I can't understand why my solution to this exercise is conceptually wrong... can you help me?
Laurent expansion of $$f(z)=\frac{z^2+2}{z^3-3z^2+2z}$$
in 1<|z|<2
My solution:
$$f(z)=\frac{1}{z}+\frac{3}{z-2}-\frac{3}{z-1}$$
$$f_1=\frac{1}{z}$$ $$f_2=\frac{3}{z-2}=\frac{3}{-2[1-\frac{z}{2}]}=-(3/2) \sum_{0}^{\infty}(\frac{z}{2})^n$$ $$f_3=\frac{3}{z-1}=-3\sum_0^{\infty}(z)^n$$
so
$$f(z)=\frac{1}{z}-(3/2) \sum_{0}^{\infty}(\frac{z}{2})^n+3\sum_{0}^{\infty}(z)^n$$
but the correct solution seems to be
$$f(z)=\frac{1}{z}-(3/2) \sum_{0}^{\infty}(\frac{z}{2})^n-3\sum_{0}^{\infty}\frac{1}{(z)^{n+1}}$$
It is "conceptually wrong" only in $f_3$ part. Your series expansion diverges for $|z|>1$. You need to use powers of $z^{-1}$ instead $$ f_3(z)=\frac{3}{z-1}=\frac{3}{z}\cdot\frac{1}{1-z^{-1}}=\frac{3}{z}\sum_{k=0}^\infty z^{-k}. $$