# Winding number in complex plane homework

If $\gamma_1,\gamma_2$ are closed paths s.t. $|\gamma_1(t)|>|\gamma_2(t)|$ for all $t$, and $\Gamma(t) := \gamma_1(t) + \gamma_2(t)$, show that $n(\Gamma,0) = n(\gamma_1,0)$, where $n$ is the winding number defined by $n(\Gamma,0) :=\frac{1}{2\pi i} \int_\Gamma \frac{dz}{z}$.

I have tried looking at $n(\Gamma,0) - n(\gamma_1,0)$ but am still stuck. Please can somebody give me a hint??

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Do your paths have to be connected? If so, then you need to modify your quantifier about when the modulus inequality holds. – Sammy Black May 30 '13 at 22:36
@SammyBlack A path is by definition a continuous function from $[0,1]$ to $\mathbb{C}$ so it must be connected? The modulus inequality holds for all $t\in [0,1]$. – user71815 May 30 '13 at 22:42
Oh, I'm sorry. I was misreading $\Gamma$ as the concatenation of the two paths $\gamma_1$ and $\gamma_2$. (This is a common alternative operation on paths that's denoted by $+$.) In that case, the endpoints would have to agree. – Sammy Black May 30 '13 at 22:48
@Sammy Of course! In that case we have $n(\Gamma,0) = n(\gamma_1,0) + n(\gamma_2,0)$ trivially. – user71815 May 30 '13 at 22:51

HINT:

Consider the homotopy $H(t,s): \mathbb{C}\times[0,1] \to \mathbb{C}$ with $H(t,0) = \gamma_1(t)$, $H(t, 1) = \Gamma(t)$, and for $0 < s < 1$ assign $H(t,s)$ to be the corresponding point on the line segment connecting $\gamma_1(t)$ and $\Gamma(t)$. The realization is that since $|\gamma_1(t)| > |\gamma_2(t)|$, this smooth deformation of $\gamma_1$ into $\Gamma$ never passes through the origin. And therefore the winding numbers are the same.

Let's restate the same idea without any algebraic topology speak. First, draw a picture. Start with $\gamma_1(t)$, some closed path around the origin. We're going to deform $\gamma_1$ into $\Gamma$ without passing through the origin, so that the winding numbers are the same. To do this, for each $t$, imagine sending $\gamma_1(t)$ to $\Gamma(t)$ by having it move down the line segment connecting the points $\gamma_1(t)$ to $\Gamma(t)$. Notice that since $|\gamma_1(t)| > |\gamma_2(t)|$, these line segments do not pass through the origin and therefore we have deformed $\gamma_1$ into $\Gamma$ without passing through the origin.

Ok - this is less of a hint and more of the entire idea behind the solution. But there are some details left to be filled in, I think.

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To develop some intuition about this problem, imagine $\gamma_1$ as the orbit of the Earth around the Sun, and $\gamma_2$ as the orbit of the Moon around the Earth. Then, $\Gamma$ describes the motion of the Moon around the Sun. (Suppose, for simplicity that a year is exactly 12 months so that these are closed curves.)

Notice that the Moon winds around the Sun once for each time that the Earth does. The motion of the Moon relative to the Earth doesn't affect these winding numbers.

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