How to find Laurent expansion of $f(z)=\frac{1}{(z+i)z^2}$ on $0<|z-i|<1$ I think the way is doing the decomposition as follows, but I just don't know how to deal with the quadratic term:
$f(z)=\frac{1}{(z+i)z^2}=-\frac{1}{z+i}+\frac{1}{z}-\frac{i}{z^2}$
where, $-\frac{1}{z+i}=-\frac{1}{2i}\frac{1}{1+\frac{z-i}{2i}}$ and $\frac{1}{z}=\frac{1}{1+\frac{z-i}{i}}$ can be expanded easily.
But how to deal with $-\frac{i}{z^2}$?
 A: With $z-i=w$ and geometric series
$$\dfrac{1}{z}=\dfrac{1}{w+i}=\dfrac{1}{i}\dfrac{1}{1+\frac{w}{i}}=\dfrac{1}{i}\left(\sum_{n=0}^\infty(-\frac{w}{i})^n\right)$$
then it's derivative
$$\dfrac{1}{(w+i)^2}=\sum_{n=0}^\infty (n+1)\left(-\dfrac{w}{i}\right)^n$$
valid for $\Big|\dfrac{w}{i}\Big|<1$. Also
$$\dfrac{1}{z+i}=\dfrac{1}{w+2i}=\dfrac{1}{2i}\dfrac{1}{1+\frac{w}{2i}}=\dfrac{1}{2i}\left(\sum_{m=0}^\infty(-\frac{w}{2i})^m\right)$$
valid for $\Big|\dfrac{w}{2i}\Big|<1$. So
\begin{align}
\dfrac{1}{z^2(z+i)}
&= \dfrac{1}{(w+i)^2(w+2i)} \\
&= \sum_{n=0}^\infty (n+1)\left(-\dfrac{w}{i}\right)^n\dfrac{1}{2i}\left(\sum_{m=0}^\infty(-\frac{w}{2i})^m\right) \\
&= \dfrac{i}{2}\sum_{n=0}^\infty\sum_{m=0}^\infty (n+1)\left(-\dfrac{w}{i}\right)^n\left(-\frac{w}{2i}\right)^m \\
&= \dfrac{i}{2}-\dfrac54w+\cdots \\
&= \dfrac{i}{2}-\dfrac54(z-i)+\cdots 
\end{align}
valid for $|z-i|<1$.
A: Let $u = z+i$. Then $z = u - i$, so
\begin{align*}
\frac{1}{z^2} = \frac{1}{(i - u)^2} = \frac{1}{-(1 - u/i)^2} = -\frac{1}{(1 + iu)^2}
\end{align*}
Since $0 < |u| < 1$, then $0 < |iu|< 1$ as well, so this can be expanded as the derivative of a geometric series:
\begin{align*}
\frac{1}{1 - x} = \sum_{n \geq 0} x^n \implies \frac{1}{(1 - x)^2} = \frac{d}{dx} \frac{1}{1-x} = \frac{d}{dx} \sum_{n \geq 0} x^n = \sum_{n \geq 1} n x^{n-1} \, .
\end{align*}
A: The body of the title has a typo I think.
Assuming the region $0<|z+i|<1$, you can write $\frac{i}{z^2}=\frac{i}{(z+i-i)^2}=\frac{-i}{(1-\frac{z+i}{i})^2}=-i[1+2(z+i)/i+3(z+i)^2/i^2+...]$
However for $0<|z-i|<1$, The function $f(z)=\frac{1}{z^2(z+i)}$ is analytic and you can look for a Taylor's series.
