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?

  • 1
    $\begingroup$ The $\frac{1}{z-5}$ term will converge as a geometric series for $|z|<5$. The $\frac{1}{2z-3}$ will not. Expand that in powers of $\frac{1}{z}$. $\endgroup$
    – sharding4
    Commented May 17, 2017 at 16:25
  • 1
    $\begingroup$ Since the question only asks you about $\frac{a_1}{a_2}$, you actually don't need to expand the function in its Laurent Series. You only need to think about which expression contributes terms with positive exponents and which contributes terms with negative exponents. $\endgroup$
    – sharding4
    Commented May 17, 2017 at 16:57

1 Answer 1


The function

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

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

  • $\begingroup$ Very Clear! Thanks! Thanks a lot! $\endgroup$
    – user444830
    Commented May 18, 2017 at 1:52
  • $\begingroup$ @Dim: You're welcome! Good to see the answer is useful. :-) $\endgroup$ Commented May 18, 2017 at 6:47

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