Q:

Winding number $=0$ imply homotopic to a point?

Suppose $\gamma$ is a smooth closed curve in  $U=\mathbb{C} - \{0\}$.   Suppose the winding number of $\gamma$ around 0 is 0.  Is $\gamma$ homotopic to a point in $U$?

A:

The curve $\gamma\colon[a,b]\to\mathbb C-\{0\}$  can be written as $\gamma(t)=r(t)e^{i\theta(t)}$ for $a\le t\le b$ with $r,\theta$ continuos and $r>0$. Wlog. $\gamma(a)=\gamma(b)=1$, i.e. $r(a)=r(b)=1$, $\theta(a)=0$ and (because the winding number is zero)  $\theta(b)=0$.
We can define $\sqrt[n]{}$ on $\gamma([a,b])$ simply by letting $\sqrt[n]{\gamma(t)}=\sqrt[n]{r(t)}\cdot e^{\frac ini\theta(t)}$.
If $n>\frac\pi2\max|\theta|$, we see that $\Re(\sqrt[n]{\gamma(t)})>0$. This allows one to easily contract $\sqrt[n]{\gamma(t)}$ to $1$ within the right half plane, e.g. by letting.
$$H(\tau,t)=(1-\tau)\left(\sqrt[n]{\gamma(t)}-1\right)+1.$$
As a consequence,
$$(H(\tau,t))^n=\left((1-\tau)\left(\sqrt[n]{\gamma(t)}-1\right)+1\right)^n$$
is a homotopy that retracts $\gamma$ to $1$.

