# Winding number of two closed courves in C with the constraint ∀t:|c1(t)−c2(t)|≠|c1(t)|+|c2(t)|.

Show that the curves $c_1,c_2$ are running in $\mathbb C^{\times}$ with $win(c_1,0)=win(c_2,0)$. Hint: How does the curve $c=\frac{c_1}{c_2}$ look like?

First I tried to find out how $c_1,c_2$ behave to eachother.

It follows from the inequation that the curves $c_1,c_2$ never run trough $0$. Suppose $c_2=0$, then we get the contradiction $|c_1|\neq |c_1|$. And for $c_1=0$ we get the contradiction $|-c_2|\neq |c_2|$ . $\Rightarrow c_1,c_2\neq 0$.

Then I wanted to show that the curves are allowed to touch/cross each other. Suppose $\exists t$ such that $c_1(t)=c_2(t)$. Then $|c_1-c_1|\neq |c_1|+|c_1|\Rightarrow 0\neq c_1$, which is true like I showed before.

In a previous task I showed that $win(c,0)=win(c_1,0)-win(c_2,0)$, with closed curves $c_1,c_2$ in $\mathbb C^{\times}$ and $c=\frac{c1}{c2}$. So I need to show that the winding number $win(c,0)$ equals zero.

I know a winding number is zero if the point (in our case $0$) is outside the winding. Unfortunatly I don't know how to show this. I have also problems to imagine how $c$ looks like. Any help appriciated.

## 1 Answer

From the constraint $$|c_1(t)-c_2(t)|\ne|c_1(t)|+|c_2(t)|$$ it follows that, for all $t$,

• $c_1(t)$ and $c_2(t)$ are nonzero.$\\[4pt]$
• $c_1(t)$ and $c_2(t)$ are not in opposite directions.$\\[4pt]$

hence, for all $t$, we have $c(t)\notin \{a\mid a \le 0\}$.

Assuming $c_1,c_2:[0,1]\to \mathbb{C}^{*}$ are loops with $c_1(0)=c_1(1)$, and $c_2(0)=c_2(1)$, it follows that $c:[0,1]\to \mathbb{C}^{*}$ is a loop with $c(0)=c(1)$.

Then, since $c$ is everywhere nonzero, and does not intersect the negative $x$-axis, it follows that ${\text{win}}(c,0)=0$.

As far as what $c$ "looks like", the only relevant features are:

• $c$ is a loop in $\mathbb{C}^*$.
• $c$ does not intersect the negative $x$-axis.
• Thanks for your help! I have still one part which I can't understand: I assume by opposite directions u mean the direction of the winding (clockwise/counterclock wise). I don't see why this follows from the inequation. – Mikemathics May 27 '18 at 19:11
• Two nonzero complex numbers (regarded as vectors), are said to be in the same direction if each is a scalar multiple of the other by a positive real scale factor. They are said to be in opposite directions if each is a scalar multiple of the other by a negative real scale factor. So saying $c_1(t),c_2(t)$ are never in opposite directions is equivalent to saying there is no value of $t$ such that they are both nonzero, and each is a negative multiple of the other. – quasi May 27 '18 at 19:19
• Note: For nonzero vectors $u,v$, by the triangle inequality, we get $$|u-v| \le |u|+|v|$$ with equality if and only if $u,v$ are in opposite directions. – quasi May 27 '18 at 19:33