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??

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.
