MHB Finding the Laurent series for $f(z)$ in an annulus containing $z = 2$

[Dustinsfl]

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?

 

[Sudharaka]

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.