Laurent series expansion of $f$ Find the Laurent series expansion of $f(z)=\dfrac{1}{2z^2-13z+15}$ about the annulus $\dfrac{3}{2}<|z|<5$.
I did like this :
$f(z)=\dfrac{2}{7}(\dfrac{3}{3-2z}-\dfrac{1}{z-5})$ 
Then I took $\dfrac{3}{3-2z}=\dfrac{1}{1-\dfrac{2z}{3}}=(1-\dfrac{2z}{3})^{-1}=1+\dfrac{2z}{3}+(\dfrac{2z}{3})^2+...$
I took $\dfrac{1}{z-5}=\dfrac{-1}{5}(1-\dfrac{z}{5})^{-1}=\dfrac{-1}{5}(1+\dfrac{z}{5}+\dfrac{z^2}{25}+...)$
I calculated the co-efficients of $z$ and $z^2$ i.e $a_1,a_2$; I found $a_1=\dfrac{47}{75}$ and $a_2=\dfrac{4}{9}-\dfrac{1}{125}$ but the answer given is $\dfrac{a_1}{a_2}=5$
Why am I wrong?Please give some details 
 A: Here is a somewhat detailed description of the Laurent expansion of $f(z)$ around $z=0$.

The function
\begin{align*}
 f(z)&=\frac{1}{2z^2-13z+15}
 =-\frac{1}{7}\frac{1}{z-\frac{3}{2}}+\frac{1}{7}\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 pole $\frac{3}{2}$ 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 $\frac{3}{2}$ 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 consider the second region
  
  
*
  
*Region 2: $\frac{3}{2}<|z|<5$
  
  
  \begin{align*}
f(z)&=-\frac{1}{7}\frac{1}{z-\frac{3}{2}}+\frac{1}{7}\frac{1}{z-5}\\
&=-\frac{1}{7z}\sum_{n=0}^{\infty}\left(-\frac{3}{2}\right)^{n}\frac{1}{(-z)^n}
+\frac{1}{7}\sum_{n=0}^{\infty}\frac{1}{(-5)^{n+1}}(-z)^n\\
&=-\frac{1}{7}\sum_{n=1}^{\infty}\left(\frac{3}{2}\right)^{n-1}\frac{1}{z^n}
-\frac{1}{7}\sum_{n=0}^{\infty}\frac{1}{5^{n+1}}z^n\\
\end{align*}

We observe $\frac{a_1}{a_2}=\frac{[z^1]f(z)}{[z^2]f(z)}=\frac{-\frac{1}{7\cdot 5^2}}{-\frac{1}{7\cdot 5^3}}=5$.
A: There is a small error in your partial fraction decomposition, it should be
$$
f(z) = \frac{1}{7}(\dfrac{2}{3-2z}+\dfrac{1}{z-5})
$$
(However, this does not change the result for $a_1/a_2$, see below.)
The main problem is here:
$$
\frac{3}{3-2z}=\frac{1}{1-\frac{2z}{3}}=(1-\frac{2z}{3})^{-1}=1+\frac{2z}{3}+(\frac{2z}{3})^2+\ldots
$$
converges for $\left|\frac{2z}{3}\right| < 1$, i.e. it is the power (or Laurent) series for $\frac{3}{3-2z}$ in the domain $|z| < \frac 32$.
For $|z| > \frac 32$ you have to develop the expression into
powers of $z^{-1}$:
$$
\frac{3}{3-2z}= -\frac{3}{2z}\frac{1}{1-\frac{3}{2z}}
= -\frac{3}{2z}-(\frac{3}{2z})^2-\ldots
$$
In particular, this Laurent series contains only negative powers
of $z$, which means that the ratio $a_1/a_2$ in the Laurent
series of $f$ can be computed from the Laurent series
of $1/(z-5)$ alone.
