0
$\begingroup$

I'm meant to find the laurent series for this function;

$$f(z) = \frac{1}{z^2-1}, 1<|z-2|<3 $$

I started with substituting $$ t=z-2 \implies f(t+2) = \frac{1}{(t+1)(t+3)}$$

And then I did used partial fraction so I have 2 series to work with. My two series looked like this $$\frac{1}{2}\sum_{n=0}^\infty(-1)^n(z-2)^n - \frac{1}{2}\sum_{n=0}^\infty(-\frac{1}{3})^n(z-2)^n$$

Is it correct? When I add them together, I get a new sum which is the wrong answer. The correct answer also involves a sum from negative to positive infinity. I don't understand how, since both of the sums I got start from n=0.

$\endgroup$
6
  • $\begingroup$ The statement of the question indicates that the series should be centered at $t=2$. You series is centered at $t=0$, which outside the given domain. Make the substitution $w = t-2$ and then try your method. $\endgroup$
    – B. Goddard
    Commented Oct 1, 2016 at 12:01
  • $\begingroup$ Doesn't the statement indicate that the series should be centered at $z=2$? This is why I substituted t with z, writing $t = z-2$. $\endgroup$
    – armara
    Commented Oct 1, 2016 at 12:07
  • $\begingroup$ I changed my t back to z-2 after the sum-sign, so that you understand how I've been thinking a little better. $\endgroup$
    – armara
    Commented Oct 1, 2016 at 12:09
  • $\begingroup$ Yep. Sorry. Still on my first cup of coffee. Stand by. $\endgroup$
    – B. Goddard
    Commented Oct 1, 2016 at 12:10
  • $\begingroup$ I asked Maple for the answer and it agrees with you. This function has no essential singularity, so the series should not be running off to negative infinity. I think you have the right answer. $\endgroup$
    – B. Goddard
    Commented Oct 1, 2016 at 12:16

1 Answer 1

4
$\begingroup$

Have a look! $$\begin{align}f(z) &= \frac{1}{z^2-1}\\ &=\frac{1}{(z-1)(z+1)}\\ &=\frac{1}{2(z-1)}-\frac{1}{2(z+1)}\\ &=\frac{1}{2[(z-2)+1]}-\frac{1}{2[(z-2)+3]}\\ &=\frac{1}{2(z-2)[1+1/(z-2)]}-\frac{1}{2.3[1+(z-2)/3]}\\ &=\frac{1}{2(z-2)}\sum _{n=0}^{\infty } \frac{(-1)^n}{(z-2)^n}-\frac{1}{2.3}\sum _{n=0}^{\infty } \frac{(-1)^n(z-2)^n}{3^n}\\ &=\frac{1}{2}\sum _{n=0}^{\infty } \frac{(-1)^{n}}{(z-2)^{n+1}}-\frac{1}{2}\sum _{n=0}^{\infty } \frac{(-1)^n(z-2)^n}{3^{n+1}}\\ &=\frac{1}{2}\sum _{n=1}^{\infty } \frac{(-1)^{n-1}}{(z-2)^{n}}-\frac{1}{2}\sum _{n=0}^{\infty } \frac{(-1)^n(z-2)^n}{3^{n+1}}\\ &=\frac{1}{2}\sum _{n=-\infty}^{-1 } (-1)^{(1-n)}(z-2)^{n}-\frac{1}{2}\sum _{n=0}^{\infty } \frac{(-1)^n(z-2)^n}{3^{n+1}}\\ &=\frac{1}{2}\sum _{n=-\infty}^{\infty } \Bigg(**\Bigg){(z-2)^n} \end{align}$$

$\endgroup$

You must log in to answer this question.

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