I need to find the Laurent series for a rational function. I have decomposed it to $$f(z) = \frac{1}{3z}-\frac{1}{z-1}+\frac{10}{6(z-3)}$$
The last two terms can be converted to the Laurent series easily through some manipulations and the geometric series. However, the first term is giving me some trouble. Using the geometric series we can write $$\frac{1}{3z}=\frac{1}{3}\sum_{n=0}^{\infty}(1-z)^n$$ as we are on the annulus $A_{0,1}(0)$. But this is not the "correct" form: $$\sum_{n=-\infty}^{\infty}a_nz^n$$ Taylor won't work because it requires the derivatives at $0$.
How can I represent $\frac{1}{3z}$ such that I may find the Laurent series?
For $A_{0,1}(0)$ you have to expand $f$ in the form $\sum\limits_{k=-\infty}^{\infty}a_kz^{k}$. [In powers of $z$, not $z-1$]. So you don't have to do anything with $\frac 1 {3z}$. Just leave it as it is.
The coefficient $a_{-1} $ is $\frac 1 3$.