Solving Trig Integrals with Residue Theorem

[dyn]

Hi.
I have looked through an example of working out a trig integral using the residue theorem. The integral is converted into an integral over the unit circle centred at the origin. The singularities are found.
One of them is z₁ = (-1+(1-a²)^1/2)/a
It then states that for |a| < 1 , z₁ lies inside the unit circle.
Should this be obvious just by looking at z₁ ? Because I can't see it. I have tried a few values and it seems to be true but that doesn't prove it. If its not obvious how do I go about trying to show it ?
Thanks

 

[mfb]

Hint: Let a=sin(x) for real |x|<pi/2 and simplify.

 

[dyn]

I must be missing something as the answer I get doesn't seem to help : -cosec x + cot x
Also I'm not sure why I can specify | x | < π/2

 

[mfb]

Hmm... I was assuming a is real but it is probably complex? It might still work but the range of x will need more thought.