Does $\frac{z^2}{(z^2-1)^{\frac{1}{2}}}$ have a simple pole at $\infty$? If we define $w=\frac{1}{z}$ and Laurent expand the function about $w=0$ we have:
$$\frac{1}{w(1-w^2)^{\frac{1}{2}}}=\frac{1}{w}+\frac{1}{2}w+\frac{3}{8}w^3+...$$
This implies that there is a simple pole about $w=0$, because the inverse powers of $w$ only extend to $-1$.
However, is we write the expansion in terms of $z$:
$$\frac{z^2}{(z^2-1)^{\frac{1}{2}}}=z+\frac{1}{2z}+\frac{3}{8z^3}+...$$ 
I'm not sure how to read this. How does this series, when written in terms of $z$, imply there being a pole at $z=\infty$?
Furthermore, I am working through a problem that asks me to use this Laurent expansion to evaluate $\int_{C_{\infty}}\frac{z^2}{(z^2-1)^{\frac{1}{2}}}$, where $C_{\infty}$ is the circle at infinity. If my expansion is correct how would I go about solving this integral? Surely the integral is infinite? 
 A: I think your error here is where you expanded the function. Notice that 
$$g(z)=\frac{z^2}{(z^2-1)^{\frac{1}{2}}}$$
has no singularity at zero, but it does have a singularity at $z=-1$ and $z=1$. You cannot use the Laurent series about $z=0$ of
$$f(z)=\frac{1}{({z^2-1})^{\frac{1}{2}}}$$
to evaluate something as $z\rightarrow\infty$. We know that the Laurent series for a function about a point $z_0$ converges everywhere inside a circle of radius equal to the distance from $z_0$ to the closest singularity to $z_0$. Thus the radius of convergence is $R=1$ for the Laurent series about $z=0$. For this reason, there is no series expansion that will work for $f(z)$ as $z\rightarrow\infty$ because the radius of convergence will always be capped by however far your $z_0$ is from $z=1$. With all of that said, there is no reason to use the Laurent series for $f(z)$ because $f(z)$ converges near infinity. We simply see that 
$$\lim_{z\rightarrow\infty}g(z)=\lim_{z\rightarrow\infty}\frac{z^2}{(z^2-1)^{\frac{1}{2}}}=\lim_{z\rightarrow\infty}z$$
which is equivalent to (if $w=z^{-1}$)
$$\lim_{w\rightarrow 0}\frac{1}{w}$$
so $g(z)$ has a pole of order $1$ as $z\rightarrow \infty$.
A: This answer is meant to address the precise notion of pole/residue at infinity, which seem to confuse the OP.
The term residue at infinity might be confusing, and it is common to think it is just the residue of the function $f\left(\frac{1}{z}\right)$ at $z=0$. This is, however, incorrect.
To talk about infinity as a point, we would like to consider a new space $\hat{\mathbb{C}}=\mathbb{C}\cup\left\{\infty\right\}$. You can think of this new space as a sphere (known as the Riemann sphere). Why is this important? Because now you can imagine that winding the origin (lower pole) counter-clockwise is identical to winding $z=\infty$ (upper pole) clockwise. This means we can write
$${\rm Res}\left(f\left(z\right),z=\infty\right)=\color{red}{-}\dfrac{1}{2\pi i}\oint_{\left|z\right|=R}f\left(z\right){\rm d}z$$
given that $f$ is holomorphic in the annulus $R<\left|z\right|<\infty$. Now let $z=\frac{1}{w}$, so ${\rm d}z=-\frac{1}{w^{2}}{\rm d}w$ and
$$=\dfrac{1}{2\pi i}\oint_{\left|w\right|=\frac{1}{R}}f\left(\dfrac{1}{w}\right)\left(-\dfrac{1}{w^{2}}\right){\rm d}w=-\dfrac{1}{2\pi i}\oint_{\left|w\right|=\frac{1}{R}}\dfrac{1}{w^{2}}f\left(\dfrac{1}{w}\right){\rm d}w=-{\rm Res}\left(\dfrac{1}{w^{2}}f\left(\dfrac{1}{w}\right),w=0\right)$$
Note the extra minus in the first integral. It is because when you change variables to $w=\dfrac{1}{z}$ the direction we trace the circle in the integration is reversed. Therefore, the residue at infinity is in fact
$${\rm Res}\left(f\left(z\right),z=\infty\right)=-{\rm Res}\left(\dfrac{1}{z^{2}}f\left(\dfrac{1}{z}\right),z=0\right)$$
and not ${\rm Res}\left(f\left(\frac{1}{z}\right),z=0\right)$ as you might think. For example, the residue of $f\left(z\right)=\frac{1}{z}$ at infinity is minus the residue of $\frac{1}{z^{2}}\frac{1}{\frac{1}{z}}=\frac{1}{z}$ at the origin, i.e. minus one.
