Let $\arg\colon\mathbb{C}\setminus\{0\}\to(-\pi,\pi]$ be the argument function. Suppose that $\gamma\colon[a,b]\to\mathbb{C}$ is a smooth closed (i.e. $\gamma(a)=\gamma(b)$) curve around the origin. Here smooth means that $\gamma$ is a $C^{\infty}$-submanifold of $\mathbb{C}\cong\mathbb{R}^{2}$.
Question: Why is $$\arg\circ\gamma\colon[a,b]\to(-\pi,\pi]$$ differentiable almost everywhere?
Attempt: The argument function is (real) differentiable on $\mathbb{C}\setminus(-\infty,0]$. I think the problematic points can be described as follows. Let's call $z_{0}\in\gamma$ a "jump" point if $z_{0}\in(-\infty,0]$ and $\gamma$ intersects $D(z_{0},\delta)\cap\{\text{Im}(z)<0\}$ for all $\delta>0$. In other words, jump points are the points on $(-\infty,0]$ where $\gamma$ moves from $\{\text{Im}(z)\geq0\}$ to $\{\text{Im}(z)<0\}$, or the other way around. Indeed, this is where the argument function makes jumps. I think $\arg\circ\gamma$ is not differentiable precisely at those $t_{0}$ for which $\gamma(t_{0})$ is special. If the latter claim is true, then it would suffice to prove that the set $$\{t_{0}\in[a,b]:\gamma(t_{0}) \ \text{is a jump point}\}$$ is a null-set.