Laurent Series and Taylor Series

Question

I am trying to find the Laurent series of $\dfrac{1}{(1+x)^3}$; would this be the same as finding the Maclaurin series for the same function?

Answer 1

It is easy to verify this function $f(z)=\dfrac{1}{(z+1)^3}$ has and only has one pole of order 3 at $z=-1$, if one wants to expand it about the pole $z=-1$, two formulas should be took into account.

1. the coefficient formula of Laurent Seriers $$ c_n=\frac{1}{2\pi i}\int_\gamma \dfrac{f(z)\mathrm{d}z}{(z+1)^{n+1}}$$
2. The well-known formula in complex analysis textbook: $$I=\oint_c\frac{1}{(z-z_0)^{m+1}}\mathrm{d}z=\left\{ \begin{array}{cc} 2\pi i,& \;\;m=0\\ 0, &\;\;m\ne 0 \end{array}\right.$$

You can get that only the coefficient of $c_{-3}\ne 0, c_{-3}=1 $, all others coefficients are zero, noting $\mathrm{res}f(-1)=c_{-1}=0$.

Therfore the Laurent Series of this fucntion is itself. $$ f(z)= (z+1)^{-3} $$

If you expand it about $ z \ne -1$, the Laurent series is the same as Taylor series because of the property of ordinary points.