Complex Analysis/Topology Proof Help

In summary: Thanks in advance for your time and all the wonderful previous answers I've received lurking on this site - its been great!In summary, the two curves cannot be homotopic over the region C - {0}, but they can be homotopic over the entire complex plane. You can check this by computing the winding numbers around the origin of the two curves.
  • #1
MurraySt
8
0
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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  • #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!
 
  • #3
Thanks for the fast reply.

By singularity you mean not defined at 0? Such as 1/z?
 
  • #4
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?
 
  • #5
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
 

Related to Complex Analysis/Topology Proof Help

1. What is the purpose of complex analysis/topology?

The purpose of complex analysis/topology is to study the properties and behavior of complex numbers and topological spaces, respectively. This includes understanding the geometric and algebraic structures of these objects and how they relate to each other.

2. What are some common techniques used in proofs for complex analysis/topology?

Some common techniques used in proofs for complex analysis/topology include using the properties of complex numbers and topological spaces, using mathematical induction, and constructing counterexamples.

3. How do I know which theorems to use in a proof for complex analysis/topology?

The best way to determine which theorems to use in a proof for complex analysis/topology is to carefully read and understand the question and then review the relevant theorems and definitions. It may also be helpful to discuss the problem with other mathematicians or consult textbooks or online resources.

4. What is the role of continuity in complex analysis/topology?

Continuity plays a crucial role in complex analysis/topology as it helps to define and understand the behavior of functions and spaces. In complex analysis, continuity is used to study the properties of complex functions, while in topology, continuity is used to understand the connectedness and compactness of topological spaces.

5. How can I improve my skills in writing proofs for complex analysis/topology?

To improve your skills in writing proofs for complex analysis/topology, it is important to practice regularly and to seek guidance from experienced mathematicians or instructors. It is also helpful to read and analyze well-written proofs and to actively engage in discussions and problem-solving sessions with others.

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