Complex Analysis/Topology Proof Help

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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?
 
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Hi MurraySt! :smile:

It seems that you need to find a complex function f:\mathbb{C}\setminus\{0\}\rightarrow \mathbb{C}, 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!
 
Thanks for the fast reply.

By singularity you mean not defined at 0? Such as 1/z?
 
MurraySt said:
Thanks for the fast reply.

By singularity you mean not defined at 0? Such as 1/z?

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?
 
MurraySt said:
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?

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

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