How to find Taylor or Laurent series of $z \mapsto \frac{z}{z^2-1}$?

Question

I have the function

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

I have to find Taylor or Laurent series of this function about the indicated point $z=2$. I tried to find Laurent series but looks like Taylor series will be much easier. So how can I find its Taylor series expansion?

Answer 1

\begin{align*} \frac{z}{z^2 - 1} &= \frac{-z}{1-z^2} \\ &= \frac{1/2}{z+1} + \frac{1/2}{z-1} \\ &= \frac{1/2}{(z-2)+2+1} + \frac{1/2}{(z-2)+2-1} \\ &= \frac{1/2}{3--(z-2)} \cdot \frac{1/3}{1/3}+ \frac{1/2}{1--(z-2)} \\ &= \frac{1/6}{1-(-1/3)(z-2)} + \frac{1/2}{1--(z-2)} \\ &= \sum_{k=0}^\infty \frac{1}{6}(-1/3)^k(z-2)^k + \sum_{k=0}^\infty \frac{1}{2}(-1)^k(z-2)^k \\ &= \sum_{k=0}^\infty \left( \frac{1}{6}(-1/3)^k+\frac{1}{2}(-1)^k \right)(z-2)^k \text{.} \end{align*}

First, separate by partial fraction decomposition. Then coerce every appearance of $z$ into the combination "$z-2$". Rearrange the denominators to be of the form "$(\text{constant, preferably $1$})-(\text{term})$" and then scale the numerator and denominator so that the constant(s) are all $1$. Now apply the standard geometric series to $\frac{c}{1-x}$. Finally, collect by like powers of $z-2$.

Answer 2

I don't think that you can compute the Laurent series. Nevertheless you can compute some termes : $$\frac{z}{z^2-1}=\frac{z-2}{(z-2)^2+4(z-2)+4}+\frac{2}{(z-2)^2+4(z-2)+4}.$$

Now, if $f(z-2)=\frac{(z-2)^2+4(z-2)}{4}$, $$\frac{1}{f(z-2)+1}=\sum_{k=0}^\infty (-1)^nf(z)^n,$$

whenever $|z-2|<1$. After, to compute $f(z-2)^n$ for all $n$ may be rather hard.

Answer 3

The function's Laurent expansion around $z=2$ will just be a Taylor expansion since it's analytic at this point.

Answer 4

Hint:

Set $u=z-2$ and use partial fractions: $$\frac z{z^2-1}=\frac{u+2}{(u+2)^2-1}=\frac{u+2}{(u+1)(u+3)}=\frac12\biggl(\frac1{u+1}+\frac1{3+u}\biggr).$$ Can you take it from there?