the n-th roots of a complex number
Asking for an n-th root of w means asking for every z with z^n = w. Over the real numbers a positive number has at most two square roots and a negative number has none; over the complex numbers the situation is far cleaner and more symmetric: every nonzero complex number has exactly n distinct n-th roots, no more and no fewer. The polar form is what makes them visible, because de Moivre's angle-multiplication runs neatly in reverse.
Here is the method, walked through in plain steps. Write w in polar form as w = R e^(i phi). An n-th root must have modulus equal to the real n-th root of R (the ordinary positive root) and an angle theta with n theta equal to phi plus any whole number of full turns. Solving, theta = (phi + 2 pi k)/n. Letting k = 0, 1, 2, up to n - 1 gives n genuinely different angles; beyond that the values just repeat. So the n roots are z_k = R^(1/n) e^(i (phi + 2 pi k)/n) for k = 0, 1, ..., n - 1.
Geometrically the picture is gorgeous: all n roots have the same modulus R^(1/n), so they sit on a single circle, and their angles are equally spaced by 2 pi / n, so they form the vertices of a perfect regular n-gon centered at the origin. This is the n-th-root machine behind the roots of unity (the special case w = 1), and it shows in one stroke why the complex numbers, unlike the reals, give every polynomial all of its roots.
The cube roots of 8: write 8 = 8 e^(i 0). Then z_k = 2 e^(i (2 pi k)/3) for k = 0, 1, 2, giving 2, 2 e^(i 2 pi/3) = -1 + (the square root of 3) i, and 2 e^(i 4 pi/3) = -1 - (the square root of 3) i — an equilateral triangle on the circle of radius 2.
n roots, same modulus, angles 2 pi / n apart: the vertices of a regular n-gon.
Avoid writing 'the' n-th root of a complex number as if it were one number — there are n of them. The symbol z^(1/n) is multivalued; pinning a single value down requires choosing a branch, exactly as for the logarithm.