Laurent series for $\frac{1}{(z - 1)(z + 1)}$
by Hidenori
Proposition
Find a Laurent series for
\[\begin{align*} \frac{1}{(z - 1)(z + 1)} \end{align*}\]centered at $z = 1$ and specify the region in which it converges.
Solution
\[\begin{align*} \frac{1}{(z - 1)(z + 1)} &= \frac{1}{(z - 1)(z - 1 + 2)} \\ &= \frac{1}{(z - 1)^2 + 2(z - 1)} \\ &= \frac{1}{2(z - 1)}\frac{1}{\frac{z - 1}{2} + 1} \\ &= \frac{1}{2(z - 1)}\frac{1}{1 - \frac{z - 1}{2}} \\ &= \frac{1}{2(z - 1)}\sum_{k=0}^{\infty} \Big(\frac{1 - z}{2}\Big)^k \\ &= \frac{1}{4}\sum_{k=0}^{\infty} \Big(\frac{1 - z}{2}\Big)^{k - 1} \\ &= \frac{1}{4}\sum_{k=-1}^{\infty} \Big(\frac{1 - z}{2}\Big)^k \\ &= \frac{1}{4}\sum_{k=-1}^{\infty} \frac{1}{2^k}(1 - z)^k \\ &= \sum_{k=-1}^{\infty} \frac{1}{2^{k + 2}}(1 - z)^k \\ &= \sum_{k=-1}^{\infty} \frac{1}{(-2)^{k + 2}}(z - 1)^k. \end{align*}\]From calculus, we know that a geometric series converges if and only if the common ratio is less than 1. Thus this converges if and only if $\abs{1 - z} < \abs{-2} = 2$. In other words, the region in which it converges is the interior of the disk with center at 1 and radius 2.
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