# Laurent series (once again)

I need to find the Laurent series expansion of $\dfrac{1}{z-1} - \dfrac{1}{z^2}$ about $z=i$.

I'm not sure how to deal with the $\dfrac{1}{z^2}$

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 – joriki Dec 13 '12 at 15:48 Do you need the series in a neighbourhood of $i$, or in the annulus $1<\lvert z-i\rvert<\sqrt2$ or in $\lvert z-i\rvert>\sqrt2$, or all of those? – Harald Hanche-Olsen Dec 13 '12 at 15:58 In $1 < |z-i| < \sqrt{2}$ – nigelvr Dec 13 '12 at 16:00

The taylor expansion of $z^{-2}$ at $i$ is $$z^{-2}=\sum_{n=0}^{\infty}\frac{(-1)^ni^{n+1}}{(n+1)!}(z-i)^n$$ for $\left|z-i\right|<1$. This can be easily shown with the $n$th derivative of $z^{-2}$. Now you just need to expand $(z-1)^{-1}$ at $i$ which I am sure you know how
When $\lvert z-i\rvert>1$, write $$\frac1{z^2}=\frac1{(z-i+i)^2}=\frac1{(z-i)^2}\frac1{\Bigl(1+\dfrac i{(z-i)}\Bigr)^2}$$ and apply the (hopefully known) Taylor series of $$\frac1{(1+\zeta)^2}$$ valid for $\lvert\zeta\rvert<1$.