Related: https://math.stackexchange.com/questions/1441725/winding-number-and-cauchy-integral-formula
Let $G$ be an open connected subset of $\mathbb{C}$.
Let $\gamma:[0,1]\rightarrow G$ be a rectifiable curve.
Then, does there exist a $C^1$-curve $\Gamma:[0,1]\rightarrow G$ such that $\gamma$ and $\Gamma$ are homotopic relative to $\{0,1\}$ in $G$?
Yes; you only need $\gamma$ to be continuous and you can get $\Gamma$ to be $C^\infty$. By compactness, we can find finitely many open disks $U_1,\dots, U_n\subseteq G$ and points $t_0=0<t_1<\dots<t_n=1$ such that $\gamma([t_i,t_{i+1}])\subset U_{i+1}$ for each $i$. Let $\Gamma$ be any smooth path such that $\Gamma([t_i,t_{i+1}])\subset U_i$ and $\Gamma(t_i)=\gamma(t_i)$ for each $i$ (for instance, take the polygonal path between the points $\gamma(t_i)$ and use bump functions to smooth it out at the corners). Then $\Gamma$ is homotopic to $\gamma$ by a straight-line homotopy; the straight-line homotopy stays in $G$ because each $U_i$ is convex.
