Is this the correct way to find the Laurent Expansion of the complex function $f(z) = \frac{z}{z^2 - 1}$?

Question

The function $$ f(z) = \frac{z}{z^2 - 1} $$ has singularities at $z = \pm 1$. I will expand about the point $z = 1$. Then, with the substitution $w = z+1$, \begin{align*} f(x) =& \frac{z}{z^2 - 1} \\[3mm] =& \frac{z}{(z+1)(z-1)} \\[3mm] =& \frac{w-1}{w(w-2)} \\[3mm] =& \frac{1-w}{2w} \cdot \frac{1}{1 - \frac{w}{2}} \\[3mm] =& \frac{1-w}{2w} \sum_{n=0}^\infty \frac{w^n}{2^n} \\[3mm] =& \sum_{n=0}^\infty \frac{w^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{w^n}{2^{n+1}} \\[3mm] =& \sum_{n=0}^\infty \frac{(z+1)^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}} \end{align*} This is almost in the form of a Laurent expansion, but not quite. How can I continue this to get the correct final result?

Answer 1

From where you left off: \begin{align} &= \sum_{n=0}^\infty \frac{(z+1)^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}}\\[4px] &=\frac{(z+1)^{-1}}{2}+ \sum_{n=1}^\infty \frac{(z+1)^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}}\\[4px] &=\frac{(z+1)^{-1}}{2}+ \sum_{n=0}^\infty \frac{(z+1)^{n}}{2^{n+2}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}}\\[4px] &=\frac{(z+1)^{-1}}{2}+\sum_{n=0}^{\infty} \left(\frac{1}{2^{n+2}}-\frac{1}{2^{n+1}}\right)(z+1)^n\\[4px] &=\frac{(z+1)^{-1}}{2}+\sum_{n=0}^{\infty} -\frac{(z+1)^n}{2^{n+2}} \end{align}

Answer 2

At the penultimate line you have $$\sum_{n=0}^\infty \frac{w^{n-1}}{2^{n+1}} -\sum_{n=0}^\infty \frac{w^n}{2^{n+1}}.$$ This is $$\frac1{2w}+\frac14+\frac w8+\frac{w^2}{16}+\cdots-\frac12-\frac w4-\frac{w^2} 8-\cdots=\frac1{2w}-\sum_{n=0}^\infty\frac{w^n}{2^{n+2}}.$$ Does this look more like a Laurent series?

Answer 3

You did the right thing, and can conclude as the others suggested, but notice that this way you are expanding around $z=-1$ and not around $z=1$.