Laurent expansion problem Expand the function $$f(z)=\frac{z^2 -2z +5}{(z-2)(z^2+1)} $$ on the ring $$ 1 < |z| < 2 $$
I used partial fractions to get the following $$f(z)=\frac{1}{(z-2)} +\frac{-2}{(z^2+1)}  $$
then 
 $$ \frac{1}{z-2} = \frac{-1}{2(1-z/2)} = \frac{-1}{2} \left[1+z/2 + (z/2)^2 + (z/2)^3 +\cdots\right]
 $$  
but I'm stuck with $$ \frac{-2} {z^2+1} $$ how can I expand it? 

Hint:
$$\frac{-2}{1+z^2} = \sum_{n=0}^\infty (-(-i)^n-i^n) z^n,\quad\mid z\mid<1$$
and:
$$\frac{-2}{1+z^2} = \sum_{n=0}^\infty  (-1+z)^n (-1+i) (2^{-1-n} ((-1-i)^n+i (-1+i)^n))$$
