$ f(z) = \frac {z^2 -z + 3} {z^3 -3z +2} $ and $A=\{z: |z|<1\}$.
I found the Laurent Series of this function in the set A:
\begin{align*}f(z) &= \frac {z^2 -z + 3} {(z-1)^2 (z+2)} =(1-z)^{-2} + \frac {1} {2}.\frac {1} {1+\frac {z}{2}} \\ &= 1+ \sum_{n=1}^{\infty} {(-1)^{n} (n+1) (-z)^{n}}+\sum_{n=0}^{\infty} \frac {(-1)^{n} (z)^{n}}{2^{n+1}}\\ &= \sum_{n=0}^{\infty} {(n+1) (z)^{n}}+\sum_{n=0}^{\infty} \frac {(-1)^{n} (z)^{n}}{2^{n+1}}. \end{align*}
And in the solutions shows : $\sum_{n=1}^{\infty} (n-(\frac {-1}{2})^n)z^{n-1}$.
Could someone help me understand what I did wrong to not come up with the intended solution? Thanks
These are the same. Note that $$\sum\limits_{n=0}^\infty(n+1)z^n=\sum\limits_{n=1}^\infty nz^{n-1},$$ and $$\sum\limits_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}z^n=\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{2^{n}}z^{n-1}=-\sum\limits_{n=1}^\infty \frac{(-1)^{n}}{2^{n}}z^{n-1}.$$ They just re-indexed the sums and put a minus in front of the last sum to get the coefficients to have the same power of $n$.