Let $U\subseteq\mathbb{C}\setminus\left\{0\right\}$ be an open set and suppose that there is a path $\gamma$ in $U$ such that $\mbox{Ind}_{\gamma}(0)=1$. Show that there is no nth roots in $U$. $(n\geq 2)$
Remark: When we say that there is no nth root in $U$, it means that the function $f(z)=z$ does not have $n$th root $U$, ie, does not exists $g:U\rightarrow \mathbb{C}$ holomorphic such that $f(z)=(g(z))^n$ for all $z\in U$.
Let $U\subseteq\mathbb{C}\setminus\left\{0\right\}$ be an open set and $\gamma: [0, 1] \to U$ be a closed path in $U$ such that $\mbox{Ind}_{\gamma}(0)=1$.
Assume that $g:U\rightarrow \mathbb{C}$ is holomorphic in $U$ and satisfies $g(z)^n = z$ for all $z \in U$, and therefore $$ n \frac{g'(z)}{g(z)} = \frac 1z \, . $$
Then $\Gamma := g \circ \gamma$ is a closed path in $U$ and $$ \mbox{Ind}_{\Gamma}(0) = \frac{1}{2 \pi i} \int_\Gamma \frac{dw}{w} = \frac{1}{2 \pi i} \int_0^1 \frac {g'(\gamma(z)) \, \gamma'(t)}{g(\gamma(t))} \, dt = \\ \frac{1}{2 \pi i} \int_0^1 \frac{\gamma'(t)}{n \, \gamma(t)} \, dt = \frac{1}{n} \frac{1}{2 \pi i} \int_\gamma \frac{dz}{z} = \frac{1}{n} \mbox{Ind}_{\gamma}(0) = \frac{1}{n} $$ which contradicts the fact that the winding number $\mbox{Ind}_{\Gamma}(0)$ is an integer.