<s>[INST] Laurent series expansion of $f(z)$ which converges for $\frac12 <|z|<1$ 
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!
 [/INST] Note: This is just an affirmation of OPs calculation.

The function
\begin{align*}
 f(z)&=\frac{1}{3}\frac{1}{1-z}-\frac{1}{3}\frac{1}{z+\frac{1}{2}}\\
\end{align*}
   has two simple poles at $z=1$ and $z=-\frac{1}{2}$.
Since we want to find a Laurent expansion with center $0$, we look at these poles and see they determine three regions.
\begin{align*}
 |z|<\frac{1}{2},\qquad\quad
 \frac{1}{2}<|z|<1,\qquad\quad
 1<|z|
 \end{align*} 
  
  
*
  
*The first region $ |z|<\frac{1}{2}$ is a disc with center $0$, radius $\frac{1}{2}$ and the pole $\frac{1}{2}$ at the boundary of the disc. In the interior of this disc all two fractions with poles $\frac{1}{2}$ and $1$  admit a representation as power series at $z=0$.
  
*The second region $\frac{1}{2}<|z|<1$ is the annulus with center $0$, inner radius $\frac{1}{2}$ and outer radius $1$. Here we have a representation of the fraction with pole $\frac{1}{2}$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $1$ admits a representation as power series.
  
*The third region $|z|>1$ containing all points outside the disc with center $0$ and radius $1$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.

A power series expansion of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{a^{n+1}}z^n
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{(-a)^n}{z^n}\\
&=\sum_{n=1}^{\infty}\frac{(-a)^{n-1}}{z^n}
\end{align*}
OPs calculation corresponds to the representation in the second region and is correct. It is sufficient to calculate the power series part for $\frac{1}{1-z}$ and the principal part of the Laurent series for $\frac{2}{2z+1}$.
</s>