Then find the value of $\frac{a_1}{a_2}$

Question

Let $\sum _{n=-\infty}^\infty a_nz^n $ be the Laurent series in the expansion of $f(z)=\dfrac{1}{2z^2-13z+15}$ in the annulus $\frac{3}{2}<|z|<5$ .

Then

find the value of $\frac{a_1}{a_2}$.

IN order to find the value I will have to find Laurent series expansion of the function which is very tough.Is there any easy way to do it ?

Answer 1

Your series must be such that the following holds:

$$(2z^2-13z+15)\sum a_nz^n = 1$$

$$\sum 2a_nz^{n+2} - \sum 13a_nz^{n+1} + \sum 15 a_n z^n = 1$$

Equate the coefficients of the LHS to the RHS and you can find all $a_n$. Do this with the equations that contains $a_1$ and $a_2$ and you should be able to find them

Also, since there is a simple pole in $z = 3/2$, there will be a term in $z^{-1}$ but you know all the other negative coefficients are $0$.

Answer 2

Hint :

$\displaystyle \frac{1}{2z^2-13z+15}=\frac{1}{(z-5)(2z-3)}=\frac{1}{7}\left(\frac{1}{z-5}-\frac{2}{2z-3}\right)$

Now expand it in the given region.