I need to find the Laurent series of $\frac{3}{2+z-z^2}$ about the following domains:
(a) $B_1(0)$ (b) ann$(0;1,2)$ (c) ann$(0;2,\infty)$
so I have
(a) Note we can write $$f(z)=\frac{3}{2+z-z^2}=\frac{3}{(1+z)(2-z)}=\frac{1}{1+z}+\frac{1}{2-z}.$$ Here, the first term has power series $$\frac{1}{1+z}=\sum_{n=0}^\infty (-1)^n z^n.$$ As for the later term, we write \begin{align*} \frac{1}{2-z}&= \frac{1}{2}(\frac{1}{1-\frac{z}{2}})\\ &=\frac{1}{2} \sum_{n=0}^\infty \frac{z^n}{2^n}\\ &=\sum_{n=0}^\infty \frac{z^n}{2^{n+1}}. \end{align*} Therefore we have as the Laurent series: $$\sum_{n=0}^\infty (-1)^n z^n + \sum_{n=0}^\infty \frac{z^n}{2^{n+1}}= \sum_{n=0}^\infty ((-1)^n+\frac{1}{2^{n+1}})z^n$$ as desired. \ \ (b) On the annulus ann$(0;1,2)$ we have that $$\vert \frac{z}{2} \vert <1, \vert \frac{1}{z} \vert <1.$$ Thus for $\frac{1}{1+z}$ we have \begin{align*} \frac{1}{1+z}&=\frac{1}{z}\frac{1}{1+\frac{1}{z}}\\ &=\frac{1}{z} \sum_{n=0}^\infty (-1)^n\frac{1}{z^n}\\ &=\sum_{n=0}^\infty (-1)^n \frac{1}{z^{n+1}}\\ &=\frac{1}{z}+\sum_{n=1}^\infty (-1)^n \frac{1}{z^{n+1}} \end{align*}
is this correct for 1(a) and part of 1(b)? And for $\frac{1}{2-z}$ we write the same thing basically but it has no singular parts? cause the $z$'s all have positive powers.
And I'm a bit stuck with the last one, I know $\infty>\vert z \vert >2$.
For $\lvert z\rvert \gt2$:
$$\dfrac 1{1+z}=\dfrac 1z\cdot \dfrac 1{1-(-\frac 1z)}=\dfrac 1z\sum_{n\ge0} (-\frac 1z)^n$$, which is valid on $\lvert z\rvert \gt1$ and which is what you already did; and
$$\dfrac 1{2-z}=-\dfrac 1z\cdot \dfrac1{1-\frac 2z}=-\dfrac 1z\cdot \sum_{n\ge0}(\dfrac 2z)^n$$.
Combining, $$\sum_{n\ge0}\dfrac {(-1)^n-2^n}{z^{n+1}}$$.