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Complex Analysis/Topology Proof Help

  1. Jun 15, 2011 #1
    Thanks in advance for your time and all the wonderful previous answers I've received lurking on this site - its been great!

    Anyway I have two functions G:[0,2pi] --> Complex Plane and
    H:[0,4pi] --> Complex Plane

    Both functions are equal to exp(it). (The complex exponential function w/ argument it).

    I am told that these functions are not homotopic over the region C - {0} and asked for a proof.

    Another hint is to use Cauchy's Theorem which says that if two curves are homotopic their integrals are equivalent.

    Any suggestions?
  2. jcsd
  3. Jun 15, 2011 #2
    Hi MurraySt! :smile:

    It seems that you need to find a complex function [itex]f:\mathbb{C}\setminus\{0\}\rightarrow \mathbb{C}[/itex], such that the two path integrals do not have the same value.

    Furthermore, something special must happen in 0. If the function f can be analytically extended to 0, then the two integrals are equal. So there must be some kind of singularity in 0.

    So, pick the easiest function which has a singularity in 0 and try it!
  4. Jun 15, 2011 #3
    Thanks for the fast reply.

    By singularity you mean not defined at 0? Such as 1/z?
  5. Jun 15, 2011 #4
    Yes, a singularity is (roughly) a point where the function is not defined. And 1/z was exactly the example I was looking for!

    What happens if you integrate 1/z with respect to these two paths?
  6. Jun 16, 2011 #5


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    Try computing the winding numbers around the origin of these two curves
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