I need another easier answer for this question

Physics
Tutor: None Selected Time limit: 1 Day

I need another easier answer for this question
http://www.cramster.com/answers-jun-11/physics/calculate-residuse-calculatenbspnbsp-residuse-theorem_1342637.aspx?rec=0

Nov 16th, 2015

Thank you for the opportunity to help you with your question!

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.


Please let me know if you need any clarification. I'm always happy to answer your questions.
Nov 16th, 2015

Are you studying on the go? Check out our FREE app and post questions on the fly!
Download on the
App Store
...
Nov 16th, 2015
...
Nov 16th, 2015
Dec 7th, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer