<s>[INST] 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 
 [/INST] 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.
</s>