I am having trouble understanding how to find a Laurent series. $$\sum\limits_{n=-\infty}^{\infty} a_{n}(z-c)^{n}$$ where $a_{n}=\frac{1}{2 \pi i} \int_{C} \frac{f(w)}{(w-c)^{n+1}}dw$, and $C$ is any circle with center $c$.

I am currently trying to solve the following problem:

Find Laurent series, about the given point $z=-2$, of $$\frac{z}{(z+1)(z+2)}$$ Hint: let $z+2=u$

I am trying to find help starting this problem and understanding Laurent series.


1 Answer 1


Following the hint yields $$\frac{u-2}{(u-1)u}=\frac{2}{u}+\frac{1}{1-u}=2u^{-1}+\sum_{n=0}^\infty u^n.$$ Now substitute $u=z+2$ to obtain $$\frac{z}{(z+1)(z+2)}=2(z+2)^{-1}+\sum_{n=0}^\infty (z+2)^n.$$

  • $\begingroup$ So how did you do that once you substituted u? $\endgroup$
    – Kim
    Commented Mar 16, 2021 at 3:26
  • 1
    $\begingroup$ The first step is partial fraction decomposition. The second step is geometric series. $\endgroup$
    – RobPratt
    Commented Mar 16, 2021 at 3:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .