What is Roots of unity? — Mathematics

Definition 28.14 High School Mathematics · Chapter 28 — Complex Numbers

For $n \geq 1$ , the $n$ -th roots of unity are the solutions of $z^n = 1$ .

The fifth roots of unity, = 2 π/5: the vertices of a regular pentagon inscribed in the unit circle (). — The fifth roots of unity, $\omega = \eu^{2\iu\pi/5}$ : the vertices of a regular pentagon inscribed in the unit circle (Exercise 28.10).

Definition 3.17 University Mathematics — Year 1 · Chapter 3 — Complex Numbers

The $n$ -th roots of unity are the solutions of $z^n = 1$ :

\mathbb{U}_n = \bigl\{\, \omega^k : k = 0, \dots, n-1 \,\bigr\}, \qquad \omega = \eu^{2\iu\pi/n}.

They form a group under multiplication (Chapter 7) and sit at the vertices of a regular $n$ -gon inscribed in the unit circle.

The fifth roots of unity, = 2 π/5: a regular pentagon on the unit circle. — The fifth roots of unity, $\omega = \eu^{2\iu\pi/5}$ : a regular pentagon on the unit circle.

Examples

Example 3.20 (Square roots in algebraic form)

To solve $z^2 = 3 + 4\iu$ without trigonometry, set $z = x + \iu y$ :

x^2 - y^2 = 3, \qquad 2xy = 4, \qquad x^2 + y^2 = \abs{3 + 4\iu} = 5 .

Adding the first and last: $x^2 = 4$ , so $x = \pm 2$ , then $y = 2/x = \pm 1$ with the same sign pairing ( $xy = 2 > 0$ ): $z = \pm(2 + \iu)$ . Combined with the usual formula, this solves every quadratic equation with complex coefficients (Exercise 3.7).

Example 3.16 (A non-real right-hand side)

Solve $z^4 = -8 + 8\iu\sqrt3$ . Exponential form of the right side: modulus $\sqrt{64 + 192} = 16$ , argument $\theta$ with $\cos\theta = -\frac12$ , $\sin\theta = \frac{\sqrt3}2$ , i.e. $\theta = \frac{2\pi}3$ . The four roots are

z_k = 2\,\eu^{\iu(\frac\pi6 + \frac{k\pi}2)}, \quad k = 0, 1, 2, 3 : \qquad z_0 = \sqrt3 + \iu,\quad z_1 = \iu z_0 = -1 + \iu\sqrt3,

z_2 = -z_0 = -\sqrt3 - \iu,\qquad z_3 = -\iu z_0 = 1 - \iu\sqrt3 .

(Check: $z_0^2 = 2 + 2\iu\sqrt3$ , so $z_0^4 = (2 + 2\iu\sqrt3)^2 = 4 - 12 + 8\iu\sqrt3 = -8 + 8\iu\sqrt3$ .) Once one root is found, the other three come free: they are its successive rotations by $\frac\pi2$ , i.e. its products with the fourth roots of unity — the general structure behind Theorem 3.14, worth exploiting before recomputing each root from scratch. No conjugate symmetry this time: the right-hand side is not real.

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