# Laurent Series (non-homework) question

• Void123
In summary: Similarly, if we set n = 1, we get a_{1} = \frac{cosh z}{z^{2}}|_{z=0} = \frac{1}{2}. So the Laurent expansion for \frac{cosh z}{z^{2}} is \frac{1}{6} + \frac{1}{2}z + \sum_{n=2}^{\infty} a_{n} z^{n}. In summary, to find the Laurent expansion for \frac{z + 1}{z^{2} - z - 6}, we use partial fraction decomposition to rewrite it as \frac{-1/5}{z - 2} + \frac
Void123

## Homework Statement

I am trying to understand Laurent series from the few and limited examples given by my texts.

I understand the basic idea, which is to expand the series about a function's pole.

So how would one go about finding the Laurent expansion for $$\frac{z + 1}{z^{2} - z - 6}$$?

Or $$\frac{cosh z}{z^{2}}$$?

## Homework Equations

$$f(z) = \sum a_{n} (z - z_{0})^{n}$$

## The Attempt at a Solution

Pretty straightforward equation. Except, could someone explain the procedure of calculating $$a_{n}$$

Thanks.

For the first equation, we can write it as \frac{z + 1}{(z - 2)(z + 3)}. Then we can use partial fraction decomposition to write it as \frac{A}{z - 2} + \frac{B}{z + 3}. Then we can solve for A and B by setting z = 2 and z = -3. This gives us A = -1/5 and B = 3/5. Now we can rewrite the equation as \frac{-1/5}{z - 2} + \frac{3/5}{z + 3}. The Laurent expansion of this equation is\frac{-1/5}{z - 2} + \frac{3/5}{z + 3} = -\frac{1}{5}\sum_{n=0}^{\infty} (z - 2)^{n} + \frac{3}{5}\sum_{n=0}^{\infty} (-1)^{n} (z + 3)^{n}.For the second equation, we can use the formula f(z) = \sum a_{n} (z - z_{0})^{n}, where z_{0} is the pole of the function. In this case, the pole is at z = 0, so we can write the equation as f(z) = \sum a_{n} z^{n}. We can find a_{n} by taking the derivative of both sides of the equation with respect to z. This gives us f'(z) = \sum na_{n} z^{n-1}. We can then equate this to the derivative of our original equation, which is f'(z) = \frac{d}{dz}\left(\frac{cosh z}{z^{2}}\right) = -\frac{2sinh z}{z^{3}}. So now we have two equations that are equal to each other, one on the left and one on the right. We can then solve for the coefficients a_{n}. For example, if we set n = 0, then we get a_{0} = \frac{sinh z}{z^{3}}|_{z=0} = \frac{1

## 1. What is a Laurent series?

A Laurent series is a representation of a complex function as an infinite sum of powers of z, including negative powers. It is similar to a Taylor series, but can be used to represent functions with singularities or poles.

## 2. How is a Laurent series different from a Taylor series?

A Taylor series is a representation of a function as an infinite sum of powers of z, but only includes non-negative powers. A Laurent series, on the other hand, includes both positive and negative powers of z, allowing for the representation of functions with singularities or poles.

## 3. What are singularities and poles?

Singularities are points in the complex plane where a function is not defined or is not analytic (i.e. does not have a well-defined derivative). Poles are a type of singularity where the function approaches infinity at that point.

## 4. How do I find the Laurent series for a given function?

To find the Laurent series for a given function, you can use various techniques such as partial fraction decomposition, substitution, or using known series expansions. It is also helpful to have a good understanding of the function's singularities and their type.

## 5. Can a function have more than one Laurent series representation?

Yes, a function can have more than one Laurent series representation. This is because the coefficients in the series depend on the chosen point of expansion, and different points may result in different series. However, all these representations will converge to the same function in their respective domains of convergence.

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