Can you Imagine the intersections?

This problem, though can be solved by other methods too, has a very elegant solution by the use of complex numbers. (This is the reason for the use of the word "Imagine" in the title of the question; I make use of imaginary numbers in my solution!)

By solving the equation $z^n -1 = 0$ , one gets n distinct n-th roots of 1 satisfying the equation. These solutions when plotted in the complex plane, give n equidistant points on $\mathbb{C}(1;O)$ , that is the unit circle centered about the origin.

I will take the following proposition for granted : The equations $z^m -1 =0$ and $z^n -1 = 0$ have the same common roots as the roots of the equation $2^d -1 =0$ , where $d = \gcd(m,n)$ .

This implies that the number of common roots between the equations will be equal to d.

Thus if the roots of the equations $z^{1982} -1 =0$ and $z^{2973}-1 = 0$ are plotted in the complex plane, they will all lie on the unit circle and give the regular polygons mentioned in the figure. Now if they have any common vertices, then the number of such common vertices will be $\gcd(1982,2973)$ = 991 .