How to show that the winding number must be $0$ or $\pm$ $1$ in this case? Prerequisite:
$K: [0,1] → \mathbb{C}$ piecewise, continuously differentiable, closed path that meets the non-positive real axis at only at one point
$K^{-1} (\{x|x \le 0\})= \{ t_0 \}$ with $0 < t_0 < 1$ and $K(t_0)=−r < 0$ and does not meet at the origin.
Show:
$\eta (0) \in \{0, \pm 1 \}$ (where $\eta$ is the winding number),
and $0, 1$, and $− 1$ occur if $K$ runs locally at $t_0$ on one side of the negative real axis or intersects the negative real axis from top to bottom or rather from bottom to top.
I tried to prove by contradiction, that if we assume, that $n=2$, then the closed path would obviously intersect the negative real axis twice at two real numbers $w=(a,0)$ and $z=(b,0)$. Would that be enough to prove the statement?
 A: Pick a continuous branch $\theta(t)$ of $\operatorname{Arg}\gamma(t)$ such that $\theta(t_0)=\pi$. By assumption, $\theta(0),\theta(1)\notin\{\ldots,-3\pi,-\pi,\pi,3\pi,\ldots\}$. By intermediate value theorem, the values of $\theta(0)$ and $\theta(1)$ must lie inside the open interval $(-\pi,3\pi)$, otherwise $\theta(t)$ will be equal to $-\pi$ or $3\pi$ for some $t\in[0,1]$. Hence $|\theta(1)-\theta(0)|<4\pi$ and it must be equal to $0$ or $2\pi$. This means the winding number of $\gamma$ about the origin is $0$ or $\pm1$.
A: Here is an approach that is geometrically evident but the analysis is tedious.
The basic idea is that either the curve $\gamma$ lies in some simply connected region (that does not include the origin) and the index is therefore zero, or the curve can be written as the 'sum' of two curves each of which lies in some simply connected region and so the index of the 'sum is zero which means that the index of one is the negative of the other. The rest is just grind.
Let $S=B(-r, {r \over 2})$, $ S_R = \{ z \in S | \operatorname{im} z \ R \ 0 \}$, where $R $ is one of $<,>$.
Some notation.
If f $\gamma$ is a curve on $[0,1]$, define $\gamma_{a,b}(t) = \gamma((1-t)a+tb$ (on $[0,1]$), and let $-\gamma = \gamma_{1,0}$.
If $\gamma_1,\gamma_2$ are curves with $\gamma_1(1) = \gamma_2(0)$, then define $(\gamma_1+\gamma_2)(t) = \begin{cases} \gamma_1(2t), & t \in [0,{1 \over2}] \\ \gamma_2({1 \over 2}+2 (t-{1 \over 2})),& t \in ({1 \over 2},1]\end{cases}$.
If $\gamma$ is a closed curve on $[0,1]$ (that is, $\gamma(0) = \gamma(1)$) then extend the domain to $\mathbb{R}$ such that $\gamma(t) = \gamma(\{t\})$ ($\{t\}$ is the fractional part of $t$).
If the curve $\gamma$ is closed, the starting point is irrelevant from the perspective of the index integral, that is, the curves $\gamma$ and $t \mapsto \gamma(t+\beta)$ will result in the same index, so I will be somewhat cavalier in dealing with the starting points of closed curves.
Choose $\delta>0$ such that $\gamma([t_0-\delta, t_0 +\delta] ) \subset S$. Let $\Gamma_- = \gamma([t_0-\delta, t_0))$, $\Gamma_+ = \gamma((t_0, t_0 +\delta] )$ and note that each of $\Gamma_-,\Gamma_+$ must lie in one of $S_-,S_+$.
There are essentially two cases to consider. (i) $\Gamma_- \subset S_-$ and $\Gamma_+ \subset S_+$  or (ii) $\Gamma_-,\Gamma_+ \subset S_-$, the other two cases are similar.
Consider (i) first. Let $c$ be the curve on $[0,1]$ defined by $c(t) = r e^{2 \pi i (t+({1 \over 2}-t_0))}$ (so that $c(t_0) = -r$).
By reducing if necessary, we can assume that the $
\delta>0$ above is such that $c(t) \in S$ for $t \in [t_0-\delta, t_0 +\delta]$, and $0<t_0-\delta, t_0+\delta < 1$.
Define $c_S = c_{t_0-\delta, t_0+\delta}$, $c_r = c_{ t_0+\delta, 1}+ c_{ 0, t_0-\delta}$ ('r' for remainder :-)). Note that $c$ is equivalent to $c_S+c_r$ in the sense mentioned above.
Define $\gamma_S = \gamma_{t_0-\delta, t_0+\delta}$ and $\gamma_r = \gamma_{ t_0+\delta, 1}+ \gamma_{ 0, t_0-\delta}$. Note the difference in direction from $c_S$.
Finally, let $b_+$ be the straight line curve on $[0,1]$ that joins $\gamma(t_0+\delta)$ to $c(t_0-\delta)$ and $b_+$ be the straight line curve on $[0,1]$ that joins $c(t_0+\delta)$ to $\gamma(t_0-\delta)$.
It might help to look at the following crude drawing.

Note that $b_++c_S+b_-+\gamma_S$ is a closed curve lying entirely in the simply connected set $S$ and so the winding number satisfies $\eta(b_++c_S+b_-+\gamma_S, 0) = 0$.
Note that $(-b_+)+\gamma_r+ (-b_-)+c_r$ is a closed curve lying entirely in the simply connected set  $(-\infty,0]^c$ and so the winding number satisfies $\eta((-b_+)+\gamma_r+ (-b_-)+c_r, 0) = 0$.
Consequently, $\eta(b_++c_S+b_-+\gamma_S + (-b_+)+\gamma_r+ (-b_-)+c_r, 0) = \eta(c_S+\gamma_S +\gamma_r+c_r, 0) = 0$ and hence
$\eta(c+\gamma) = 0$ from which we get $\eta(\gamma,0) = -1$.
Case (ii) is a little simpler. Let $R_k = (-\infty, -r-{1 \over k}] \cup \{ z | |z-(-r)| = {1 \over k}, \operatorname{im} z \ge 0 \} \cup [-r+{1 \over k},0]$ and note that there is some $k$ such that $R_k$ does not intersect the range of the curve $\gamma$. Hence the curve lies entirely in the simply connected set $R_k^c$ and hence $\eta(\gamma,0) = 0$.
