Laurent Series expansion of $\frac{z-1}{z+1}$ about $z=-1$

Question

I am unsure how to make a Laurent series expansion about $z=-1$ for $f(z)=\frac{z-1}{z+1}$ as $f(z)$ has a discontinuity at that point. I think I have some fundamental misunderstanding of Laurent series and would like that to be cleared up. Note $z$ is a complex variable.

Answer 1

A Laurent series comes with an annulus of definition. From what you've said here, you want the closest-in annulus $0<|z+1|<R$ for some $R$. The point $z=-1$ is a pole, so we will have finitely many negative-degree terms. We just have to subtract them off.

Dividing through to get a polynomial plus a proper rational function, $$f(z)=1-\frac{2}{z+1}$$ And... that's it. That's the complete Laurent series. All but two coefficients are zero, so it converges for all $z$ other than $-1$.

If there were something else in the denominator, we'd run partial fractions and then construct a geometric series from the other part - but there isn't. It's just the power of $z+1$ in the denominator, and that makes things simple.

Answer 2

Just put $z'=z+1$, and you get $\dfrac{z'-2}{z'}=1-\dfrac{2}{z'}=1-\dfrac{2}{z+1}$.