Given $f(z)=(z^2-1)^{\frac{1}{2}}$ find the first three terms of the Laurent expansion

Question

Given

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

which has a branch cut for $|z|<1$. Find the first three terms of the Laurent expansion.

I proceed by factoring out a $z^2$.

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

where I know the following expansion

$$\frac{1}{z^2}=\sum_{n=0}^{\infty} (n+1)(z+1)^n=1+2(z+1)+3(z+1)^3+ \ldots$$

However, although plugging this in results in a cancellation of the 1's, I do not know how to deal with the 1/2 power.

$$z(2(z+1)+3(z+2)^3+\ldots)^{\frac{1}{2}}$$

Advice, and also how might I deal with the branch cut?

Answer 1

This function has branch points at $z=+1$ and at $z=-1$ and a branch line connecting these two points. Because it is a square-root singularity, the branch line for $|z|>1$ cancels, as may be seen by considering the net phase change when both branch points are encircled: \begin{align*} \left.\arg\left(\sqrt{z^2-1}\right)\right|_{\arg z =0}^{\arg z=2\pi}=0\qquad \mod 2\pi \end{align*}

So, the Laurent expansion for $|z|>1$ is using the binomial series expansion \begin{align*} \left(z^2-1\right)^{\frac{1}{2}}&=z\left(1-\frac{1}{z^2}\right)^{\frac{1}{2}}\\ &=z\sum_{n=0}^{\infty}\binom{\frac{1}{2}}{n}\left(-\frac{1}{z^2}\right)^n\\ &=z\left[\binom{\frac{1}{2}}{0}-\binom{\frac{1}{2}}{1}\frac{1}{z^2}+\binom{\frac{1}{2}}{2}\frac{1}{z^2}\mp\cdots\right]\\ &=z-\frac{1}{2z}-\frac{1}{8z^3}\mp\cdots \end{align*}

Note: This is example 6.3.1 found here.