<s>[INST] Find the Taylor series about $z=0$ and the Laurent series about $z=-3$ 
  
*
  
*Let $f(z)=\frac{10z}{z^2+z-6}$, find the coefficient of $z$ in the Taylor series of $f$ expanded about $z=0$ and state the open set in $\mathbb C$ where the series converges.
  
*Find the Laurent series of $f$ about $z=-3$ and state the open set where this series converges.



*

*For the first part: 


$$\begin{align}f(z)&=\frac{6}{z+3}+\frac{4}{z-1}\\
&=\frac{6}{3}\frac{1}{1+\frac{z}{3}}-4\frac{1}{2-z}\\
&=2\frac{1}{1-(-\frac{z}{3})}-2\frac{1}{1-\frac{z}{2}}\end{align}$$
So Taylor series would be 
$$2\sum_{n=0}^{\infty}\left(-\frac{z}3\right)^n-2\sum_{n=0}^{\infty}\left(\frac{z}2\right)^n$$
Where $|-\frac{z}{3}|<1$ and $|\frac{z}{2}|<1$, so $|z|<3$ and $|z|<2$, is this correct so far and how do I deduce the open set from this? Would it be $B_3(0)\setminus B_2(0)$?


*

*For the second part:


Let $z=-3+w$ and input into $f$,
$$\begin{align}f(z)&=\frac{6}{(-3+w)+3}+\frac{4}{(-3+w)-2} \\
&=\frac{6}{w}+\frac{4}{-5+w}\\
&=\frac{6}{1-1+w}+\frac{-4}{5-w}\end{align}$$
....and then I get a little lost from there.
Any corrections, tips or solutions would be great!!
 [/INST] HINT: let $g(z):=\frac1{z-1}$. Then there are two distinct Laurent expansions of $g$ around $z=-3$ depending of the considered domain.
Observe that
$$g(z)=\frac1{z-1}=\frac1{(z+3)-4}=\begin{cases}\frac1{z+3}\cdot\frac1{1-\frac4{z+3}}=\frac1{z+3}\sum_{k=0}^\infty\left(\frac4{z+3}\right)^k,& |z+3|>4\\
-\frac14\cdot\frac1{1-\frac{z+3}4}=-\frac14\sum_{k=0}^\infty\left(\frac{z+3}4\right)^k,&|z+3|<4\end{cases}$$
</s>