# Complex Analysis/Topology Proof Help

1. Jun 15, 2011

### MurraySt

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. Jun 15, 2011

### micromass

Staff Emeritus
Hi MurraySt!

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!

3. Jun 15, 2011

### MurraySt

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

4. Jun 15, 2011

### micromass

Staff Emeritus
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. Jun 16, 2011

### lavinia

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