# Thread: Laurent series of a square root function example

1. ## Laurent series of a square root function example

Hello,

I'm trying to figure out this example:

\begin{equation*}\begin{split} 1 - 1/z^2 &= x \\ 1 - x &= 1/z^2 \\ 1/z^2 &\geq 1 \:\: z \in [-1,1] \\ \implies g(z) &= z\sqrt{1 - 1/z^2} \:\: z \not\in [-1,1] \\ f(z)^2 &= z^2(1-1/z^2)=z^2-1. \end{split}\end{equation*} So $f(z)=\text{SQRT}{(z^2-1)}$ and $f$ has a square root on $\mathbb{C} \backslash [-1,1]$.

Using the binomial theorem, for $(1/z)^k<1$, we find $$z\sqrt{1-1/z^2} = z \sum_{n=0}^\infty {{1/2}\choose{k}} (-1)^k \Big(\frac{1}{z^{2k}}\Big)^k = \sum_{n=0}^\infty {{1/2}\choose{k}} (-1)^k \frac{1}{z^{2k-1}}.$$

If $r>1$ then $$\frac{1}{2 \pi i} \int_{|z|=r} \text{SQRT}(z^2-1)dz = \frac{1}{2 \pi i} \int_{|z|=r} \sum_{n=0}^\infty {{1/2}\choose{k}} (-1)^k \frac{1}{z^{2k-1}} dz \overset{?}{=} \frac{1}{2 \pi i} \int_{|z|=r} {{1/2}\choose{1}} (-1) \frac{1}{z^{2-1}} dz = -1/2$$

The problem I'm having is justifying the equality with the question mark over it.

The example also includes the information $\lim_{z \to \infty}\frac{f(z)}{z}=\sqrt{1}=1 \neq 0$. I haven't seen how this can be related/applies (if it does at all.)

Any ideas?

2. ## Re: Laurent series of a square root function example

The only pole with a non-zero residue at $0$ occurs when $k=1$

That's why they can go from the summation to the single value of $k=1$

3. ## Re: Laurent series of a square root function example

OK I think I understand.

$$\frac{1}{2 \pi i} \int_{|z|=r} \text{SQRT}(z^2-1)dz = \frac{1}{2 \pi i} \int_{|z|=r} \sum_{n=0}^\infty {{1/2}\choose{k}} (-1)^k \frac{1}{z^{2k-1}} dz = \frac{1}{2 \pi i} \sum_{n=0}^\infty \int_{|z|=r} {{1/2}\choose{k}} (-1)^k \frac{1}{z^{2k-1}} dz.$$

Now I think you use the fact that $\text{Res}_{z_0} f = \frac{1}{2 \pi i} \int_\gamma f(z) dz = 0$ when $\gamma$ is a simple counter clockwise curve around $z_0$ and $z_0$ is a removable singularity of $f$.

So like for this example,

https://en.wikipedia.org/wiki/Residu...Series_methods

The only term of the series expansion which has a non-zero residue occurs at $\frac{\sin{1}}{z-1}$. Since that's the only term that has a non-removable singularity.

And for this example

$$\sum_{n=0}^\infty {{1/2}\choose{k}} (-1)^k \Big(\frac{1}{z}\Big)^k = \sum_{n=0}^\infty {{1/2}\choose{k}} (-1)^k \frac{1}{z^{k-1}}$$

the only non-removable singularity would occurs when $k=2$.