function (z^2-1)^(-2) near singular points by using both positive and negative powers of z. It differs from a Taylor Series Expansion in that it includes negative powers of z. Its purpose is to calculate the behavior of the function near singular points and to find its residues. The expansion is calculated using partial fractions and the binomial theorem, and can be used to approximate the values of the function at singular points, although convergence may not always occur.
jjangub

## Homework Statement

1) Find the Laurent series for (z^2)*cos(1/3z) in the region $$\left|z\right|$$
2) Find the Laurent series expansion of (z^2 - 1)^(-2) valid in the following region
a) 0 < $$\left|z - 1\right|$$ < 2
b) $$\left|z + 1\right|$$ > 2

## The Attempt at a Solution

I did all the work and got the answer, so I basically just want to check that I got it right.
It is pretty hard to write all the steps that I did, so I will briefly write it down.

1) Since z takes infinitely many inverse within the range, we use w = 1/z
then (z^2)*cos(1/3z) becomes cos(w/3)/(w^2).
i) we find the taylor series of cos(w/3),
1 - (w^2)/((3^2)*(2!)) + (w^4)/((3^4)*(4!)) - (w^6)/((3^6)*(6!)) + ...

ii) multiply the series in i) by 1/(w^2),
1/(w^2) - 1/((3^2)*(2!)) + (w^2)/((3^4)*(4!)) - (w^4)/((3^6)*(6!)) + ...

iii) since the series starts at third term, we can just simplify first and second term, and find series,
-1/18 + z^2 + $$\sum$$( n=2 to infinity) ((-1)^n)*(w^n)) / ((3^2n)*(2n!))

iiii) now sub back w = 1/z, then the Laurent Series is,
f(z) = -1/18 + z^2 + $$\sum$$( n=2 to infinity) ((-1)^n) / ((z^n)*(3^2n)*(2n!))

2)a) since z converges everywhere in this range, the Laurent Series is simply
f(z) = 1 / (4*(z+1)) + 1 / (4*(z+1)^2) - 1 / (4*(z - 1)) + 1 / (4*(z-1)^2)

b) Since z takes infinitely many inverse within the range, I did same way as 1).
let w = 1/z and find the taylor series of each term in a) (4 total) and sub back w = 1/z at the last step. After all this, I got,

f(z) = $$\sum$$(n=1 to infinity) ((-1)^(n+1)) / (4*z^n) + $$\sum$$(n=2 to infinity) ((n-1)*(-1)^n) / (4*z^n) - $$\sum$$(n=1 to infinity) (1/(4*z^n)) + $$\sum$$(n=2 to infinity) (n-1) / (4*z^n)

Sorry about the mass...I tried to use the code and symbol, but I just failed to do like other people...

Please tell me if I did something wrong.
Thank you.

Your solution for both parts looks correct to me. Your approach for finding the Laurent series in the given regions is sound and your final answers match with what I have obtained as well. Good job! Keep up the good work.

## 1. What is the Laurent Series Expansion of (z^2-1)^(-2)?

The Laurent Series Expansion of (z^2-1)^(-2) is a mathematical representation of the function (z^2-1)^(-2) in terms of powers of z. It is a series that consists of both positive and negative powers of z, and is used to approximate the function near singular points (such as z=1 or z=-1) where the function is not defined.

## 2. How is the Laurent Series Expansion of (z^2-1)^(-2) different from a Taylor Series Expansion?

A Taylor Series Expansion is a type of power series that represents a function as a sum of powers of z, where all the powers are positive. On the other hand, a Laurent Series Expansion includes both positive and negative powers of z, making it a more general representation that can approximate functions near singular points.

## 3. What is the purpose of finding the Laurent Series Expansion of (z^2-1)^(-2)?

The Laurent Series Expansion of (z^2-1)^(-2) is useful for calculating the behavior of the function near singular points, as well as for finding the residues of the function at those points. It can also be used to approximate the function in a region where it is otherwise difficult to evaluate.

## 4. How is the Laurent Series Expansion of (z^2-1)^(-2) calculated?

The Laurent Series Expansion of (z^2-1)^(-2) can be calculated using the method of partial fractions, where the function is broken down into simpler terms. Then, the coefficients of the resulting series can be found by expanding each term using the binomial theorem and comparing it to the original function.

## 5. Can the Laurent Series Expansion of (z^2-1)^(-2) be used to find the values of the function at singular points?

Yes, the Laurent Series Expansion of (z^2-1)^(-2) can be used to find the values of the function at singular points. By substituting a singular point into the series, we can obtain an approximation of the function's value at that point. However, it is important to note that the series may not converge at the singular point, in which case the approximation may not be accurate.

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