# Laurent series question

#### dwsmith

##### Well-known member
Find the Laurent series of the form $\sum\limits_{n = -\infty}^{\infty}c_nz^n$ for $f(z) = \dfrac{z^2}{(z - 1)(z - 3)}$ that converges in an annulus containing the point $z = 2$, and state precisely where this Laurent series converges.}

By the method of partial fractions, (how does the 1 effect the solution?)
$$f(z) = \frac{z^2}{(z - 1)(z - 3)} = 1 + \frac{9}{2z - 6} - \frac{1}{2z - 2}.$$
So $f(z)$ has simple poles at $z = 3$ and $z = 1$.
$f(z)$ will be analytic when
$$|z| < 1,\quad 1 < |z| < 3,\quad\text{and}\quad 3 < |z|.$$
However, $z = 2$ is only in the annulus $1 < |z| < 3$.
The series will converge when
$$\left|\frac{1}{z}\right| < 1\quad\text{and}\quad \left|\frac{z}{3}\right| < 1.$$

Continuing like so what do I do to account for the constant 1? Would I just proceed by expression each fraction of the expansion as the appropriate geometric series?

So I would have:
$$\frac{1}{2z - 2} = \frac{1}{2z}\frac{1}{1 - \frac{1}{z}} = \frac{1}{2}\sum_{n = 0}^{\infty}\left(\frac{1}{z}\right)^{n + 1}$$

Then doing the other fraction so the solution would be

$$f(z) = 1 - \frac{1}{2}\left[9\sum_{n = 1}^{\infty}\left(\frac{z}{3}\right)^n + \sum_{n = 1}^{\infty}\left(\frac{1}{z}\right)^{n}\right]$$

I know how to solve the other series that isn't the question so don't present the solution. The question is in regards to how 1 would or wouldn't affect writing f as a Laurent series. Would this be correct?

Last edited:

#### Sudharaka

##### Well-known member
MHB Math Helper
Find the Laurent series of the form $\sum\limits_{n = -\infty}^{\infty}c_nz^n$ for $f(z) = \dfrac{z^2}{(z - 1)(z - 3)}$ that converges in an annulus containing the point $z = 2$, and state precisely where this Laurent series converges.}

By the method of partial fractions, (how does the 1 effect the solution?)
$$f(z) = \frac{z^2}{(z - 1)(z - 3)} = 1 + \frac{9}{2z - 6} - \frac{1}{2z - 2}.$$
So $f(z)$ has simple poles at $z = 3$ and $z = 1$.
$f(z)$ will be analytic when
$$|z| < 1,\quad 1 < |z| < 3,\quad\text{and}\quad 3 < |z|.$$
However, $z = 2$ is only in the annulus $1 < |z| < 3$.
The series will converge when
$$\left|\frac{1}{z}\right| < 1\quad\text{and}\quad \left|\frac{z}{3}\right| < 1.$$

Continuing like so what do I do to account for the constant 1? Would I just proceed by expression each fraction of the expansion as the appropriate geometric series?

So I would have:
$$\frac{1}{2z - 2} = \frac{1}{2z}\frac{1}{1 - \frac{1}{z}} = \frac{1}{2}\sum_{n = 0}^{\infty}\left(\frac{1}{z}\right)^{n + 1}$$

Then doing the other fraction so the solution would be

$$f(z) = 1 - \frac{1}{2}\left[9\sum_{n = 1}^{\infty}\left(\frac{z}{3}\right)^n + \sum_{n = 1}^{\infty}\left(\frac{1}{z}\right)^{n}\right]$$

I know how to solve the other series that isn't the question so don't present the solution. The question is in regards to how 1 would or wouldn't affect writing f as a Laurent series. Would this be correct?
Hi dwsmith, I think the Laurent series that you have obtained contains a few errors.

$f(z) = \frac{z^2}{(z - 1)(z - 3)} = 1 + \frac{9}{2z - 6} - \frac{1}{2z - 2}$

$\Rightarrow f(z) =1-\frac{3}{2\left(1-\frac{z}{3}\right)}-\frac{1}{2z(1-\frac{1}{z})}$

$\Rightarrow f(z)=1-\frac{3}{2}\sum_{n=0}^{\infty}\left( \frac{z}{3}\right)^{n}-\frac{1}{2z}\sum_{n=0}^{ \infty}\left(\frac{1}{z}\right)^{n}\mbox{ where }1<|z|<3$

$\Rightarrow f(z)=-\frac{1}{2}-\frac{3}{2}\sum_{n=1}^{\infty}\left( \frac{z}{3}\right)^{n}-\frac{1}{2}\sum_{n=1}^{ \infty}\left(\frac{1}{z}\right)^{n}\mbox{ where }1<|z|<3$

This is the Laurent series expression for $$f$$ in the annulus $$1<|z|<3$$. Comparing the above expression with the standard form $$\sum\limits_{n = -\infty}^{\infty}c_nz^n$$ note that $$c_{0}=-\frac{1}{2}$$.

Kind Regards,
Sudharaka.