Complex Integral

  • Thread starter Tangent87
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  • #1
Tangent87
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In our Complex Methods lecture today, our lecturer went through the example of evaluating the integral [tex]\int_{-1}^1 \frac{(1-x^2)^{1/2}}{x^2+1}dx[/tex] and then proceeded to do the whole contour calculation using the complex function [tex]\frac{(z^2-1)^{1/2}}{z^2+1}[/tex]. I'm concerned that the answer will be a factor of i out because [tex](1-z^2)^{1/2}=i(z^2-1)^{1/2} [/tex] but he said in the lecture it didn't matter, can anyone explain why?
 

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  • #2
dingo_d
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Maybe because in the end you are only interested in real solution (part)?

Just guessing here...
 
  • #3
Tangent87
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Maybe because in the end you are only interested in real solution (part)?

Just guessing here...

You're probably on the right lines but the real part of [tex](z^2-1)^{1/2}[/tex] is def not [tex](1-x^2)^{1/2}[/tex] because at the very least there'll be loads of y's involved as well since z=x+iy?
 
  • #4
dingo_d
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Oh, I was thinking that at the end of the integration you get something real and sth imaginary and then you take the real part...

Don't take that as a concrete answer, I too don't see how that wouldn't matter at first...

Because it really is not the same integral (by that factor).

Altho the residues do seem to cancel each other, so maybe there's the catch :\
 

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