Let $\gamma:[0,1]\to \mathbb C$ be continuous, and not passing through $0$. How can we prove that, using complex analysis, there is a continuous $G:[0,1]\to \mathbb C$ so that $\gamma=e^G$ ?
This can be done in the $C^1$ case, using $G(x)=\int_0^x\gamma'(t)/\gamma(t)dt$. To pass to the more general continuous case, one can convole $\gamma$ with a mollifier, $\gamma_r=\gamma*\phi_r$, so that $\gamma_r=e^{G_r}$. $\gamma_r$ converges to $\gamma$ uniformly, as $r\to 0$, but I do not know if the $G_r$ will also converge uniformly. If they did, then the uniform limit would be a continuous function $G$, and $\gamma=e^G$.