closed path, winding number, Jordan contour If $ D$ is a domain in $\Bbb C$, $z_0\in \Bbb C\setminus D$, and $\gamma$ is a closed, piecewise smooth path in $ D$ for which the winding number $n(\gamma, z_0)\ne0$, show that there is a Jordan contour $\gamma_1$ in $D$ for which $n(\gamma_1, z_0)=1$
Here, Jordan contour: positively oriented, simple, closed, piecewise smooth path.
Bruce P.Palka, An Introduction to Complex Function Theory, P183， The author leave as an exercise 

I think, $C\setminus \gamma$ is the disjoint union of domains, the components of $C\setminus \gamma$, and $z_0$ in a bounded component $U$ of $C\setminus \gamma$.  then,  the boundary of $ U$ is a Jordan contour. But I don't know  how to prove this(if correc).
 A: Sketch of a constructive proof/algorithm: Continuous deformations (free homotopies) do not change winding numbers, so you may assume that $\gamma$ is a path on some $\epsilon$-grid, i.e., made up of horizontal and vertical segments of length $\epsilon > 0$. Now start traversing $\gamma$, and any time you hit a grid vertex you already visited, calculate the winding number of the loop. If it is $1$, stop right there. If it is $-1$, you can reverse it and stop as well. If it is $0$, discard (remove from $\gamma$) and keep going. Since the winding number of $\gamma$ is the sum of the winding numbers of all these loops, at least one of them has to have a non-zero winding number, so the algorithm terminates and you get a simple loop (i.e., a Jordan curve) in $D$ with winding number $1$. (Note that the reduction to a grid is only there to make the self-intersections nice and easy to handle.)
A: We may assume that $D$ is connected and $z_0=0$. 
The paths in $D$ can be reversed and concatenated, resulting the negatives and sums of their winding numbers. Hence, if the gcd of the possible winding numbers is $d>0$, then there is a path in $D$ with winding number $d$. So, we need to prove $d=1$.
If $d\ge2$ then all winding numbers of paths in $D$ is divisible by $d$; it follows that there is a branch of $\root{d}\of{z}$ over $D$.
From this point you can follow a similar question that was asked here:
If there is a branch of $\sqrt{z}$ on an open set $U$ with $0 \notin U,$ then there is also a branch of $arg$ $z.$
The first answer works without any change; in the second answer the curve $-\gamma$ must be replaced by $e^{2\pi i/d}\cdot\gamma$.
Update: the assumption $g\ge2$ and the existence of $\arg z$ together provides contradiction. Therefore, $g\ge2$ is not possible.
A: Redefining the Concept of "Path"
I'll answer the question under the assumption that the definition of path includes the requirement that it self-intersects at most a finite number of times. As I pointed out in this comment, I believe that in the context of Bruce Palka's textbook this assumption does not detract from the generality of my answer.

Some Preliminary Notations
For every path $\gamma : [a,b]\rightarrow\mathbb{C}$ ($a,b\in\mathbb{R}$, $a < b$) and every $c \in \mathbb{C}$ define
$$
\begin{align*}
\Xi(\gamma) &:= \big\{x\in[a,b] : \exists y\in[a,b]\setminus\{x\}.\ \gamma(x)=\gamma(y)\big\}\\
\xi(\gamma) &:= \big|\Xi(\gamma)\big|\\
\Omega(\gamma,c) &:= \big\{x\in[a,b] : \gamma(x)=c\big\}\\
\omega(\gamma,c) &:= \big|\Omega(\gamma,c)\big|
\end{align*}
$$
With these notations, we can state our assumption on paths formally: for every path $\gamma$, $\Xi(\gamma)$ is finite, or equivalently $\xi(\gamma) \in \{0,1,2,\dots\}$.
Note that $1\notin\operatorname{Rng}\xi$, and that for every closed path $\gamma:[a,b]\rightarrow\mathbb{C}$, $\xi(\gamma)\geq 2$, since $a, b \in \Xi(\gamma)$. A closed path $\gamma:[a,b]\rightarrow\mathbb{C}$ is simple iff $\xi(\gamma) = 2$.

A Restatement of the Problem
Let $D$ be a domain in $\mathbb{C}$, and let $z_0 \in \mathbb{C}\setminus D$.
It suffices to show that for every piecewise smooth path $\gamma$ in $D$ such that $n(\gamma,z_0)\neq0$, there is a simple, closed, piecewise smooth path $\gamma_1$ in $D$ such that $n(\gamma_1,z_0)\neq0$.
In fact, if we show this, then by Lemma 2.1 (ii) and (iii) on p. 157 of Palka's textbook, $n(\gamma_1,z_0)\in\{-1,1\}$. If $n(\gamma_1,z_0)=1$, we're done. If $n(\gamma_1,z_0) = -1$, then $-\gamma_1$ is a simple, closed, piecewise smooth path in $D$ such that $n(-\gamma_1,z_0) = -n(\gamma_1,z_0) = 1$, as desired.

