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jjangub
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Homework Statement
Evaluate [tex]\int_{0}^{2\pi}[/tex] (cos^4[tex]\theta[/tex] + sin^4[tex]\theta[/tex]) d[tex]\theta[/tex] by converting it to a complex integral over the unit circle and applying the Residue Theorem.
Homework Equations
The Attempt at a Solution
First, I switch (cos^4[tex]\theta[/tex] + sin^4[tex]\theta[/tex]) to 1-2cos^2[tex]\theta[/tex]+2cos^4[tex]\theta[/tex] (I will skip this one, this is not hard)
and then coverting to a complex integral by inputting cos[tex]\theta[/tex] = (z+1/z)/2 and
multiply cos terms by 1/iz, so 1 - 1/iz(((-2)(z+1/z)/2)^2 + ((2)(z+1/z)/2)^4)
this is how prof. taught us...we have to use this formula
so if I do the work and make it simpler, then I get equal to
= (-z^8+2z^6+8iz^5+z^4+2z^2-1)/(8iz^5) in integral(-infinity to infinity)
= 2[tex]\pi[/tex]i [tex]\sum[/tex] (Residue of f on upper half plane)
= 2[tex]\pi[/tex]i(-1/8i)Res0
= -[tex]\pi[/tex]i/4Res0
since Resa = h(z)/(z-a)^n = h^(n-1)(a)/(n-1!)
so in this case, Res0 = h^4(0)/(4!)
h(z) = z^8-2z^6-8iz^5-z^4-2z^2+1
if we do h^4(0), we get -12, so
= (-[tex]\pi[/tex]i/4)*(-12/4!) = [tex]\pi[/tex]/8
I did it in this way and I got this answer.
Did I do anything wrong?
Thank you.