My task is to find the Laurent series expansion of $$f(z)=\frac{1}{2z^2-z-1},$$ which converges for $\displaystyle \frac12 <|z|<1$.
I proceeded by doing the following: $$\frac{1}{2z^2-z-1} = \frac{1}{(2z+1)(z-1)} = \frac13 \left( \frac{1}{z-1}-\frac{2}{2z+1} \right).$$
For the first term: $$\frac13 \frac{1}{z-1} = \frac{1}{3z} \frac{1}{1-\frac1z}=\frac{1}{3z}\sum_{n=0}^\infty \frac{1}{z^n},$$ for $\displaystyle \left| \frac1z \right|<1 \Rightarrow |z|>1.$ Hence, for $|z|<1$ (using geometric series properties): $$\frac13 \frac{1}{z-1} =-\frac{1}{3z}\sum_{n=1}^\infty z^n=-\frac13 \sum_{n=1}^\infty z^{n-1}.$$ For the second term: $$-\frac13\frac{2}{2z+1}=-\frac23\frac{1}{1+2z}=-\frac23\sum_{n=0}^\infty\left( -2z \right)^n,$$ for $\displaystyle |2z|<1 \Rightarrow |z|<\frac12$. Hence, for $\displaystyle |z|>\frac12$: $$-\frac13\frac{2}{2z+1}=\frac23 \sum_{n=1}^\infty\frac{1}{(-2z)^n}=\frac13 \sum_{n=1}^\infty(-1)^{-n}2^{1-n}z^{-n}.$$ Hence, $$\frac{1}{2z^2-z-1}=\frac13 \sum_{n=1}^\infty(-1)^{-n}2^{1-n}z^{-n}-\frac13 \sum_{n=1}^\infty z^{n-1}.$$
I think this is the correct approach. Some further guidance/clarification would be great!