Question

Find two Laurent series in powers of z gor the function f defined by f(z)=1/z^2(1-z) and specify the regions in which the series converges to f(z).

Answer 1

We remind the formula for geometric progression:

$\frac{1-z^{n+1}}{1-z}=1+z+z^2+\ldots +z^n,\quad z\neq1,\quad n\in{\mathbb{N}}.$ (1)

If $|z|<1$ , we can take the limit $n\rightarrow\infty$ and obtain

$\frac{1}{1-z}=1+z+z^2+z^3+\ldots =\sum\limits_{n=0}^{+\infty}z^n.$

If we multiply (1) by $z^{-n}$ , we get

$\frac{z^{-n}-z}{1-z}=z^{-n}+z^{1-n}+z^{2-n}+\ldots +1=1+z^{-1}+z^{-2}+\ldots+z^{-n},\quad n\in{\mathbb{N}}.$

If $|z|>1$ we take the limit $n\rightarrow\infty$ and get

$\frac{z}{z-1}=1+z^{-1}+z^{-2}+z^{-3}+\ldots$

Thus, we obtain two Laurent series

$f(z)=\frac{1}{z^2(1-z)}=z^{-2}+z^{-1}+1+z+z^2+\ldots=\sum\limits_{n=-2}^{+\infty}z^n,\qquad 0<|z|<1,$

$f(z)=\frac{1}{z^2(1-z)}=-z^{-3}-z^{-4}-z^{-5}-\ldots=-\sum\limits_{n=-\infty}^{-3}z^n,\qquad |z|>1.$