Let $\gamma,\sigma: [0,1]\mapsto \mathbb{C}$ be differentiable closed curves. Prove that if $|\gamma(t)-\sigma(t)|<|\gamma(t)|$. Then $n(\gamma;0)=n(\sigma;0)$ where $n(\gamma;0)$ denote the winding number of $\gamma$ around $0$
I tried to prove that $\int_\sigma \frac{1}{z} = \int_\gamma\frac{1}{z}$ by showing that $|\int_\sigma\frac{1}{z} - \int_\gamma\frac{1}{z}|=0$. However, I failed and I did not see how to use the condition $|\gamma(t)-\sigma(t)|<|\gamma(t)|$.
Note that neither $\gamma$ nor $\sigma$ can vanish at any point $t$. Let $\tau = \frac {\sigma} {\gamma}$. Then $\frac {\sigma '} {\sigma}=\frac {\gamma '} {\gamma}+\frac {\tau '} {\tau}$. Since $\int_{\gamma} \frac 1 z \, dz=\int \frac {\gamma ' (t)} {\gamma (t)} dt$, etc we only have to show that $\int \frac {\tau '} {\tau}=0$. The hypothesis gives $|\tau -1| <1$. Since there is an analytic branch of logarithm on the disk $\{z:|1-z|<1\}$ we can write $\tau = e^{\xi}$ which makes $\int \frac {\tau '} {\tau}=\int \xi '=0$.