Finding the complete Taylor expansion of $\frac{1}{1+z^2}$ around $z=0$. For an exercise, I need to find the complete Taylor expansion for $(1+z^2)^{-1}$ around $z=0$. I have tried decomposing first $(1+z^2)^{-1}$ into partial fractions. Since $1+z^2=0$ gives $z=\pm i$, the partial fractions are:
$$\frac{1}{1+z^2} = \frac{1}{(z+i)(z-i)} = \frac{i}{2(z+i)} - \frac{i}{2(z-i)} = \frac{i}{2} \Big{(}\frac{1}{z+i} - \frac{1}{z-i}\Big{)}$$
Therefore, my idea was to find the much easier Taylor expansion of both fractions, and add them together. I have worked out each Taylor expansion, since the $n$-th derivative of each of the fractions can be easily found:
$$\frac{1}{z+i}=\sum_{n=0}^\infty -i^{n+1}z^n, \hspace{25px} \frac{1}{z-i}=\sum_{n=0}^\infty \frac{-z^n}{i^{n+1}}$$
Typing each sum in WolframAlpha returns the original fraction, so I think they are okay. Now, I replace both sums into the partial fractions:
$$\frac{1}{1+z^2} = \frac{i}{2} \Big{(}\sum_{n=0}^\infty -i^{n+1}z^n - \sum_{n=0}^\infty \frac{-z^n}{i^{n+1}}\Big{)}$$
But I always get a sum that does not return the correct Taylor expansion for $1/(1+z^2)$. Where am I getting this wrong?

Your approach is all correct. Note that your sum can be rewritten as $$\frac{1}{2}\sum_{n=0}^\infty i^n ((-1)^n+1)z^n.$$
For an odd index value $n$, the term equals zero, and for even - it equals $2(-1)^kz^{2k}$, $k\in \mathbb{Z}$.
Therefore
$$\frac{1}{2}\sum_{n=0}^\infty i^n ((-1)^n+1)z^n=\frac{1}{2} ( 2-2z^2+2z^4-2z^6+\ldots)=1-z^2+z^4-z^6+\ldots$$ 
