$\int \frac{dz}{z\sqrt{(1-{1}/{z^2})}}$ over $|z|=2$ I need help in calculating the integral of $$\int \frac{dz}{z\sqrt{\left(1-\dfrac{1}{z^2}\right)}}$$ over the circle $|z|=2$. (We're talking about the main branch of the square root).
I'm trying to remember what methods we used to calculate this sort of integral in my CA textbook. I remember the main idea was using the remainder theorem, but I don't remember the specifics of calculating remainders....
Thanks you for your help!

$$\frac{1}{z \sqrt{1-\frac{1}{z^{2}}}} =  \frac{1}{z} \Big( 1 - \frac{1}{2z^{2}} + O(z^{-4}) \Big) \text{for} \ |z| >1 \implies \int_{|z|=2} \frac{1}{z \sqrt{1-\frac{1}{z^{2}}}} \ dz = 2 \pi i (1) = 2 \pi i $$
