Boundary of Polygon as Contour

Theorem
Let $P$ be a polygon embedded in the complex plane $\C$.

Denote the boundary of $P$ as $\partial P$.

Then there exists a simple closed contour $C$ such that $\operatorname{Im} \left({C}\right) = \partial P$, where $\operatorname{Im} \left({C}\right)$ denotes the image of $C$.

Proof
Let $n \in \N$ be the number of sides of $P$.

Denote the vertices of $P$ as $A_1, \ldots, A_n$.

From Complex Plane is Metric Space, it follows that $\C$ is homeomorphic to $\R^2$.

Then, we can consider $\partial P$ as a subset of $\R^2$.

From Boundary of Polygon is Jordan Curve, it follows that there exists a Jordan curve $\gamma: \left[{0\,.\,.\,1}\right] \to \R^2$ such that $\operatorname{Im} \left({\gamma}\right) = \partial P$.

The same theorem shows that $\gamma$ is a concatenation of $n$ line segments, parameterized as:


 * $\gamma_k \left({t}\right) = \left({1-t}\right) A_k + tA_{k+1}$

where $k \in \left\{ {1, \ldots, n}\right\}$, and we identify $A_{n+1}$ with $A_1$.

Then, $\gamma_k: \left[{a_{k-1}\,.\,.\,a_k}\right] \to \R^2$, where $a_{k-1}, a_k \in \left[{0\,.\,.\,1}\right]$.

As $\gamma$ is a concatenation of $\gamma_1, \ldots, \gamma_n$, it follows that $\left\{ {a_0, a_1, \ldots, a_n}\right\}$ is a subdivision of $\left[{0\,.\,.\,1}\right]$.

We have $\dfrac{ \mathrm d}{ \mathrm dt } \gamma_k \left({t}\right) = A_{k+1} - A_k \ne \mathbf 0$, as $A_k$ and $A_{k+1}$ are two different vertices.

As $\C$ is homeomorphic to $\R^2$, we can consider $\gamma$ as a continuous complex function $\gamma: \left[{0\,.\,.\,1}\right] \to \C$.

Then $\gamma_k$ is complex-differentiable for all values of $t \in \left({a_{k-1}\,.\,.\,a_k}\right)$ with its derivative $\gamma_k'$ defined by:


 * $\gamma_k' \left({t}\right) = x \left({ \dfrac{ \mathrm d}{ \mathrm dt } \gamma_k \left({t}\right) }\right) + i y \left({ \dfrac{ \mathrm d}{ \mathrm dt } \gamma_k \left({t}\right) }\right)$

where:


 * $x \left({ \dfrac{ \mathrm d}{ \mathrm dt } \gamma_k \left({t}\right) }\right)$ is the real part of $\dfrac{ \mathrm d}{ \mathrm dt } \gamma_k \left({t}\right)$
 * $y \left({ \dfrac{ \mathrm d}{ \mathrm dt } \gamma_k \left({t}\right) }\right)$ is the imaginary part of $\dfrac{ \mathrm d}{ \mathrm dt } \gamma_k \left({t}\right)$

As shown above, $\gamma_k' \left({t}\right) \ne 0$.

By definition of smooth path, it follows that $\gamma_k = \gamma{ \restriction_{ \left[{a_{k-1}\,.\,.\,a_k}\right] } }$ is a smooth path.

From Path as Parameterization of Contour, it follows that there exists a simple closed contour $C$ with $\operatorname{Im} \left({C}\right) = \partial P$.