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?
Best Answer
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*}