Integrate over complex contour

  • Thread starter beefcake24
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  • #1
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Evaluate the integral over the contour C when:

f(z) = 1/z and C = {z(t) = sin(t) + icos(t) | 0 <= t <= 2*pi}


I know f(z) = 1/r*e^(-it) = 1/r(cos(t) + isin(t)). But, when I try to take the contour integral by integrating f[z(t)]*z'(t), I get really messy formulas ((1/r*cos(sin(t)) + i*1/r*sin(cos(t)))*(cos(t) + -i*sin(t)), which makes me think I'm missing something.

Can anyone help me out on this? I have a midterm tomorrow and this was one of the practice questions.
 

Answers and Replies

  • #2
lanedance
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what does the contour look like? do you think you could use the residue theorem?
 
  • #3
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I don't know, we haven't covered the residue theorem yet, and we do not need to know the residue theorem for our midterm, so I think we should be able to solve this without it.
 
  • #4
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The contour is just the unit circle, but starting at (0,i) when t=0 instead of (0,0) and traversed clockwise. Does this mean I can just take the negative of the integral around the unit circle? I know traversing a path in the opposite direction just changes the sign of the integral, and I don't think having a different starting point should affect the value, because you are traversing the same path. Can someone confirm this, or if I'm wrong, point me in the right direction?
 
  • #5
HallsofIvy
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Yes, so it would be better to write z in polar form. [itex]z= cos(t)+ i sin(t)= e^{it}[/itex]
What are 1/z and dz?

That is pretty close to trivial to integrate.

Just as in real integration, integrating over the same path in the opposite direction (you would be integrating from [itex]t= 2\pi[/itex] to 0 rather than vice-versa, changes the sign.

If you want to do it more "directly", remember that [itex]sin(z)= (e^{iz}- e^{-iz})/2i= -i(e^{iz}- e^{-iz})/2[/itex] and [itex]cos(x)= (e^{iz}+ e^{-iz})/2[/itex] so that [itex]sin(z)+ icos(z)= -i(e^{iz}- e^{-iz})/2+ i(e^{iz}+ e^{-iz})/2= e^{-iz}[/itex]. Again, just the previous formula with -iz rather than iz so that [itex]dz= -ie^{-iz}dz[/itex], and now you integrate from 0 to [itex]2\pi[/itex].
 
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  • #6
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Perfect, that was just what I needed to finish the problem. Thanks!
 

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