Laurent series for $\frac{1}{z(z - 2)^2}$
by Hidenori
Proposition
Find a Laurent series for
\[\begin{align*} \frac{1}{z(z - 2)^2} \end{align*}\]centered at $z = 2$ and specify the region in which it converges.
Solution
\[\begin{align*} \frac{1}{z(z - 2)^2} &= \frac{1}{(z - 2 + 2)(z - 2)^2} \\ &= \frac{1}{(z - 2)^3 + 2(z - 2)^2} \\ &= \frac{1}{2(z - 2)^2}\frac{1}{1 + \frac{z - 2}{2}} \\ &= \frac{1}{2(z - 2)^2}\sum_{i=0}^{\infty} \Big(\frac{z - 2}{-2}\Big)^i \\ &= \frac{1}{8}\sum_{i=0}^{\infty} \Big(\frac{z - 2}{-2}\Big)^{i - 2} \\ &= \sum_{i=0}^{\infty} \frac{-1}{(-2)^{i + 3}} (z - 2)^i. \end{align*}\]Since a geometric series converges if and only if the absolute value of the common ratio is less than 1, the series converges if and only if $\abs{z - 2} < 2$.
Subscribe via RSS