Consider an $n$-sided polygonal curve in the plane, represented by an ordered set of points $(x_0, x_1, \ldots, x_{n-1})$; line segments connect consecutive points and also $x_{n-1}$ to $x_0$. It is convenient to think of each of these points as lying in $\mathbb{C}$, and so a polygonal curve with n vertices can be thought of as a point in $\mathbb{C}^n$.

Now consider the discrete Fourier transform of this polygon:

\begin{equation} X_k = \sum_{j=0}^{n-1}x_j \cdot e^{-2\pi ijk/n}, k \in \mathbb{Z}. \end{equation}

Is it possible to determine directly from these Fourier coefficients $X_k$ whether our original polygonal curve is convex or not? or whether it is simple (non-self-intersecting) or not?

Consider an $n$-sided polygonal curve in the plane, represented by an ordered set of points $(x_0, x_1, \ldots, x_{n-1})$; line segments connect consecutive points and also $x_{n-1}$ to $x_0$. It is convenient to think of each of these points as lying in $\mathbb{C}$, and so a polygonal curve with n vertices can be thought of as a point in $\mathbb{C}^n$.

Now consider the discrete Fourier transform of this polygon:

\begin{equation} X_k = \sum_{j=0}^{n-1}x_j \cdot e^{-2\pi ijk/n}, k \in \mathbb{Z}. \end{equation}

Is it possible to determine directly from these Fourier coefficients $X_k$ whether our original polygonal curve is convex or not? or whether it is simple (non-self-intersecting) or not?

[Three short comments added in edit]

1. The same question can also be asked for smooth curves and their Fourier transforms; perhaps more is known in that case.
2. Changing the first Fourier coefficient $X_0$ only translates the curve, so does not affect convexity or simplicity.
3. There is some ambiguity in defining convexity (for curves) that is relevant to this question. In particular, if $X_k = 0$ for all but one $k \neq 1$ that divides $n$, then the curve is a $k$-cover of the regular $n/k$-gon. It is not clear whether such curves should be considered convex or not.