Proof
Denote by $M$ the set of those $m \in \{2,3,\dots\}$ satisfying that for all closed, piecewise smooth paths $\gamma$ in $D$ for which $n(\gamma,z_0)\neq0$ and $\xi(\gamma) = m$, there is a simple, closed, piecewise smooth path $\gamma_1$ in $D$ such that $n(\gamma_1,z_0)\neq0$. We will now prove the claim if we show that $\{2,3,\dots\}\subseteq M$. We will do so by complete induction.
Base case: Let $\gamma$ be closed, piecewise smooth path in $D$ such that $n(\gamma,z_0)\neq0$ and $\xi(\gamma) = 2$. Set $\gamma_1 := \gamma$. Then $\gamma_1$ is a simple, closed, piecewise smooth path in $D$ such that $n(\gamma_1,z_0)\neq0$. Then $2\in M$.
Inductive case: Assume that for some $m\in\{2,3,\dots\}$ it is the case that $\{2,3,\dots,m\}\subseteq M$, and let $\gamma:[a,b]\rightarrow\mathbb{C}$ be a closed, piecewise smooth path in $D$ such that $n(\gamma,z_0) \neq 0$ and such that $\xi(\gamma) = m+1$.
If $\omega\big(\gamma,\gamma(a)\big) > 2$, define
$$
\begin{align*}
c &:= \max\Big(\Omega\big(\gamma,\gamma(a)\big)\setminus\{b\}\Big)\\
\alpha &:= \gamma\big|_{[a,c]}\\
\beta &:= \gamma\big|_{[c,b]}
\end{align*}
$$
Then $\gamma = \alpha+\beta$, and both $\alpha$ and $\beta$ are closed, piecewise smooth paths in $D$. Then $n(\gamma,z_0) = n(\alpha,z_0) + n(\beta,z_0)$. Since by assumption $n(\gamma,z_0)\neq0$, either $n(\alpha,z_0)\neq0$ or $n(\beta,z_0)\neq0$. Say w.l.g. that $n(\alpha,z_0)\neq0$. Since $\xi(\alpha) < \xi(\gamma)$, by the induction hypothesis there is a simple, closed, piecewise smooth path $\gamma_1$ in $D$ such that $n(\gamma_1,z_0)\neq0$. Then $m+1\in M$.
So assume henceforth that $\omega\big(\gamma,\gamma(a)\big) = 2$.
If $\gamma$ is simple, set $\gamma_1 := \gamma$. Then $\gamma_1$ is a simple, closed, piecewise smooth path in $D$ such that $n(\gamma_1,z_0)\neq0$. Then $m+1\in M$.
So assume henceforth that $\gamma$ is not simple.
Then there is a finite sequence $x_0, x_1, \dots, x_k \in [a,b]$, for some odd $k \in \{3,4,\dots\}$, such that $a=x_0 < x_1 < \cdots < x_k=b$, and such that the following conditions are satisfied with $\alpha_i := \gamma\big|_{[x_{i-1},x_i]}$, $i \in \{1,2,\dots,k\}$:

*

*For every $i \in \{1,2,\dots,k\}$, $\alpha_i$ is a piecewise smooth path in $D$.

*$\gamma = \alpha_1 + \alpha_2 + \cdots + \alpha_k$.

*For every even $i \in \{2,3,\dots,k\}$, $\alpha_i$ is closed, and $\xi(\alpha_i) \leq m$.

*For every odd $i \in \{1,2,\dots,k-2\}$, $\alpha_i(x_i) = \alpha_{i+2}(x_{i+1})$, and $\alpha_1 + \alpha_3 + \cdots + \alpha_k$ is simple and closed and $\xi(\alpha_1 + \alpha_3 + \cdots + \alpha_k) \leq m$.

(I leave the verification of these facts to the reader. The fact that $\alpha_1 + \alpha_3 + \cdots + \alpha_k$ is simple will not be used below.)
Define $\ell := \frac{k-1}{2}$, and
$$
\begin{align*}
\beta_0 &:= \alpha_1 + \alpha_3 + \cdots + \alpha_k\\
\beta_1 &:= \alpha_2\\
\beta_2 &:= \alpha_4\\
&\vdots\\
\beta_{\ell} &:= \alpha_{k-1}
\end{align*}
$$
Then

*

*For every $i \in \{0,1,\dots,\ell\}$, $\beta_i$ is a closed, piecewise smooth path in $D$ with $\xi(\beta_i) \leq m$.

*$n(\gamma,z_0) = n(\beta_0,z_0) + \cdots n(\beta_{\ell},z_0)$.

(I leave the verification of these facts to the reader.)
Since by assumption $n(\gamma,z_0)\neq0$, there is some $i\in\{0,1,\dots,\ell\}$ such that $n(\beta_i,z_0)\neq0$. Then by the induction hypothesis there exists some simple, closed, piecewise smooth path $\gamma_1$ in $D$ such that $n(\gamma_1,z_0)\neq0$. We conclude that $m+1 \in M$.
$\square$
