1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Complex line integrallll

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data

    for [itex]n=1,2,3,...[/itex] , evaluate the integral,

    [itex]I= \int_C\frac{e^{iz}}{z^n}dz[/itex]
    where [itex]C [/itex] is a curve like [itex]z(t)=e^{it}[/itex] and [itex]0 \leq t \leq 2\Pi[/itex]

    2. Relevant equations



    3. The attempt at a solution
    I tried to use Cauchy integral formula; that

    [itex]f^{(n)}(z)=\frac{n!}{2 \Pi i}\int_C\frac{f(\zeta)}{(\zeta-z)^{n+1}}d\zeta[/itex]

    then we can obtain,

    [itex]f^{(n-1)}(z)=\frac{(n-1)!}{2 \Pi i}\int_C\frac{f(\zeta)}{(\zeta-z)^{n}}d\zeta[/itex]

    [itex]f^{(n-1)}(z)=\frac{(n-1)!}{2 \Pi i}\int_C\frac{f(\zeta)}{(\zeta-z)^{n}}d\zeta[/itex]

    [itex](e^{iz})^{(n-1)}(z) \Big\vert_{z=0}=\frac{(n-1)!}{2 \Pi i}\int_C\frac{e^{iz}}{z^{n}}d\zeta[/itex]

    [itex]i^{n-1}=\frac{(n-1)!}{2 \Pi i}\int_C\frac{e^{iz}}{z^{n}}d\zeta [/itex]

    [itex]\int_C\frac{e^{iz}}{z^{n}}d\zeta=\frac{i^{n-1} 2 \Pi i} {(n-1)!} [/itex]

    [itex]~~~~~~~~=\frac{i^n 2 \Pi} {(n-1)!} [/itex] .

    can you check, is it right??????
     
  2. jcsd
  3. Sep 14, 2011 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yes, that's correct.

    You could also expand the integrand as a Laurent series and use the residue theorem.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook