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[SOLVED] Laurent series 2

dwsmith

Well-known member
Feb 1, 2012
1,673
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

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
340
Both are correct.