I had an exam question that has been troubling me. It looks simple, but I cannot seem to be able to figure this out.
How can I find the Laurentian series expansion of $\frac1{1-z}$ in the region $|z+2|<3$?
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1$\begingroup$ If a series in powers of $z+2$ confuses you, change variables to $w=z+2$ do it in powers of $w$. Then at the end change back. $\endgroup$– GEdgarMar 23, 2017 at 17:54
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2 Answers
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Well, try the following:
\begin{align*} \frac{1}{1-z} &= \frac{1}{3 - (z+2)} \\ &= \frac{1}{3} \frac{1}{1- (z+2)/3} \\ &= \frac{1}{3} \sum_{n=0}^\infty \left(\frac{z+2}{3} \right)^n \end{align*}
This is a standard trick by the way, in the end you only need to rewrite it in order to make use of the geometric series.
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$\frac {1}{1-z}$ centered at $z = -2$
$\frac {1}{3-(z+2)}\\ \frac {\frac 13}{1-\frac {(z+2)}{3}}$
$\frac 13 \sum_\limits{i=0}^{\infty} (\frac {z+2}{3})^i$
