Find the Laurent Series on the annulus $1 < |z| < 4$ for

$$R(z) = \frac{z+2}{(z^2-5z+4)}$$

So I am having a few issues with this. I know there are two poles in this problem particulaly $z = 1$ and $z = 4$, so if I factor out I get it into a form as:

$$ \frac{z+2}{(z-1)(z-4)} $$

and here is where it get's a little hazy. I know there is a relationship in which I would have to split this expression into partial fractions:

$$ \frac{z+2}{(z^2-5z+4)} = \frac{2}{z-4} + \frac{-1}{z-1} $$

now the textbook goes on about using their geometric series, which I somewhat see, but I cannot understand how to get the coefficients. I was trying to use the method of treating each of the numerators in the partial fractions as power series and then solving for coeffecients, but that resulted to no avail. Then I tried not even expanding the expression into partial fractions and attempting to solve for the coefficients by accounting for each singularity and using the remaining part as a power series i.e

$ \frac{z+2}{(z-4)} $ as one power series and then $ \frac{z+2}{(z-1)} $ as the other.

Still not working out. Perhaps my ideas are scattered.

  • $\begingroup$ Your "\begin"s and "\end"s were replaced by dollar signs. Click on "edit" and see. $\endgroup$
    – Ron Gordon
    Commented Apr 1, 2015 at 19:23

2 Answers 2


You need put each partial fraction into the form $\frac{1}{1-w}$ where $|w| \lt 1$ in order to use the geometric series expansion.

$\frac{2}{z-4}$ is analytic in $|z| \lt 4$ and $\Big|\frac{z}{4}\Big| \lt 1$ so we have: $$ \frac{2}{z-4} = -\frac{1}{2}\cdot\frac{1}{1-\frac{z}{4}} = -\frac{1}{2}\sum_{n=0}^{\infty}\frac{z^n}{4^n}=\sum_{n=0}^{\infty}-\frac{z^n}{2^{2n+1}} $$

$-\frac{1}{z-1}$ is analytic in $|z| \gt 1$ and $\Big|\frac{1}{z}\Big| \lt 1$ so we have: $$ -\frac{1}{z-1} = -\frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}} = -\frac{1}{z}\sum_{n=0}^{\infty}\frac{1}{z^n} = \sum_{n=0}^{\infty}-\frac{1}{z^{n+1}} $$

So in total we have: $$ R(z) = \dots -\frac{1}{z^3}-\frac{1}{z^2}-\frac{1}{z} -\frac{1}{2}-\frac{z}{8}-\frac{z^2}{32}-\dots $$


Look carefully at your conditions..

|z|>1 and |z|<4

So for the first fraction

2/(z-4) take -4 outside from the denominator. You will get (-1/2)*(1-z/4)^-1 and expand binomially.

For the second fraction take z outside from the denominator. You will get (-1/z)*(1-1/z)^-1 and expand binomially.

Add both of them finally to get the final laurent series.

PS: I am still learning tex..so just write it down on a page for it to make more sense.

  • $\begingroup$ I understand how to turn them into those two series, specifically: $$\frac{-1}{2} * \sum (\frac{z}{4})^2$$ and $$\sum (\frac{1}{z})^j$$ What I'm having trouble with is trying to get the coefficients. Also since is there another way to find each expressions power series if I don't see the "geometric series" trick? $\endgroup$ Commented Apr 1, 2015 at 23:30
  • $\begingroup$ use two poles as $$\frac{x^7}{8192 (x-4)}-\frac{x^7}{x-1}+\frac{8191 x^6}{8192}+\frac{2047 x^5}{2048}+\frac{511 x^4}{512}+\frac{127 x^3}{128}+\frac{31 x^2}{32}+\frac{7 x}{8}+\frac{1}{2}$$ Mittag-Leffler's Theorem $\endgroup$
    – user167276
    Commented Dec 1, 2016 at 10:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .