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 Variables integration formulas

  1. May 6, 2008 #1
    1. The problem statement, all variables and given/known data
    Let C. denote the circle |z-z.|= R, taken counter clockwise. use the parametric representation z= z. + Re^(io) (-pi </= o </= pi) for C. to derive the following integration formulas:
    integral C. (dz/(z-z.)) = 2ipi

    2. Relevant equations
    note: z. and C. represent z knot and c knot , and </= represents less than or equal to
    and o represents theta

    3. The attempt at a solution

    i was able to do some work to it, but i eventually came out with (integral of) (R i e^io)/R which goes to integral of (i e ^(io))from -pi to pi, which comes out to e^(ipi) - e^(-ipi), which my teacher showed me comes out to zero. (i didnt quite understand his method, but i do know 0=/=2ipi
  2. jcsd
  3. May 6, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    It's a little hard to follow your notation. But for simplicity put 'z.'=0 and R=1. Then you are integrating (1/z)*dz where z=exp(i*theta). You are not integrating exp(i*theta)*d(theta) which it looks like you are doing. exp(i*pi)=(-1). exp(-i*pi)=(-1), so difference is zero. Your teacher was right about that.
    Last edited: May 6, 2008
  4. May 6, 2008 #3
    hmm. so it seems to work if 'z.' = 0, but how can i say that? same goes for R = 1
    is it because z. is just some arbitrary point i can pick?
    we've done integral of (1/z)*dz, where z= R(exp(i*theta)) and dz = R*i(exp(i*theta))d(theta)
    that comes out to be 2i*pi, but we were also doing over the interval of (0 to 2pi)

    we're supposed to prove it in general i believe, so i would like to try and leave the z. and R in there.

    i think im seeing the relationship now...do i just say let z= z-z. and that yields the same (1/z) integral ive done before?

    sorry bout sorta train of thought post...just trying to keep you updated with what im doing :)
  5. May 6, 2008 #4


    User Avatar
    Science Advisor
    Homework Helper

    If you did it with z.=0 and R=1 I think you'll have no problems working it out in general. Yes, you'll get the same thing. You should notice the exponential part and the R just plain cancel. Welcome to PF.
  6. May 6, 2008 #5
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook