1
$\begingroup$

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 $|z+2|<3$?

$\endgroup$
  • 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. $\endgroup$ – GEdgar Mar 23 '17 at 17:54
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

$\frac {1}{1-z}$ centered at $z = -2$

$\frac {1}{3-(z+2)}\\ \frac {\frac 13}{1-\frac {(z+2)}{3}}$

$\frac 13 \sum_\limits{i=0}^{\infty} (\frac {z+2}{3})^i$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.