How can I find the Laurent series expansion for
$$\frac{z^2-1}{z^2+1} $$ with $|z|>2$.
$$\frac{z^2-1}{z^2+1} =\frac{z^2+1-2}{z^2+1}=1-\frac2{z^2+1}$$
The fraction on the right-hand-side can be expressed as $$\frac1{z^2+1}=\frac i{2(z+i)}+\frac{-i}{2(z-i)}$$ also $$\frac 1{z+i}=\frac1{z}\frac1{1-(-\frac i z)}=\sum_{k=-\infty}^{-1}(-1)^{-k+1}i^{-k+1}z^k $$ $$\frac1{z-1}=\frac 1 z \frac1{1-\frac i z }=\frac 1 z \sum_{k=0}^\infty (\frac i z)^k $$
And here I don't know how to combine these two series.
Many exercises concerning the computation of Laurent expansions can simply be solved by making use of the geometric series $$\sum_{k=0}^\infty x^k = \frac{1}{1-x}.$$ However, the important point is that the above identity is valid only for $x \in \mathbb{C}$ satisfying $|x| <1$. Since you are working under the condition that $|z|>2$, you thus cannot apply the identity directly, but it works if you somehow manage to replace $z$ by $1/z$.
We have $$\frac{1}{z^2+1} = \frac{1}{z^2} \cdot \frac{1}{1+\frac{1}{z^2}} = z^{-2} \cdot \frac{1}{1 - (-\frac{1}{z^2})} = z^{-2} \cdot \sum_{k=0}^\infty (-z^{-2})^k = \sum_{k=0}^\infty (-1)^k z^{-2k-2}.$$ Hence $$\frac{z^2-1}{z^2+1} = (z^2-1) \cdot \frac{1}{z^2+1} = z^2 \cdot \Big( \sum_{k=0}^\infty (-1)^k z^{-2k-2}\Big) - \Big( \sum_{k=0}^\infty (-1)^k z^{-2k-2}\Big)$$ $$= \Big( \sum_{k=0}^\infty (-1)^k z^{-2k}\Big) - \Big( \sum_{k=0}^\infty (-1)^k z^{-2k-2}\Big) = \Big( \sum_{k=-1}^\infty (-1)^{k+1} z^{-2k-2}\Big) - \Big( \sum_{k=0}^\infty (-1)^k z^{-2k-2}\Big)$$ $$= 1 + \sum_{k=0}^\infty ((-1)^{k+1} - (-1)^k) z^{-2k-2} = 1 - 2 \sum_{k=0}^\infty (-1)^k z^{-2k-2}.$$
The manipulations which I used (e.g., taking the difference of two (infinite) series) are valid because of the absolute convergence of the geometric series.