Find the Laurant expansion Find the Laurent expansion for the following function $$g(z) = \dfrac{z^4}{(z^2+1)^2}$$ in a region nearby $$z=i$$
Attempt: I tried to use the idea of finding the series for $$\dfrac{1}{z^2}$$ and the fact that the representation is valid for $$|z-i|<2$$ A quick check on WolframAlpha shows that the representation I arrived at is wrong. How should I go about finding the expansion? 
 A: Note that $z^2+1=(z-i)(z+i)$, so that
$$
\frac{z^4}{(z^2+1)^2}=\frac{1}{(z-i)^2}\cdot\frac{z^4}{(z+i)^2}.
$$
Now, you can write
$$\tag{1}
\frac{z^4}{(z+i)^2}=\frac{((z-i)+i)^4}{((z-i)+2i)^2}=\sum_{k=0}^{4}\binom{4}{k}\frac{(z-i)^k(2i)^{4-k}}{((z-i)+2i)^2}.
$$
Now,
$$
\frac{1}{((z-i)+2i)^2}=\frac{1}{(2i)^2}\cdot\frac{1}{\left(1+\frac{z-i}{2i}\right)^2}=-\frac{1}{4}\cdot\frac{1}{\left(1-\left(-\frac{z-i}{2i}\right)\right)^2}.
$$
But, we know that
$$
\frac{1}{(1-x)^2}=\sum_{n=0}^{\infty}(n+1)x^n,\qquad\lvert x\rvert<1,
$$
so that
$$\tag{2}
\frac{1}{((z-i)+2i)^2}=-\frac{1}{4}\sum_{n=0}^{\infty}\left(-\frac{1}{2i}\right)^n(n+1)(z-i)^n.
$$
Note that our "$x$" here is $-\frac{z-i}{2i}$; so, this series converges if
$$
\left\lvert-\frac{z-i}{2i}\right\rvert=\frac{\lvert z-i\rvert}{2}<1,\qquad\text{or}\qquad\lvert z-i\rvert<2.
$$
From here, it is a matter of combining (1) and (2) and index manipulation to find the series representation of 
$$
\frac{z^4}{(z+i)^2},
$$
then multiplying by $(z-i)^{-2}$.
