Laurent series of $z^{-3}$ at $z_0 = i$. Is there a way to do this by hand or is the question just evil? I have to find the two Laurent series expansions of $\frac{1}{z^3}$ about $i$. The only approach I can think of is to do:
$$\frac{1}{z^3} = \frac{1}{(z-i)^3} \left( \frac{z-i}{z} \right) ^3 = \frac{1}{(z-i)^3} \left( 1 - \frac{i}{i+(z-i)} \right) ^3 = \frac{1}{(z-i)^3} \left( 1 - \frac{1}{1-i(z-i)} \right) ^3 $$
This expansion will work in the disk $|z-i|<1:$
$$=\frac{1}{(z-i)^3} \left( 1 - \sum_{n=0}^{\infty}i^n(z-i)^n \right) ^3  =i \left(  \sum_{n=0}^{\infty}i^{n}(z-i)^{n} \right) ^3 .$$
But this is nasty, because I don't know how to cube the series! And I don't think anything nice would come out of it, anyway.
The second expansion is in the annulus $|z-i|>1:$
$$\frac{1}{z^3} = ... =  \frac{1}{(z-i)^3} \left( 1 - \frac{1}{z-i}\frac{1}{ \frac{1}{z-i}-i} \right) ^3 = \frac{1}{(z-i)^3} \left( 1 - \frac{1}{z-i}\frac{i}{ 1-\frac{-i}{z-i}} \right) ^3 =$$
$$=  \left( \frac{1}{z-i} - i \sum_{n=0}^{\infty} \frac{(-i)^n}{(z-i)^{n-2}} \right) ^3$$
Same problem, except now it's even worse... How do I deal with this?
 A: I'm sure you have been taught about this before. $\frac{1}{z^3}$ is regular at 
$z = i$, so the Laurent series there is simply the ordinary Taylor series expansion (in the vicinity of $i$).
For any $\alpha > 0$ and $a \in \mathbb{C}$ not on the negative real axis, we have
$$\begin{align}
\frac{1}{z^\alpha} 
= & \sum_{k=0}^{\infty} \frac{1}{k!} \left( \left.\frac{d^k}{dz^k} \frac{1}{z^{\alpha}}\right|_{z=a}\right) (z-a)^k\\
= & \sum_{k=0}^{\infty} \frac{1}{k!} \left( \frac{(-\alpha)(-\alpha-1)\cdots(-\alpha-k+1)}{a^{\alpha+k}}\right) (z-a)^k\\
= & \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \left( \frac{\alpha(\alpha+1)\cdots(\alpha+k-1)}{a^{\alpha+k}}\right) (z-a)^k\\
\end{align}$$
When $\alpha$ is a positive integer $n$, this reduces to
$$
\frac{1}{z^{n}} 
= \sum_{k=0}^{\infty} (-1)^k \binom{n+k-1}{k}   \frac{(z-a)^k}{a^{n+k}}
= \sum_{k=0}^{\infty} (-1)^k \binom{n+k-1}{n-1} \frac{(z-a)^k}{a^{n+k}}
\tag{*1}$$
In particular, when $n = 3$ and $a = i$, one has
$$\frac{1}{z^3} = \sum_{k=0}^{\infty} (-1)^k \binom{k+2}{2} \frac{(z-i)^k}{i^{3+k}}
=\sum_{k=0}^{\infty} i^{k+1}\frac{(k+1)(k+2)}{2} (z-i)^k
\tag{*2}$$
Update
To derive the other Laurent series (i.e, the one in the vicinity of $\infty$),
one can let $u = \frac{1}{z-i}$ and rewrite 
$$\frac{1}{z^3} \quad\text{ as }\quad\frac{1}{(i + \frac{1}{u})^3} = \frac{i u^3}{(u-i)^3}.$$
One then consider the Taylor series expansion of it at $u = 0$. Compare this with $(*1)$ and $(*2)$, one find $u$ is now taking the role of $z-i$ and $-i$ is taking the role of $a$ there. As a result,
$$\frac{1}{z^3} = i u^3 \sum_{k=0}^{\infty}(-1)^k\binom{k+2}{k} \frac{u^k}{(-i)^{3+k}}
= \sum_{k=0}^{\infty}(-i)^k \frac{(k+1)(k+2)}{2}\frac{1}{(z-i)^{k+3}}
$$
A: Here is an approach
$$ \frac{1}{z^3}= \frac{1}{((z-i)+i)^3}=\frac{1}{(z-i)^3}\frac{1}{\left( 1+\frac{i}{(z-i)}\right)^3} $$
$$ \frac{1}{(z-i)^3} \sum_{k=0}^{\infty} {-3\choose k}\frac{i^{3k}}{(z-i)^{3k}},$$
by the Newton's generalized binomial theorem. Then we have

$$\frac{1}{z^3} = \sum_{k=0}^{\infty} {-3\choose k}\frac{i^{3k}}{(z-i)^{3k+3}}\quad |z-i|>1. $$

