Laurent expansion problem

Question

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?

Answer 1

Hint: Find $A,B$ such that $$\frac{-2}{z^2+1}=\frac{A}{z+i}+\frac{B}{z-i},$$ then proceed in a similar fashion to what you did with the term $\frac1{z-2}.$ Alternately, note that $1<|z|$ if and only if $1<|z|^2=|z^2|$ if and only if $|\frac1{z^2}|<1,$ and just work with $\frac{-2}{z^2+1}.$

Answer 2

So, on $|z|>1$ we have $$\frac1{z^2+1}=\frac1{z^2}\frac1{1+\frac1{z^2}}=\frac1{z^2}\left(1-\frac1{z^2}+\frac1{z^4}-\dots\right)\,.$$

Answer 3

Hint:

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

and:

$$\frac{-2}{1+z^2} = \sum_{n=0}^\infty (-1+z)^n (-1+i) (2^{-1-n} ((-1-i)^n+i (-1+i)^n))$$