TransWikia.com

Closed curve not passing origin and winding number (complex analysis)

Mathematics Asked by 10understanding on December 6, 2021

I am pretty sure this is elementary, but I am having a hard time understanding winding number (although intuitively I think I get it). My reference is Brown & Churchill book, section 86 on argument principle.

My main concern is only about the independent of choice of starting point. Why is winding number independent of the choice of the starting point? Shouldn’t it include modulo $2pi$?

My understanding is that a closed curve $w(t)$ not passing the origin, say $0leq tleq 2$ (so that $w(0)=w(2)$), can be thought as a curve with parameter over any real $t$ such that $w(t)=w(t+2)$. This means that $w(1)=w(3)$, etc, does it not? If we fix an argument value for $arg(w(0))$ (since else it is multi-valued), I get why $A:=arg(w(2))-arg(w(0))$ is fixed. Similarly, I get why $B:=arg(w(3))-arg(w(1))$ is fixed, etc for any period $2$.

However, using equality of complex numbers in polar coordinates, I cannot deduce that $A=B$, but only $A=B+2pi k$ for some $k$. In order to continue with the definition, I must make sure that $k=0$ always, right?

If I imagine some pictures, I surely understand the equality, but I cannot seem to convince myself about it rigorously. Maybe it has something to do with the continuity of the parameter?

Any suggestion is appreciated. Thank you very much

One Answer

Let $gamma:[0,1]tomathbb C$ be a piecewise smooth, closed curve. Let $Tin[0,1]$ and $gamma_1:=gammavert_{[0,T]}$ and $gamma_2:=gammavert_{[T,1]}$. Basically just cut the curve into two parts at some arbitrary $T$. Now the curve with the original choice of starting point is $gamma=gamma_2gamma_1$, where $betaalpha$ denotes the curve which first follows $alpha$ and then $beta$. With an arbitrary choice of $T$, we can now write the curve with a different choice of starting point as $tildegamma=gamma_1gamma_2$. It's starting at $gamma(T)$, the starting point of $gamma_2$, and then follows the original curve from there.

Clearly, we have $int_{betaalpha}f(z)mathrm dz=int_alpha f(z)mathrm dz+int_beta f(z)mathrm dz$, and then we can just switch the order of summation however we want, so it doesn't matter wether we integrate over $gamma_2gamma_1$ or $gamma_1gamma_2$. This makes any integral over a closed curve independent of the choice of starting point.

Answered by Vercassivelaunos on December 6, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP