Laurent series expansion of a given function 
Let $\sum_{-\infty}^{\infty} a_n z^n$ be the Laurent series expansion of $f(z)=\frac{1}{2z^2-13z+15}$ in an annulus $\frac{3}{2}< \vert z \vert < 5$.
What is the value of $\frac{a_1}{a_2}$?

My attempt:
First note that $f$ is analytic in the given annulus.
$f(z)=\frac{1}{2z^2-13z+15}=\frac{1}{(2z-3)(z-5)}$ ,
$\quad$ $\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ $=\frac{\frac{-2}{7}}{2z-3}+\frac{\frac{1}{7}}{z-5}$  , by Partial Fraction.
How to goes further to find out the required function in the power series form and find $\frac{a_1}{a_2}$?
I'm Confused with which term can be taken outside of the expression.
Any help?
 A: 
The function
\begin{align*}
 f(z)&=\frac{1}{2z^2-13z+15}\\
 &=-\frac{1}{7}\cdot\frac{1}{z-\frac{3}{2}}+\frac{1}{7}\cdot\frac{1}{z-5}\\
\end{align*}
   has two simple poles at $\frac{3}{2}$ and $5$.
Since we want to find a Laurent expansion with center $0$, we look at the poles $\frac{3}{2}$ and $5$ and see they determine three regions.
\begin{align*}
 |z|<\frac{3}{2},\qquad\quad
 \frac{3}{2}<|z|<5,\qquad\quad
 5<|z|
 \end{align*} 
  
  
*
  
*The first region $ |z|<\frac{3}{2}$ is a disc with center $0$, radius $\frac{3}{2}$ and the pole $\frac{3}{2}$ at the boundary of the disc. In the interior of this disc all two fractions with poles $\frac{3}{2}$ and $5$  admit a representation as power series at $z=0$.
  
*The second region $\frac{3}{2}<|z|<5$ is the annulus with center $0$, inner radius $\frac{3}{2}$ and outer radius $5$. Here we have a representation of the fraction with poles $\frac{3}{2}$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $5$ admits a representation as power series.
  
*The third region $|z|>5$ containing all points outside the disc with center $0$ and radius $5$ 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}{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=0}^{\infty}\frac{a^n}{(-z)^{n+1}}\\
&=-\sum_{n=1}^{\infty}\frac{a^{n-1}}{(-z)^n}
\end{align*}

We can now obtain the Laurent expansion of $f(z)$ at $z=0$ for all three regions. Here we have to consider 
  
  
*
  
*Region 2: $\frac{3}{2}<|z|<5$
  
  
  Since we only need to calculate $\frac{a_1}{a_2}$ it is sufficient to expand the power series part only. We obtain
\begin{align*}
f(z)&=-\frac{1}{7}\cdot\frac{1}{z-\frac{3}{2}}+\frac{1}{7}\cdot\frac{1}{z-5}\\
&=-\frac{1}{7}\cdot\frac{1}{z-\frac{3}{2}}+\frac{1}{7}\sum_{n=0}^\infty \frac{1}{(-5)^{n+1}}(-z)^n\\
&=-\frac{1}{7}\cdot\frac{1}{z-\frac{3}{2}}-\frac{1}{7}\sum_{n=0}^\infty \frac{1}{5^{n+1}}z^n\\
\end{align*}
We conclude since $a_1=-\frac{1}{7}\cdot\frac{1}{5^2}$ and $a_2=-\frac{1}{7}\cdot\frac{1}{5^3}$
  \begin{align*}
\color{blue}{\frac{a_1}{a_2}=5}
\end{align*}

