The problem is:
Prove: There exist constants $a$, $b$ such that $\frac{z^3-5z^2+3z}{(z+2)^3}=1+\frac{a}{z}+\frac{b}{z^2}+O(\frac{1}{z^3})$ as $z\rightarrow \infty$ and find an explicit values for $a$ and $b$.
My thought: we know the power series of $\frac{1}{1-z}$ so by taking derivative we can get the power series of $\frac{1}{(1-z)^3}=\sum_{n=0}^{\infty }\frac{(n+1)(n+2)z^n}{2}$. Can we adapt this to get a power series for $(2+z)^{-3}$? Or: I have tried to divide $z^3$ both at the numerator and denominator of $\frac{z^3-5z^2+3z}{(z+2)^2}$, and get $\frac{1-\frac{5}{z}+\frac{3}{z^3}}{(1+\frac{2}{z})^3}$, so we can use the power series for $(1+\frac{2}{z})^{-3}$, which is $\sum_{n=0}^{\infty }\frac{(n+1)(n+2)(-2z)^n}{2}$, but how to continue?
Thanks in advance.
Take the series for $(1+\frac{2}{z})^{-3}$, write out the terms up to $(1/z)^3$, multiply each of the terms by $1-5/z+3/z^3$ and collect like terms (for the terms after $(1/z)^3$, note that if you hit them with a 3rd degree polynomial in $1/z$, their contributions are at most $(1/z)^4$ (more precisely, if you hit $(1/z)^n$ with $1-5/z+3/z^3$, you get a $(1/z)^{n+3}, (1/z)^{n+2},(1/z)^{n)}$ term, so for $n\geq 4$, the largest term is $(1/z)^n$ which is $O(1/z^3)$), so you only need to worry about coefficients up to $(1/z)^3$ for doing the multiplication (the rest will be captured by the $O$)).