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Topos Uranos
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The nth roots of a complex number and the roots of unity

Summary

This lesson solves the equation zn=wz^n = w in the complex numbers. With the polar form and De Moivre’s formula it proves that every nonzero complex number has exactly nn nnth roots, which share the modulus and have arguments separated by 2πn\frac{2\pi}{n}, so that they form a regular polygon centered at the origin. It then studies the roots of unity: they are the powers of a single one of them, they multiply as the remainders of division by nn add, their powers with exponent mm add up to nn or to 00 according as nn divides mm or not, and they factor the polynomial zn−1z^n - 1. It characterizes by means of the greatest common divisor the primitive roots, whose powers run through all the roots, and, by grouping conjugate roots, it obtains the real factorization of xn−1x^n - 1 and of xn+1x^n + 1 into linear and quadratic factors, with which the value of cos⁡2π5\cos \frac{2\pi}{5} reappears.

Learning objectives

  1. Find the nn nnth roots of a nonzero complex number with the polar form, prove that there are exactly nn of them, and describe them as the vertices of a regular polygon.
  2. Work with the roots of unity as powers of one of them, compute the sums of their powers, and factor the polynomial zn−1z^n - 1.
  3. Decide which roots of unity are primitive by means of the greatest common divisor, and prove the criterion with Bézout’s identity.
  4. Factor the polynomials xn−1x^n - 1 and xn+1x^n + 1 over the reals by grouping conjugate roots, and apply the roots of unity to problems on distances, trigonometric values and equations.

The nnth roots of a complex number

In the lesson of the course Algebra and Geometry I on the complex numbers, the equation z3=1z^3 = 1 was solved by factoring a difference of cubes, and it was announced that the roots of index nn of an arbitrary complex number would require the polar form. The factoring method does not generalize: for the equation z5=1+iz^5 = 1 + i there is no factor in sight. The polar form, studied in the lesson on the polar form of complex numbers, instead turns the equation into two real conditions, one on the moduli and one on the arguments, because, by the formula proved in the lesson on De Moivre’s formula, raising a complex number to the power nn raises its modulus to the power nn and multiplies its argument by nn.

Let ww be a complex number and nn a natural number. A complex number zz is an nnth root of ww when zn=wz^n = w. The roots of order two are called square roots, and those of order three, cube roots.

The only complex number whose nnth power is 00 is 00 itself, because a product of complex numbers vanishes only if one of its factors vanishes; the interesting case is therefore w≠0w \neq 0. One then writes ww in polar form, w=R(cos⁡φ+isin⁡φ)w = R(\cos \varphi + i \sin \varphi), with R=∣w∣>0R = \abs{w} > 0 and φ\varphi any one of its arguments. The symbol Rn\sqrt[n]{R} keeps the meaning it has in the Course I lesson on powers with rational exponents and radicals: it denotes the principal root of the positive real number RR, the only positive real number whose nnth power is RR. The radical is reserved for the principal root of a nonnegative real number and, as in that lesson, for the real root of odd index of a negative real number, such as −83=−2\sqrt[3]{-8} = -2; applied to any other complex number, it will be used only with an explicit warning that it denotes any one of its roots, because, as will be seen shortly, none of its nn roots stands out from the others, and choosing one by convention breaks the laws of radicals: if the square root of −1-1 were taken to be the number ii, the product of that root by itself would be −1-1, and not the square root of (−1)(−1)=1(-1)(-1) = 1.

Finally, the theorem on equality in polar form, proved in that same lesson, is used: two nonzero complex numbers are equal if and only if they have the same modulus and their arguments differ by an integer multiple of 2π2\pi.

Theoremnnth roots

Let w=R(cos⁡φ+isin⁡φ)w = R(\cos \varphi + i \sin \varphi) be a nonzero complex number, with R=∣w∣R = \abs{w}, and nn a natural number. The equation zn=wz^n = w has exactly nn distinct solutions, which are the complex numbers zkz_k below, for k=0,1,…,n−1k = 0, 1, \ldots, n - 1. All of them have modulus Rn\sqrt[n]{R}, and the argument of each one exceeds that of the one before it by 2πn\frac{2\pi}{n}.

zk=Rn(cos⁡φ+2kπn+isin⁡φ+2kπn)z_k = \sqrt[n]{R} \left( \cos \frac{\varphi + 2k\pi}{n} + i \sin \frac{\varphi + 2k\pi}{n} \right)

The solution is sought in polar form, and the polar forms of znz^n and of ww are equated. The figure follows the proof with w=−8=8(cos⁡π+isin⁡π)w = -8 = 8(\cos \pi + i \sin \pi) and n=3n = 3, whose cube roots have modulus 22.

Proof

  1. z=r(cos⁡θ+isin⁡θ)→zn=rn(cos⁡nθ+isin⁡nθ)z = r(\cos \theta + i \sin \theta) \rightarrow \resaltar{z^n} = \dato{m}{\resaltar{r^n (\cos n\theta + i \sin n\theta)}}

    A solution cannot be zero, because 0n=0≠w0^n = 0 \neq w. We write it in polar form, with r>0r > 0, and raise it to the power nn with De Moivre’s formula.

  2. zn=w↔rn=R∧nθ=φ+2kπ,k∈Z\resaltar{z^n = w} \leftrightarrow \dato{c}{\resaltar{r^n = R \wedge n\theta = \varphi + 2k\pi}}, \quad k \in \mathbb{Z}

    By the theorem on equality in polar form, the equation zn=wz^n = w is equivalent to two real conditions: the moduli are equal and the arguments differ by an integer multiple of 2π2\pi.

