Laurent series of $\frac{1}{z^2(z-1)}$ when $0<\lvert z\rvert<1$

Question

$\frac{1}{z^2(z-1)} = -\left(\frac{1}{z}+\frac{1}{z^2}+\frac{1}{1-z}\right)$.

I know that $\frac{1}{1-z}=\sum\limits_{n=0}^\infty z^n$, but what about the other two terms, should they be left as they are, since we can already think of $\frac{1}{z}$ and $\frac{1}{z^2}$ as Laurent series where $b_1 = 1$ in the first case, and the rest of $b_n=0$, and $b_2 = 1$ in the 2nd case, with all other $b_n=0$? Please let me know if there is a better way to find Laurent series.

Answer 1

Your approach is right, at the end you just add the three series to obtain the Laurent series of $\frac{1}{z^2(z-1)}$ : $$-(\sum\limits_{n=0}^\infty z^n+\frac{1}{z}+\frac{1}{z^2})=-\sum\limits_{n=-2}^\infty z^n$$

Answer 2

Your answer $$\frac{1}{z^2(z-1)} = -\left(\frac{1}{z}+\frac{1}{z^2}+\sum\limits_{n=0}^\infty z^n\right), \quad 0<\lvert z\rvert<1,$$ is fine. The first two terms on the right hand side are already the principal part of the sought Laurent series expansion.