# Laurent series of $1\over (z - i)$

I need help in finding the Laurent series of the following complex function: $$f(z) = {1\over (z-i)}$$ around $z_0 = i$, over the whole complex plane.

The part I'm getting confused with is the $z_0 = i$ condition, I'm not sure how I should adjust the normal Taylor series expansion to account for this. Any help is appreciated.

• Hint: how does a Laurent series around $i$ look like. Now compare this to the given $f(z)$... – Fabian Dec 1 '15 at 18:12

Your function is equal to its Laurent series. The series around $i$ looks like the following
$$\ldots+\frac{a_{-2}}{(z-i)^2}+{\color{red}{\frac{a_{-1}}{(z-i)}}}+a_0+a_1(z-i)+a_2(z-i)^2+\ldots$$
In your case $$a_n=\begin{cases}1,&\text{for }n=-1\\0,&\text{for }n\ne -1\end{cases}$$