Evaluate Integral with Residue Theorem: Residue Theorem Homework

[jjangub]

Homework Statement

Evaluate \int_{0}^{2\pi} (cos^4\theta + sin^4\theta) d\theta 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\theta + sin^4\theta) to 1-2cos^2\theta+2cos^4\theta (I will skip this one, this is not hard)
and then coverting to a complex integral by inputting cos\theta = (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\pii \sum (Residue of f on upper half plane)
= 2\pii(-1/8i)Res₀
= -\pii/4Res₀
since Res_a = 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
= (-\pii/4)*(-12/4!) = \pi/8
I did it in this way and I got this answer.
Did I do anything wrong?
Thank you.

 

[vela]

Two mistakes. First, you have an algebra mistake somewhere. Your integrand should be

\frac{z^8+6z^4+1}{8iz^5}

Second, your contour is wrong. You're using the substitution z=e^iθ. As θ runs from 0 to 2π, z moves along the unit circle. That should be your contour.

 

[jjangub]

yes, I made a mistake.
Now, I got that equation. I know this quesetion sounds silly, but I am kind of lost on cauchy integral and residue theorem. so do we have different method for when we have 1 order of pole, 2 order of pole or higher, and for cos and sin?
Is there any good site which is well and easily organized about this?
And for this question, I got c₀=e^(pi*i/5) and c₁=3(pi*i/5)
but I can't get exact value of pi/5 for cos and sin...
maybe I did wrong...
Thank you.

 

[vela]

yes, I made a mistake.
Now, I got that equation. I know this quesetion sounds silly, but I am kind of lost on cauchy integral and residue theorem. so do we have different method for when we have 1 order of pole, 2 order of pole or higher, and for cos and sin?

The easiest, and probably the least error-prone, method is to expand the function as a Laurent series about each pole and then pick off the coefficient of the (z-z₀)^-1 term. In this problem, you have a (fifth-order) pole inside the contour at z=0, and the integrand is already a Laurent series about z=0:

-\frac{i}{8} z^{-5} -\frac{3i}{4} z^{-1} - \frac{i}{8} z^3

so the residue is -3i/4. This method, of course, requires proficiency in expanding functions as series, but once you acquire the skill, I think you'll find it's the easiest and quickest way to find the residue.

You can also find the residue of an n-th order pole using the formula

\frac{1}{(n-1)!} \lim_{z \to z_0} \frac{d^{n-1}}{dz^{n-1}} [(z-z_0)^n f(z)]

For this problem, this expression is pretty easy to evaluate, but when you have a more complicated function, you end up having to apply the product rule, chain rule, quotient rule, etc., and the algebra can easily become messy.

That said, either method is pretty straightforward conceptually. It's just a matter of doing the actual calculation correctly. That skill is something you'll develop with time and practice doing these problems.

Is there any good site which is well and easily organized about this?
And for this question, I got c₀=e^(pi*i/5) and c₁=3(pi*i/5)
but I can't get exact value of pi/5 for cos and sin...
maybe I did wrong...
Thank you.

Sorry, I'm not sure what c₀ and c₁ are.

 

[jjangub]

maybe this will be the last question...I hope
so if I convert polar integral (with cos and sin) to complex integral, then
it changes from integral(zero to 2pi) to integral(negative infinity to positive infinity)?
or it works only if it is unit circle?
so i end up with Res(z=-3i/4)f = (2*pi*i)(-3i/4) = 3pi/2
I really appreciate your help...
Thank you so much!

 

[vela]

maybe this will be the last question...I hope
so if I convert polar integral (with cos and sin) to complex integral, then
it changes from integral(zero to 2pi) to integral(negative infinity to positive infinity)?

The substitution you use to convert the integral is z=e^iθ. What path is traced out in the complex plane as θ runs from 0 to 2π?

or it works only if it is unit circle?
so i end up with Res(z=-3i/4)f = (2*pi*i)(-3i/4) = 3pi/2
I really appreciate your help...
Thank you so much!

 

[jjangub]

so...path is 0 to infinity?
so I have to multiply by 1/2 to 3pi/2?
well...this cauchy and residue theorem parts are so confusing for me...
in class, prof gave us the easy example, so I understood, but the question for assignment and tests are hard...
I have test next week...how should I study cauchy and residue theorem?
just memorize theorem and do examples?
Thank you...

 

[vela]

No. Where are you getting infinity from?

 

[jjangub]

well, if i do it again, i got
z = e^i(0) = 1 and z = e^i(2pi) = cos(2pi) + isin(2pi) = 1
is this right?

 

[vela]

You have z=e^iθ=cos θ+i sin θ. What path does this function map the interval [0,2π] onto?