Recipe

Find the nth roots of a complex number

Use when: you need every complex number w with w^n = z, not just the single real root a calculator's root button would return

Goal: all n roots, each with the correct modulus and angle, evenly spaced around one circle

Taught in nth roots of complex numbers

1. 1

   Find the modulus r=\sqrt{a^2+b^2} and the argument \theta (placed in the correct quadrant) for the complex number z=a+bi you're taking roots of.

   Example

   For z=-8: r=\sqrt{(-8)^2+0^2}=8, and \theta=\pi (180°), since -8 sits on the negative real axis.

   Why: the roots formula only works on z written as r(\cos\theta+i\sin\theta); nothing about the roots can be read off directly from rectangular form.

2. 2

   Every one of the n roots shares the identical modulus, s = r^{1/n}, the ordinary positive real nth root of r.

   Example

   8^{1/3}=2, since 2^3=8.

   Why: raising s to the nth power must reproduce r exactly, and there's only one positive real number that does that.

3. 3

   Compute \varphi_k = \dfrac{\theta+2\pi k}{n} for each integer k from 0 up to n-1, then stop; k=n would just repeat k=0's angle.

   • working in degrees — use \varphi_k=\dfrac{\theta+360°k}{n} throughout instead.
   • working in radians — use \varphi_k=\dfrac{\theta+2\pi k}{n} throughout.

   Why: \theta, \theta+2\pi, \theta+4\pi,\dots all name the exact same direction for z itself, but dividing each of those by n lands on a genuinely different angle — that's where the extra roots come from.

4. 4

   Each root is w_k = s(\cos\varphi_k+i\sin\varphi_k). Write out all n of them, and confirm there are exactly n, evenly spaced \dfrac{360°}{n} apart around a circle of radius s.

   Why: a degree-n equation w^n-z=0 has exactly n complex solutions by the Fundamental Theorem of Algebra, and checking the even spacing is what confirms no root got computed twice or skipped.