I'm given the function
$$f(z)=\frac{(z^2-1)^{1/2}}{z^2+1}$$
where $-\pi < arg(z \pm 1) \leq \pi$ and the only branch cut required is the section $[-1,1]$ of the real axis.
I'm required, using the asymptotic form of $f(z)$, to find $\oint f(z) \,dz$ around a circle of large radius. Firstly, i'm unsure what "the asymptotic form of $f(z)$" is referring to. I took this to mean a series expansion for $f$, so as $z \rightarrow 0$
$$f(z)=i-\frac{3i}{2}z^2+\frac{11i}{8}z^4-\frac{23i}{16}z^6+\cdots$$
Assuming this is correct, how would I start going about integrating this around a circle of large radius? If this isn't the asymptotic form of $f$, what is or at least how do I go about finding the asymptotic form? Thanks in advance.
Imagine a large circle of radius $R$. The integral you seek is
$$i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \frac{(R^2 e^{i 2 \theta}-1)^{1/2}}{R^2 e^{i 2 \theta}+1} $$
To get the asymptotic form, just pull out factors of $R$:
$$\begin{align}i \int_{-\pi}^{\pi} d\theta \, \frac{\displaystyle \left (1-\frac1{R^2 e^{i 2 \theta}} \right )^{1/2}}{\displaystyle 1+\frac1{R^2 e^{i 2 \theta}}}&= i \int_{-\pi}^{\pi} d\theta \, \left (1-\frac1{2 R^2 e^{i 2 \theta}}-\cdots \right ) \left (1-\frac1{R^2 e^{i 2 \theta}}+\cdots \right )\\ &= i 2 \pi - i \frac{3}{2 R^2} \int_{-\pi}^{\pi} d\theta \, e^{-i 2 \theta} - \cdots\end{align}$$
Note that all higher order contributions are equal to zero because of the integral of a single harmonic of an integer frequency over the unit circle. Thus, the integral over the circle is simply $i 2 \pi$, for sufficiently large radius. This is the so-called "residue at infinity."