If $\frac{z^4+z^3+z^2}{(z-2)^3} = \sum_{n=-\infty}^{n=\infty} a_n(z-2)^n$, how can I find $a_{-2}$? If $\sum_{n=-\infty}^{n=\infty} a_n(z-2)^n$  is  the laurent series  of the function $f(z)=\frac{z^4+z^3+z^2}{(z-2)^3}$  for  $z \in \mathbb{C} \setminus\{2\}$ ,then find $a_{-2}   ?$
My attempt :By  Taylor  theorem, we have $f(z)=\sum_{-\infty}^{\infty}a_n(z-z_0)^n$  where $$a_n= \frac{1}{2\pi i}\int_C \frac{f(z)}{(z-z_0)^{n+1}} dz$$
$$\implies a_{-2}= \frac{1}{2\pi i}\int_C \frac{f(z)}{(z-2)^{{-2}+1}} dz=\frac{1}{2\pi i}\int_C \frac{f(z)}{(z-2)^{-1}} dz$$
$$=\frac{1}{2\pi i}\int_C \frac{\frac{z^4+z^3+z^2}{(z-2)^3}}{(z-2)^{-1}} dz$$
$$=\frac{1}{2\pi i}\int_C \frac{z^4+z^3+z^2}{(z-2)^2} dz$$
$$=\frac{1}{2\pi i} 2\pi i({4z^3+3z^2+2z})|_{z=2} =48$$
Therefore $a_{-2}=48$

Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series, we obtain by applying the binomial theorem

\begin{align*}
\color{blue}{[(z-2)^{-2}]}&\color{blue}{\frac{z^4+z^3+z^2}{(z-2)^3}}\\
&=[(z-2)^{-2}]\frac{((z-2)+2)^4+((z-2)+2)^3+((z-2)+2)^2}{(z-2)^3}\\
&=\binom{4}{1}2^3+\binom{3}{1}2^2+\binom{2}{1}2^1\\
&=32+12+4\\
&\,\,\color{blue}{=48}
\end{align*}

