Undergrad Solving Trig Integrals with Residue Theorem

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The discussion focuses on solving a trigonometric integral using the residue theorem, specifically examining the singularity z1 = (-1+(1-a^2)^(1/2))/a. It is noted that for |a| < 1, z1 lies within the unit circle, but the reasoning behind this is questioned. The user struggles to see why this is obvious and seeks clarification on how to demonstrate it. A hint is provided to let a = sin(x) for real |x| < π/2, but the user expresses confusion about the implications of a being real or complex. The conversation emphasizes the need for a deeper understanding of the conditions under which the singularity lies inside the unit circle.
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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
 
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Hint: Let a=sin(x) for real |x|<pi/2 and simplify.
 
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
 
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.
 

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