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 Analysis Proof

  1. Apr 26, 2009 #1
    1. The problem statement, all variables and given/known data
    C = positively oriented simple closed piecewise smooth path

    Prove that:


    is the area enclosed by C.

    2. Relevant equations

    *I know that the curve C is piecewise smooth so that it can be broken up into finitely many pieces so that each piece is smooth.

    *Cauchy's integral formula

    3. The attempt at a solution

    I think that I want to let some function f(t) be a continuous complex-valued function on the path C. Let g(t) be a parametrization of the curve. Frankly, I'm not sure where to go from here. I've just started doing contour integration problems and I have problems knowing where to start with proofs.

    I'm just hoping someone can point me in the right direction on where I might want to begin. Thanks.
  2. jcsd
  3. Apr 26, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    You can compute the area inside a contour using Green's theorem as (1/2) integral xdy-ydx. Break your complex contour into real and imaginary parts.
  4. Apr 26, 2009 #3


    User Avatar
    Science Advisor
    Homework Helper

    A consequence of Green's Theorem is that the area enclosed by C is:

    (1/2)\int_C xdy - ydx

    which is

    (1/2)\int _0 ^1 x(t)y'(t)dt - y(t)x'(t)dt

    where g : [0,1] -> C, g(t) = x(t) + iy(t) is your parametrization of C.
  5. Apr 26, 2009 #4
    thank you very much for your help.
  6. Apr 27, 2009 #5
    Thank you again for the assistance. I have almost gotten to the conclusion that I need but I'm not sure how to proceed.

    I have shown that:

    (1/2)\int_{0}^{1} \bar{z}dz = (1/2)\int_{0}^{1} x(t)y'(t)dt-y(t)x'(t)dt + (1/2i)\int_{0}^{1} x(t)x'(t)dt-y(t)y'(t)dt

    I am assuming that the integral \int_{0}^{1} x(t)x'(t)dt-y(t)y'(t)dt is equal to zero because if so I will obtain the result I am looking for, namely that:

    (1/2)\int_{0}^{1} \bar{z}dz = (1/2)\int_{0}^{1} x(t)y'(t)dt-y(t)x'(t)dt

    I'm not sure how to evaluate \int_{0}^{1} x(t)x'(t)dt-y(t)y'(t)dt.

    Any ideas would be much appreciated.
  7. Apr 27, 2009 #6


    User Avatar
    Science Advisor
    Homework Helper

    Use Green's theorem on it.
  8. Apr 27, 2009 #7
    Thank you sir! Solved! Not sure how to update thread title.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook