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Complex Variables integration formulas

  • Thread starter MadCow999
  • Start date
  • #1
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Homework Statement


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


Homework Equations


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

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
T_T
thanks!
 

Answers and Replies

  • #2
Dick
Science Advisor
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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:
  • #3
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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 :)
 
  • #4
Dick
Science Advisor
Homework Helper
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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.
 
  • #5
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okiedokie
thanks!
 

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