Roots of complex numbers

The equation to solve

• The standard names equations of the form:

$$z^n = r\,\text{cis}\,\theta \qquad \text{or} \qquad z^n = a + bi$$

where $a$, $b$ are real and $n$ is a positive integer.

• The central fact: an equation $z^n = w$ has exactly $n$ solutions, no more and no fewer.
• $z^3 = 8$ has three solutions, not one. Only one of them is the real number 2.

Where the extra roots come from

• A complex number does not have one argument; it has infinitely many, all differing by $2\pi$.
• So $w = r\,\text{cis}\,\theta$ is equally well written as $r\,\text{cis}(\theta + 2k\pi)$ for any integer $k$.
• Taking the $n$th root divides that argument by $n$:

$$z = r^{1/n}\,\text{cis}\!\left(\frac{\theta + 2k\pi}{n}\right), \qquad k = 0, 1, 2, \ldots, n-1$$

• Dividing by $n$ means the $2\pi$ jumps become $\frac{2\pi}{n}$ jumps — small enough that $n$ different answers fit inside one full turn before they start repeating.
• Note why $k$ stops at $n - 1$: putting $k = n$ adds a whole $2\pi$ to the argument and returns you to the $k = 0$ root.

What the roots look like

• The $n$ roots always:

  • lie on a circle of radius $r^{1/n}$ centred on the origin — every root has the same modulus
  • are spaced equally around that circle, $\dfrac{2\pi}{n}$ apart
  • form the vertices of a regular $n$-sided polygon

• These three facts are a free check on your answer, and they are also a fast way to find the remaining roots once you have one: just keep adding $\frac{2\pi}{n}$ to the argument.

The procedure

• Convert the right-hand side to polar form. Sketch it, because a wrong argument here ruins everything.

• Take the $n$th root of the modulus: $r^{1/n}$.

• Write the general argument as $\dfrac{\theta + 2k\pi}{n}$.

• Substitute $k = 0, 1, \ldots, n-1$ in turn.

• Reduce each argument into $(-\pi, \pi]$.

• Convert to rectangular form if the question asks for it.

• Worked through for $z^3 = 8$:

  • $$8 = 8\,\text{cis}\,0$$
  • Modulus of each root: $8^{1/3} = 2$
  • Arguments: $\dfrac{0 + 2k\pi}{3}$ for $k = 0, 1, 2$, giving $0$, $\dfrac{2\pi}{3}$, $\dfrac{4\pi}{3}$
  • Reduce the last: $\dfrac{4\pi}{3} - 2\pi = -\dfrac{2\pi}{3}$
  • Roots: $2\,\text{cis}\,0 = 2$, $2\,\text{cis}\,\dfrac{2\pi}{3} = -1 + \sqrt{3}i$, $2\,\text{cis}\left(-\dfrac{2\pi}{3}\right) = -1 - \sqrt{3}i$
  • Note the second and third are a conjugate pair, exactly as they must be for a real-coefficient equation.

Choosing values of $k$ that land in range

• Using $k = 0, 1, \ldots, n-1$ always works, but the last few arguments then need reducing.
• A neater alternative is to choose $k$ symmetrically about zero — for $n = 5$, take $k = -2, -1, 0, 1, 2$. Every argument then comes out principal already.
• Either approach is acceptable; pick one and be consistent.

Roots of unity

• The solutions of $z^n = 1$ are the $n$th roots of unity.
• They lie on the unit circle, at arguments $\dfrac{2k\pi}{n}$, and always include $z = 1$ itself.
• Their sum is zero for $n \geq 2$, because the vertices of a regular polygon centred on the origin cancel out.