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!!
 A: We consider the function
\begin{align*}
f(z)&=\frac{10z}{z^2+z-6}\\
&= \frac{6}{z+3} +\frac{4}{z-2}\\
\end{align*}

First part: The function $f$ is to expand around the center $z=0$ as Taylor series.
Since there are simple poles at $z=-3$ and $z=2$ we have to distinguish three regions of convergence
  \begin{align*}
D_1:&\quad 0\leq  |z|<2\\
D_2:&\quad 2<|z|<3\\
D_3:&\quad  |z|>3
 \end{align*} 
  
  
*
  
*The first region $D_1$ is a disc with center $z=0$, radius $2$ and the pole at $z=2$ at the boundary of the disc. It admits for both fractions a representation as power series. 
  
*The region $D_2$ is an annulus containing all points outside the closure of $D_1$ and the closure of $D_3$. It admits for the fraction with pole  at $z=2$ a representation as principal part of a Laurent series and for the fraction with pole at $z=3$ a power series.
  
*The region $D_3$ contains all points outside the disc with center $z=0$ and radius $3$. It admits for both fractions a representation as principal part of a Laurent series.

We are interested in an expansion as Taylor series and observe the region $D_1$ is the corresponding open set of convergence with the representation of $f$ as Taylor series as stated by OP:
\begin{align*}
f(z)&= \frac{6}{z+3} + \frac{4}{z-2}\\
&= 2\sum_{n=0}^\infty\left(-\frac{z}{3}\right)^n-2\sum_{n=0}^\infty\left(-\frac{z}{2}\right)^n
\end{align*}

Second part: The function $f$ is to expand around the center $z=-3$ as Laurent series and the region of convergence is to determine.
Since there are simple poles at $z=-3$ and $z=2$ we have to distinguish two regions of convergence when expanding around $z=-3$:
  \begin{align*}
D_4:&\quad 0<  |z+3|<5\\
D_5:&\quad  |z+3|>5
 \end{align*} 
  
  
*
  
*The  region $D_4$ is a punctured disc with center $z=-3$, radius $5$ and the pole at $z=2$ at the boundary of the disc. It admits for the fraction with pole  at $z=-3$ a representation as principal part of a Laurent series and for the fraction with pole at $z=2$ a power series.
  
*The region $D_5$ contains all points outside the disc with center $z=-3$ and radius $5$. It admits for both fractions a representation as principal part of a Laurent series.

The expansion of $f$ as Laurent series in $D_4$ is
\begin{align*}
f(z)&= \frac{6}{z+3} + \frac{4}{z-2}\\
&= \frac{6}{z+3} +\frac{4}{(z+3)-5}\\
&= \frac{6}{z+3} -\frac{4}{5}\cdot\frac{1}{1-\frac{z+3}{5}}\\
&= \frac{6}{z+3} -\frac{4}{5}\sum_{n=0}^\infty\frac{1}{5^{n}}(z+3)^{n}\\
\end{align*}
The expansion of $f$ as Laurent series in $D_5$ can be calculated similarly.
A: 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
$$g(z)=\frac1{z-1}=\frac1{(z+3)-4}=\begin{cases}\frac1{z+3}\cdot\frac1{1-\frac4{z+3}}=\frac1{z+3}\sum_{k=0}^\infty\left(\frac4{z+3}\right)^k,& |z+3|>4\\
-\frac14\cdot\frac1{1-\frac{z+3}4}=-\frac14\sum_{k=0}^\infty\left(\frac{z+3}4\right)^k,&|z+3|<4\end{cases}$$
