# I need another easier answer for this question

Anonymous
account_balance_wallet \$5

### Question Description

I need another easier answer for this question

anusha_reddy
School: UC Berkeley

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.

flag Report DMCA
Review

Anonymous
Top quality work from this guy! I'll be back!

Brown University

1271 Tutors

California Institute of Technology

2131 Tutors

Carnegie Mellon University

982 Tutors

Columbia University

1256 Tutors

Dartmouth University

2113 Tutors

Emory University

2279 Tutors

Harvard University

599 Tutors

Massachusetts Institute of Technology

2319 Tutors

New York University

1645 Tutors

Notre Dam University

1911 Tutors

Oklahoma University

2122 Tutors

Pennsylvania State University

932 Tutors

Princeton University

1211 Tutors

Stanford University

983 Tutors

University of California

1282 Tutors

Oxford University

123 Tutors

Yale University

2325 Tutors