Finding the Laurent series of $\frac{1}{z^{2}(1-z)}, 1<|z|<\infty $

Question

Finding the Laurent Series of

$\frac{1}{z^{2}(1-z)}, 1<|z|<\infty $

I tried to divide both numerator and denominator by $z^2$, so that there is a term of $\frac{1}{(1-z)}$.

I am tempted to use the formula for the sum of a geometric series, except the condition that |z|>1.

Is there an apparent way to resolve this, since there is no poles in the interval?

Thanks for your help.

Answer 1

Let $u = \frac{1}{z}$.

Then $$f(u) = \frac{u^3}{u-1}$$ which, since $|u|< 1$ can be expanded as $$ f(u) = -\sum_{n=3}^\infty u^n $$ Then $$ f(z) = -z^{-3} - z^{-4} - z^{-5} \cdots $$ which is the desired Laurent series.

Answer 2

For $1<|z|<\infty$, we have $0<\left|\frac{1}{z}\right|<1$, and so $$\frac{1}{{{z^2}\left( {1 - z} \right)}} = \frac{1} {{{z^2}}}.\frac{1} {{ - z\left( {1 - \frac{1} {z}} \right)}} = - \frac{1} {{{z^3}}}\frac{1} {{1 - \frac{1} {z}}} = - \frac{1} {{{z^3}}}\sum\limits_{n = 0}^{ + \infty } {\frac{{{{\left( {\frac{1} {z}} \right)}^n}}} {{n!}}} = - \sum\limits_{n = 0}^{ + \infty } {\frac{{{z^{ - \left( {n + 3} \right)}}}} {{n!}}}$$