Apologies if this is a simple question. I've just read a chapter on Laurent series which seems to indicate that in practice we don't calculate the coefficients of a Laurent series using the integral, instead we often manipulate Taylor series, adjust a function so a geometric expansion may be used etc.
For the function:
$ \begin{align} f(z) = \frac{1}{z-3i} \end{align} $
How do we obtain the Laurent series at about 3i, i.e. how do we obtain:
$ \begin{align*} \frac{1}{z-3i} + 0+ 0(z-3i) + 0(z-3i)^{2} + \cdots \quad z \in \mathbb{C}-\{3i\} \end{align*} $
Many thanks,
John