1. ## Contour Integration

Hi, I need some help with my Complex analysis homework.
It is required to evaluate the integral

in TWO different ways using contour integration.

I have to consider the function

with a branch cut on [-1,1] Re(z) given the choice of

I need to find the value of:
around a circle of large radius

Which I am struggling with.
So far, I know it has cut from -1 to +1, and there are poles at +i and -i
Any help is appreciated.
Thanks.

2. First thing to know is that the multi-function $\sqrt{1-z^2}$ is analytic outside it's hull of zeros (just any contour larger than the branch cut from -1 to 1). So the large black contour in the figure is certainly outside the hull of zeros so we can evaluate it by taking the residue at infinity since the integration is over a function analytic outside of some domain (the hull of zeros). But we can integrate the remaining contours in the figure over this sheet and therefore the value of the contour integrals over the internal contours would be equal to that over the black contour:

$\mathop\int\limits_{Blue} \frac{\sqrt{1-z^2}}{1+z^2}dz+\mathop\int\limits_{Green} \frac{\sqrt{1-z^2}}{1+z^2}dz+\mathop\int\limits_{\substack{branc h \\ points }}f(z)dz+2\pi i (r_1+r_2)=\mathop\oint\limits_{Black}\frac{\sqrt{1-z^2}}{1+z^2}dz$

Where $r_1,r_2$ are the residues at the poles and I included the integrals around the branch points for completeness but these tend to zero as the contour radius around them goes to zero. But that expression is dependent on the direction of the integration (may need some negative signs in there). Also, the residues at each pole are multi-valued since this is a multi-function and therefore are dependent on which sheet the integration is being performed over. So in the case of a square root, the two residues at each pole are just the negative of one another right? So, can you convert that expression into one which contains your real integral?