Vrftsjtryk

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?

The function

begin{align*}
f(z)&=frac{1}{2z^2-13z+15}\
&=-frac{1}{7}cdotfrac{1}{z-frac{3}{2}}+frac{1}{7}cdotfrac{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},qquadquad
frac{3}{2}<|z|<5,qquadquad
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}cdotfrac{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}cdotfrac{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}cdotfrac{1}{z-frac{3}{2}}+frac{1}{7}cdotfrac{1}{z-5}\
&=-frac{1}{7}cdotfrac{1}{z-frac{3}{2}}+frac{1}{7}sum_{n=0}^infty frac{1}{(-5)^{n+1}}(-z)^n\
&=-frac{1}{7}cdotfrac{1}{z-frac{3}{2}}-frac{1}{7}sum_{n=0}^infty frac{1}{5^{n+1}}z^n\
end{align*}

We conclude since $a_1=-frac{1}{7}cdotfrac{1}{5^2}$ and $a_2=-frac{1}{7}cdotfrac{1}{5^3}$
begin{align*}
color{blue}{frac{a_1}{a_2}=5}
end{align*}