Homotopy between closed curves. Exercise: if two closed paths of $\mathbb{C}-\{0\}$ have same index with respect to zero, show that they are homotopic as closed paths in $\mathbb{C}-\{0\}$.
My question: why does this prove
depends on the value of the index as one could obtain a function such as $h(s,t)=(1-s)\gamma_{1}(t)+s\gamma_{2}(t)$ that works as an homotopy for both paths. 
 A: The idea here is to take a logarithm of both curves. 
Let $\gamma_0, \gamma_1: [0, 1] \to \mathbb{C} - \{0\}$ be two closed curves, such that $\gamma_i(j) = 1$ for $i, j = 0, 1$. We need to construct $\omega_1, \omega_2: [0, 1] \to \mathbb{R}$ such that $\exp(2 \pi i \omega_i(t)) = \gamma_i(t)$. To construct them, we start from a logarithm branch $\log_0: \mathbb{C} - (-\infty, 0] \to \mathbb{C}$ such that if $z = |z|(\cos(\alpha) + i \sin(\alpha))$ for $\alpha \in (-\pi, \pi)$, then $\log_0(z) = \log|z| + i \alpha$. We take logarithm of $\gamma_0(t)$ for $t$ starting from 0, and before we hit $t$ such that $\gamma_0(t) \in (-\infty, 0)$, we change the logarithm branch that matches $\log_0$, but has $(-\infty, 0)$ in its domain. This way we can logarithm the whole curve. By construction, $\omega_0(0) = 0$, and as $\gamma_0(1) = 1$, $\omega_0(1) = k$ for some integer $k$.  
We do the same with $\gamma_1$, and since they have the same index, $\omega_1(1) = k$ for the same $k$ as $\omega_1$. Thus $\omega_i$ share the same beginning and the end.
Since the codomain of $\omega_i$ is whole $\mathbb{R}$, we can easily find the homotopy that connects $\omega_0$ with $\omega_1$ that keeps the ends constant.  Taking $\exp$ of that homotopy gives a homotopy between $\gamma_i$.
