Finding the Laurent series and residue of a function

[scamuicune]

Homework Statement

Find the Laurent series for the given function about the specified point. Also, give the residue of the function at the point.

$$ \frac{z^2}{z^2 - 1}, z_0 = 1 $$

Homework Equations

A Laurent expansion is comparable to a power series, except that it includes negative powers as well:
$$ \sum_{n = - \infty}^\infty a_k (z - z_0)^n $$
Where $a_k$ are the coefficients of the expansion.

The residue of f at $z_0$ is the coefficient of $(z-z_0)^{-1}$ in the Laurent expansion.

The Attempt at a Solution

I was able to find the residue with moderate success by simply expanding the fraction:
$$ \frac{z^2}{z^2 - 1} = \frac{z^2 - 1 + 1}{z^2 -1} = \frac{(z+1)(z-1) + 1}{(z+1)(z-1)}$$

Afterwards I simplified this fraction by splitting it and then using partial fraction decomposition to give me:
$$ -\frac{1}{2(z+1)} + \frac{1}{2(z-1)} + 1 $$
Since we are finding the residue at $z_0 = 1$ of this function, then we look at the term that has $(z - 1)$ in its denominator, which would be $\frac{1}{2}$.

The next part would be to find the Laurent expansion of this equation, to which neither I nor my classmates had much luck. One example in the textbook suggested that I expand the numerator in powers of (z-1), which would also be something I don't understand how to do.

 

[Dick]

Homework Statement

Find the Laurent series for the given function about the specified point. Also, give the residue of the function at the point.

$$ \frac{z^2}{z^2 - 1}, z_0 = 1 $$

Homework Equations

A Laurent expansion is comparable to a power series, except that it includes negative powers as well:
$$ \sum_{n = - \infty}^\infty a_k (z - z_0)^n $$
Where $a_k$ are the coefficients of the expansion.

The residue of f at $z_0$ is the coefficient of $(z-z_0)^{-1}$ in the Laurent expansion.

The Attempt at a Solution

I was able to find the residue with moderate success by simply expanding the fraction:
$$ \frac{z^2}{z^2 - 1} = \frac{z^2 - 1 + 1}{z^2 -1} = \frac{(z+1)(z-1) + 1}{(z+1)(z-1)}$$

Afterwards I simplified this fraction by splitting it and then using partial fraction decomposition to give me:
$$ -\frac{1}{2(z+1)} + \frac{1}{2(z-1)} + 1 $$
Since we are finding the residue at $z_0 = 1$ of this function, then we look at the term that has $(z - 1)$ in its denominator, which would be $\frac{1}{2}$.

The next part would be to find the Laurent expansion of this equation, to which neither I nor my classmates had much luck. One example in the textbook suggested that I expand the numerator in powers of (z-1), which would also be something I don't understand how to do.

If you want to expand in powers of (z-1) write u=(z-1). Then z=u+1. Substitute that into the function so you can write it totally in terms of u. Then try and expand around u=0.