# I need another easier answer for this question

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I need another easier answer for this question

Nov 16th, 2015

We will use the semicircle as the contour of integration. Just for simplicity, to justify this integration, the integral on the arc going from : π to -π, vanishes, so I only need to evaluate the integral on the x-axis, i.e. real line:

The function of choice should be :

I need to find the roots of the following equation:Let z = (re^iθ)

write -1 as: -1 = e^(iπ +i2πn), (z^4) = = -1, becomes: (r^4)(e^(i4θ)) = (e^i(π+4nπ))

Equate magnitude to magnitude, to get : r^4 = 1, hence r = 1.

Equate argument to argument to get: , let n = 0,1,2,3 to get all the roots:

They are:

Let the zeros be denoted:

Rewrite f(z) as followed: f(z) =

Only the first two zero are inside the semicircle:

The residue at , the math is messy, but it's:

The residue at

Then 2πi times by the sum of the residue at z1 and z2 gives you the desired answer:

This is basically Cauchy's integral formula, which is a special case of the Residue theorem.

Nov 16th, 2015

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Nov 16th, 2015
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Nov 16th, 2015
Nov 17th, 2017
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