What is the Laurent expansion of $f(z)=\dfrac1{z-3}$?
In the region, $|z-3|>0$ ?
I just computed the Laurent expansion in the region $|z|>3$ by dividing the denominator by $\dfrac1z$ and making it as a geometric series.
I am not sure how to manipulate $f(z)$ so that this function can be somehow expressed as a laurent series in above region.
The expression $f(z)=\frac1{z-3}$ is already the Laurent series around $|z-3|>0$.