Given 3 complex numbers a, b, and c, does there always exist another number N where taking the cube root of N gives a, b, and c? If not, when is that possible? How do I find N? How does this extend to n complex numbers and nth root rather than 3?
I have a programming background much more than specifically a math background and I can only find sources talking about how to find the roots of a complex number rather that the reverse.
On
The answer is no, because of DeMoivre's theorem. If $z_1, z_2, z_3 \in \mathbb C$ are distinct complex numbers, then we may express them in polar form as $$z_k = \rho_k e^{i\theta_k}, \quad \rho_k = |z_k| > 0, \quad \theta_k = \arg(z_k), \quad k \in \{1,2,3\}.$$ Then the theorem states that $$z_k^3 = \rho_k^3 e^{3\theta_k i},$$ and in particular, $$|z_k^3| = \rho_k^3.$$ For the requirement that $z_1, z_2, z_3$ are the cube roots of the same number, say $w$, this would imply that $z_k^3 = w$ for each such $k$. But in general, there is no requirement that the magnitudes $\rho_k$ are equal; therefore, no such $w$ exists if $|z_k| \ne |z_j|$ for some $k \ne j$. Another condition following from the theorem is that we also require $\arg(3z_k) \equiv \arg(3z_j) \pmod {2\pi}$; that is to say, the difference in arguments of distinct cube roots must be some integer multiple of $2\pi/3$. Geometrically, these two conditions require $z_1, z_2, z_3$ to form an equilateral triangle centered at the origin.
The $n$th roots of any complx number will always form $n$ evenly spaced points around a circle, centred at the origin - or in other words they all have the same modulus, and their arguments differ by multiples of $\frac{2 \pi}{n}$. That means that if you have three arbitrary complex numbers they will not all have the same cube.
If $a$, $b$ and $c$ are not colinear (i.e. you can't draw a straight line through all three) then it is possible to draw a circle through them, but there's still no guarantee they will be evenly spaced, or even that the spacing between them will be rational multiples of $\pi$.