Finding Laurent Series With Square Denominator

Question

I'm lost. We very, very briefly went over Laurent series in our last class and I have no idea how to deal with them. How might one find:

$f(z)=\frac{1}{(z^2-1)^2}$

Find the Laurent series in the annulus $0 < |z-1| < 2$.

I'm not even really sure how to begin. Is there anyone who can direct me or walk me through it?

Thanks!

Answer 1

You want a series in powers of $z-1$ that works when $z-1$ is small (so only finitely many negative powers). It helps to do a change of variables $z - 1 = t$, so it's in powers of $t$. With $z = 1 + t$ you have $$f(z) = \dfrac{1}{(t+1)^2-1} = \dfrac{1}{t^2 + 2 t} = \dfrac{1}{t (t+2)}$$ Do you know how to expand $1/(t+2)$ in a Taylor series? Then divide each term by $t$, and convert back to $z$.

Answer 2

This illustrates the standard technique in pendantic detail: $$ \begin{align} \frac{1}{(z^{2}-1)^{2}} & =\frac{1}{(z-1)^{2}}\frac{1}{(z+1)^{2}}\\ & =\frac{1}{(z-1)^{2}}\frac{1}{((z-1)+2)^{2}} \\ & =\frac{1}{(z-1)^{2}}\frac{1}{4(1+\frac{z-1}{2})^{2}} \\ & =-\frac{1}{(z-1)^{2}}\frac{1}{2}\frac{d}{dz}\frac{1}{1+\frac{z-1}{2}} \\ & =-\frac{1}{(z-1)^{2}}\frac{1}{2}\frac{d}{dz}\sum_{n=0}^{\infty}(-1)^{n}\left(\frac{z-1}{2}\right)^{n}, \;\;\;\; 0 < |\frac{z-1}{2}| < 1, \\ & =-\frac{1}{(z-1)^{2}}\frac{1}{2^{2}}\sum_{n=1}^{\infty}(-1)^{n}n\left(\frac{z-1}{2}\right)^{n-1} \\ & =\frac{1}{(z-1)^{2}}\frac{1}{2^{2}}\sum_{n=0}^{\infty}(-1)^{n}(n+1)\left(\frac{z-1}{2}\right)^{n} \\ & =\sum_{n=0}^{\infty}\frac{(-1)^{n}(n+1)}{2^{n+2}}(z-1)^{n-2}. \end{align} $$ The series converges absolutely for $0 <|z-1|< 2$.