$\gamma_1,\gamma_2:[0,1] \to \Bbb C - \{0\}$ are closed paths and have the same index I need to prove that if $\gamma_1,\gamma_2:[0,1] \to \Bbb C - \{0\}$ are closed paths and have the same index ( or winding number) around $0$ , then they are homotopic.
So, I think I have an idea on how to show it:
we have $g_1,g_2:[0,1] \to \Bbb C$ continuous logarithms for $\gamma_1 , \gamma_2$ , that is $\gamma_i (t) = e \ ^ {g_i (t) } $ for $i=1,2$.
So it is enough to show that $g_1,g_2$ homotopic.
It seems easy to show that because $\Bbb C$ is convex, so i thought the homotopy would be :
$H(s,t) = (1-s) g_1(t) + sg_2(t)$
We need to show that $H(0,t) = \gamma_1(t) , H(1,t) = \gamma_2(t) $
 (which is easy) and that $H(s,0) = H(s,1) $ . to show this :
Suppose $k= n(\gamma_i , 0)$ is the index of the paths then we know that $k = \dfrac{g_i(1) - g_i(0)}{2\pi i}$
So $2\pi i k = g_1(1) - g_1(0) = g_0(1) - g_0(0)$
So $H(s,0) = H(s,1) $ iff $2 \pi i k = 0$ , and this is not required in the question.
What am I missing ? 
Thanks for helping.
 A: Take $H(s,t) = \exp\Big((1-s)g_1(t) + sg_2(t)\Big)$. I think this, with the assumption of the same winding number should solve it.
More info:
$Im(H) \subset \mathbb{C} - \{0\}$, this is obvious.
Set $s \in [0,1]$. Then $H(s,0) = exp(g_1(0))exp(s[g_2(0) - g_1(0)]) = exp(g_1(1))exp(s[g_2(1) - g_1(1)]) =  H(s,1)$ where the second equality is due to the paths being closed, $g_1$ being a logaritm, and the same winding number around  $0$.
The fact that $H(0,t) = \gamma_1$, $H(1,t) = \gamma_2$ is obvious.
This is then a homotopy in the wanted domain.
This gives a direct homotopy between the paths $\gamma_1, \gamma_2$. Using the exponential in the homotopy will result in a closed loop for all $s$ (use the assumption of the same winding number), which doesn't pass through $0$ (as opposed to not using the exponential as you did in your attempt).
A: Observe that for every $s$, the function $H\left(s,\cdot\right):\left[0,1\right]\rightarrow\mathbb{C}$ should also map under $\exp$ to a closed path with index $k$. Ergo, exactly as with the $g$'s
$$\frac{H\left(s,1\right)-H\left(s,0\right)}{2\pi i}=k$$
is the winding number of the path $\exp\Big(H\left(s,\cdot\right)\Big):\left[0,1\right]\rightarrow\mathbb{C}\setminus\left\{0\right\}$. Therefore, by requiring
$$H\left(s,1\right)=H\left(s,0\right)$$
you are unintentionally demanding that $k=0$.
