Find Laurent Series for $\frac{1}{z^2(1 - z)}$ in $0 < |z| < 1$ & $|z| > 1$

[Dustinsfl]

Find the Laurent series for $1/z^2(1 - z)$ in the regions

$0 < |z| < 1$
$$
\frac{1}{z^2(1 - z)} = \frac{1}{z^2}\frac{1}{1-z}
$$
Since $|z| < 1$, the geometric series will converge so
$$
\frac{1}{z^2}\sum_{n = 0}^{\infty}z^n = \sum_{n = -2}^{\infty}z^n.
$$$|z| > 1$
The geometric series will converge when $\left|\dfrac{1}{z}\right| < 1$.
So
$$
\frac{1}{z^2}\frac{1}{1-z} = \frac{-1}{z^3}\frac{1}{1 - \frac{1}{z}} = \frac{-1}{z^3}\sum_{n = 0}^{\infty}\left(\frac{1}{z}\right)^n = -\sum_{n = 3}^{\infty}\left(\frac{1}{z}\right)^n.
$$

 

[Fantini]

Both are correct.

 

[math771]

Thank you for sharing this solution! It is very clear and helpful. I also noticed that the series for the region $|z| > 1$ has a negative sign in front, which makes sense since the function has a pole at $z = 0$. This means that the series will have a negative power term. Great job!