Develop into Laurent series around $-i$: $\frac{z +1}{(z-i)^2} .$ Develop into Laurent series around $-i$:
$$\frac{z +1}{(z-i)^2} .$$
I don't even know how to start. Any hint helps! Thanks.
 A: Since it's supposed to be around $z={-i}$, it means that it needs to have form
$$ \sum_{n=-\infty}^\infty c_n (z-(-i))^n = \sum_{n=-\infty}^\infty c_n (z+i)^n $$
Let's then start with writing the function in the form using $z+i$ expressions:
 $$ \frac{z+1}{(z-i)^2}= \frac{(z+i)+(1-i)}{((z+i)-2i)^2}$$
Let $w=z+i$, $a=1-i$, $b=2i$. We need to find the expansion $$ \frac{w+a}{(w-b)^2} = \sum_{n=-\infty}^\infty c_n w^n$$ 
We'll do it by calculating the Laurent series of $\frac{1}{(w-b)^2}$ first and then multiplying it by $w+a$. There will be two expansions, one for $|w|<|b|$ and another for $|w|>|b|$.
For $|w|<|b|$ we have (from the formula on the sum of a geometric series)
$$\frac{1}{b-w} = \frac{1}{b} \frac{1}{1-\frac w b}= \frac1 b \sum_{n=0}^\infty\left( \frac{w}{b}\right)^n = \sum_{n=0}^\infty \frac{w^n}{b^{n+1}}$$
We then get $$ \frac{1}{(b-w)^2} = \frac{d}{dw} \frac{1}{b-w} = \frac{d}{dw} \sum_{n=0}^\infty \frac{w^n}{b^{n+1}} = \sum_{n=1}^\infty \frac{nw^{n-1}}{b^{n+1}} $$
(the term for $n=0$ can be skipped in the last sum, as it would be $0$ anyway). Finally we have
$$ \frac{w+a}{(b-w)^2} = \sum_{n=1}^\infty \frac{nw^{n-1}(w+a)}{b^{n+1}} = \sum_{n=1}^\infty \Big(\frac{nw^n}{b^{n+1}} + \frac{anw^{n-1}}{b^{n+1}}\Big) = \\
= \sum_{n=1}^\infty \frac{nw^n}{b^{n+1}} + \sum_{n=0}^\infty \frac{a(n+1)w^n}{b^{n+2}} = \frac{a}{b^2}+ \sum_{n=1}^\infty \Big(\frac{n}{b^{n+1}}+\frac{a(n+1)}{b^{n+2}}\Big)w^n$$
We have then (after substituting $w=z+i$, $a=1-i$, $b=2i$):
$$ \frac{z+1}{(z-i)^2}= \frac{1-i}{(2i)^2}+ \sum_{n=1}^\infty \Big(\frac{n}{(2i)^{n+1}}+\frac{(1-i)(n+1)}{(2i)^{n+2}}\Big)(z+i)^n = \\
= \frac{-1+i}{4}+ \sum_{n=1}^\infty \frac{(n+1)+i(n-1)}{(2i)^{n+2}}(z+i)^n$$
This expansion is valid for $|w|<|b|$, that is $|z+i|<2$.
The other expansion for $|w|>|b|$ we get by expanding first
$$\frac{1}{b-w} = -\frac{1}{w} \frac{1}{1-\frac b w}= -\frac1 w \sum_{n=0}^\infty\left( \frac{b}{w}\right)^n = \sum_{n=0}^\infty \frac{-b^n}{w^{n+1}} = -\sum_{n=-\infty}^{-1} b^{-n+1} w^n$$
Then we follow as before:
$$ \frac{1}{(b-w)^2} = \frac{d}{dw} \frac{1}{b-w} = -\sum_{n=-\infty}^{-1} nb^{-n+1} w^{n-1} $$
$$ \frac{w+a}{(b-w)^2} = -\sum_{n=-\infty}^{-1} nb^{-n+1} w^{n-1}(w+a) = -\sum_{n=-\infty}^{-1} nb^{-n+1} (w^n+aw^{n-1}) = \\ = -\sum_{n=-\infty}^{-1} nb^{-n+1} w^n-\sum_{n=-\infty}^{-2} (n+1)b^{-n}aw^n = \\
= w^{-1}-\sum_{n=-\infty}^{-2} \frac{nb+(n+1)a}{b^{n}}w^n$$
so
$$ \frac{z+1}{(z-i)^2}= (z+i)^{-1}-\sum_{n=-\infty}^{-2} \frac{(n+1)+i(n-1)}{(2i)^{n}}(z+i)^n$$
This expansion is valid for $|z+i|>2$.
