Solving for All Roots of z^6 = 64

This medium-difficulty pure mathematics question asks candidates to find every complex solution of a specific polynomial equation involving a power of a complex variable. The setup is a standard "roots of a complex number" problem, where a simple real number on the right-hand side hides a full constellation of complex roots spread around the origin. It is a classic exercise in complex analysis and algebra, often used in interviews or exams to check whether someone can move comfortably between algebraic and geometric viewpoints in the complex plane.

To handle it, a candidate must recognize how to express a complex (or real) number in polar form and then apply the general theory of nth roots of complex numbers. The solution leans on Euler's formula, the periodicity of arguments, and the idea that multiplication by a complex number is a rotation and scaling. Interviewers watch for clear use of modulus and argument, correct enumeration of all distinct roots without duplication, and clean reasoning that explains why no further solutions are possible.

What it tests

The general principle at work is that the equation $z^n = r$ (for any nonzero complex or real $r$) has $n$ distinct complex solutions, each corresponding to a different angle spaced evenly around a circle in the complex plane. This arises because the complex exponential $e^{i\theta}$ is periodic with period $2\pi$, so when taking the $n$th root, you must account for all possible angles that differ by integer multiples of $2\pi$ divided by $n$. The modulus (distance from the origin) of each root is the $n$th root of the modulus of $r$, and the arguments (angles) are given by $\frac{\arg(r) + 2\pi k}{n}$ for $k = 0, 1, ..., n-1$. This structure is a consequence of the fundamental theorem of algebra and the nature of complex multiplication as rotation and scaling.