1. ## Laurent Expansions

I am stuck at two questions about Laurent expansion.

Below are the problems:

1.

Determine the Laurent expansion of (z^2 + 2z)/(z^3 -1) in the annulus 0 < | z - 1 | < R about z = 1. Find the large R one can use.

2.

Obtain a Laurent expansions of f(z) =1/[ (z-j) (z-2) ] in the region 0 < | z - j | < square root of 5

Any help would be appreicated.

2. Originally Posted by progrocklover

1. Determine the Laurent expansion of (z^2 + 2z)/(z^3 -1) in the annulus 0 < | z - 1 | < R about z = 1. Find the large R one can use...
Setting $z-1=s$ the complex function becomes...

$\displaystyle f(s)= \frac{1}{s}\ \frac{3+4 s + s^{2}}{3+3 s + s^{2}} = \frac{1}{s}\ (1+ \frac{s}{3+3 s + s^{2}}) = \frac{1}{s} + \frac{1}{3+3 s + s^{2}}$ (1)

Now you have to find the Laurent expansion of (1) around $s=0$. The first term is the 'non analythic part' of the expansion, the second in the 'analythic part'. Are You able to proceed?...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
Setting $z-1=s$ the complex function becomes...

$\displaystyle f(s)= \frac{1}{s}\ \frac{3+4 s + s^{2}}{3+3 s + s^{2}} = \frac{1}{s}\ (1+ \frac{s}{3+3 s + s^{2}}) = \frac{1}{s} + \frac{1}{3+3 s + s^{2}}$ (1)

Now you have to find the Laurent expansion of (1) around $s=0$. The first term is the 'non analythic part' of the expansion, the second in the 'analythic part'. Are You able to proceed?...

Kind regards

$\chi$ $\sigma$
Thank you for your reply, but I have problem in finding the Laurent expansion of (1).

Actually, I tried to transfer (1) into the form of f(s)[1/(1-s)], but I failed. So, I cannot use the Binomial Theorem of (1+z)^-1. I also tried other methods such as appiling the equation of Taylor series directly on to (1) which does not work as well.

It would be kind of you if you could give me some idea on this.

4. All right!... so that the next step is to find the Taylor expansion of...

$\displaystyle f(s)= \frac{1}{3+3\ s + s^{2}}$ (1)

Because f(s) is analytic in s=0 we can write $\displaystyle f(s)= \sum_{n=0}^{\infty} a_{n}\ s^{n}$ and the $a_{n}$ can be computed imposing the condition...

$(a_{0} + a_{1}\ s + a_{2}\ s^{2} + ...) (3 + 3\ s + s^{2})=1$ (2)

From (2) we derive first...

$\displaystyle a_{0}= \frac{1}{3}$

$\displaystyle a_{1}= - a_{0}= - \frac{1}{3}$ (3)

... and for n>1 the $a_{n}$ satisfy the 'recurrence relation'...

$\displaystyle a_{n}= - a_{n-1} - \frac{a_{n-2}}{3}$ (4)

... with the 'initial conditions (3). The 'formal' solution is possible but a little 'uncorfortable' ... some of the $a_{n}$ derived 'directly' from (4) are...

$\displaystyle a_{2}= \frac{2}{9}$

$\displaystyle a_{3}= - \frac{1}{9}$

$\displaystyle a_{4}= \frac{1}{27}$

...

Kind regards

$\chi$ $\sigma$

5. A better way to expand $\displaystyle\frac{1}{s^2+3s+3}$ to a power series around 0 would be:

let $a,b$ be the roots of $s^2+3s+3$. Then

$\displaystyle{
\frac{1}{s^2+3s+3} = \frac{1}{(s-a)(s-b)}=\frac{1}{a-b}\left(\frac{1}{s-a}-\frac{1}{s-b}\right)
}$

$\displaystyle{
=\frac{1}{a-b}\left(\frac{-\frac{1}{a}}{1-\frac{s}{a}}+\frac{\frac{1}{b}}{1-\frac{s}{b}}\right)
}$

I believe you can take it from here.

6. A seen before, the coefficients of the Taylor expansion of $\displaystyle f(s)= \frac{1}{3 + 3\ s + s^{2}}$ are solution of the 'recurrence relation' ...

$\displaystyle a_{n} = -a_{n-1} - \frac{a_{n-2}}{3}\\ , \\ a_{0}=\frac{1}{3}, a_{1}= -\frac{1}{3}$ (1)

The (1) can be solved finding the roots of the characteristic equation...

$\displaystyle x^{2} + x + \frac{1}{3}=0$ (2)

... that are $\displaystyle \frac{1}{\sqrt{3}}\ e^{\pm i\ \frac{\pi}{6}}$ so that is...

$\displaystyle a_{n} = (\frac{1}{\sqrt{3}})^{n}\ (c_{1}\ \cos n\ \frac{\pi}{6} + c_{2}\ \sin n\ \frac{\pi}{6})$ (3)

... where $c_{1}$ and $c_{2}$ can be derived from the 'initial conditions'. It is obvious that the $a_{n}$ can be easily iteratively computed from (1) but (3) is important in order to extablish the radious of convergence of the Taylor series that is $r=\sqrt{3}$...

Kind regards

$\chi$ $\sigma$