  3. r=Rn,θk=φ+2kπn,k∈Zr = \resaltar{\sqrt[n]{R}}, \qquad \dato{t}{\theta_k = \resaltar{\frac{\varphi + 2k\pi}{n}}}, \quad k \in \mathbb{Z}

    The positive real number whose nnth power is RR is unique, the principal root, and the argument is determined by the integer kk. Conversely, every complex number with that modulus and that argument is a solution, because it satisfies both conditions. With w=−8w = -8 and n=3n = 3, we get r=2r = 2 and θ=π+2kπ3\theta = \frac{\pi + 2k\pi}{3}.

  4. θqn+j=φ+2(qn+j)πn=θj+2qπ→zqn+j=zj\theta_{qn + j} = \resaltar{\frac{\varphi + 2(qn + j)\pi}{n}} = \resaltar{\theta_j + 2q\pi} \rightarrow z_{qn + j} = z_j

    The integers kk are infinitely many, but the solutions are not. By division with remainder, every integer can be written as k=qn+jk = qn + j, with qq an integer and 0≤j≤n−10 \leq j \leq n - 1; the argument θk\theta_k then exceeds θj\theta_j by an integer multiple of 2π2\pi, and the two solutions coincide. In the figure, z3=z0z_3 = z_0.

  5. 0<θl−θj=2(l−j)πn<2π→zl≠zj0 < \theta_l - \theta_j = \resaltar{\frac{2(l - j)\pi}{n}} < 2\pi \rightarrow \resaltar{z_l \neq z_j}

    The solutions z0,z1,…,zn−1z_0, z_1, \ldots, z_{n - 1} are distinct from one another: if 0≤j<l≤n−10 \leq j < l \leq n - 1, their arguments differ by a number lying strictly between 00 and 2π2\pi, which is not an integer multiple of 2π2\pi. There are therefore exactly nn solutions. With w=−8w = -8, they are 1+i31 + i\sqrt{3}, −2-2 and 1−i31 - i\sqrt{3}.

In practice, the first root has argument φn\frac{\varphi}{n}, and each of the others is obtained by adding 2πn\frac{2\pi}{n} to the argument of the one before it. Replacing the argument φ\varphi of ww by another one, φ+2π\varphi + 2\pi, does not change the set of roots, only their numbering: what used to be z1z_1 is now called z0z_0. For n=2n = 2, the arguments of the two roots differ by π\pi, so that they are opposite, z1=−z0z_1 = -z_0; this is the double sign that appears in the square root of a complex number, computed in the Course I lesson on the complex numbers by equating real and imaginary parts. The number of roots also agrees with what is known about polynomials: by the lemma on distinct roots from the Course I lesson on the algebra of polynomials, whose proof holds unchanged over the complex numbers, the polynomial zn−wz^n - w, of degree nn, cannot have more than nn distinct roots; the theorem shows that it has all of them.

The regular polygon of the roots

The nn roots have the same modulus, so that they lie on the circle with center 00 and radius Rn\sqrt[n]{R}, and their arguments are equally spaced. The following theorem expresses this arrangement through the product of complex numbers, which in polar form multiplies the moduli and adds the arguments, as was proved in the lesson on the polar form.

TheoremThe polygon of the roots

Let ww be a nonzero complex number, with R=∣w∣R = \abs{w}, and n≥3n \geq 3 a natural number. Each nnth root of ww is obtained from the one before it by multiplying it by the number ω\omega below, which amounts to rotating it through an angle of 2πn\frac{2\pi}{n} about the origin; the nn roots are the vertices of a regular polygon with nn sides inscribed in the circle with center 00 and radius Rn\sqrt[n]{R}, and its side has length 2Rnsin⁡πn2\sqrt[n]{R} \sin \frac{\pi}{n}.

zk+1=zk ω,ω=cos⁡2πn+isin⁡2πnz_{k + 1} = z_k \, \omega, \qquad \omega = \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n}

The figure follows the proof with w=64iw = 64i, whose cube, fourth and sixth roots lie on circles of radii 643=4\sqrt[3]{64} = 4, 644=22\sqrt[4]{64} = 2\sqrt{2} and 646=2\sqrt[6]{64} = 2.

Proof

  1. zk ω=Rn(cos⁡(θk+2πn)+isin⁡(θk+2πn))=zk+1z_k \, \omega = \sqrt[n]{R} \left( \cos \resaltar{\left( \theta_k + \frac{2\pi}{n} \right)} + i \sin \left( \theta_k + \frac{2\pi}{n} \right) \right) = \dato{g}{\resaltar{z_{k + 1}}}

    Since ∣ω∣=1\abs{\omega} = 1, the product zkωz_k \omega has the modulus of zkz_k and its argument increased by 2πn\frac{2\pi}{n}: it is the root that comes after it. By induction on kk, each root is the first one multiplied by a power of ω\omega, zk=z0ωkz_k = z_0 \omega^k.

  2. ∣uω−vω∣=∣(u−v)ω∣=∣u−v∣∣ω∣=∣u−v∣\abs{u\omega - v\omega} = \abs{\resaltar{(u - v)\omega}} = \abs{u - v} \abs{\omega} = \dato{d}{\resaltar{\abs{u - v}}}

    Multiplying by ω\omega is therefore rotating through an angle of 2πn\frac{2\pi}{n} about the origin, and this rotation preserves distances, because the modulus of a product is the product of the moduli.

  3. ∣zk+2−zk+1∣=∣zk+1ω−zkω∣=∣zk+1−zk∣\abs{z_{k + 2} - z_{k + 1}} = \abs{\resaltar{z_{k + 1} \omega - z_k \omega}} = \resaltar{\abs{z_{k + 1} - z_k}}

    The rotation carries each side of the polygon to the one after it, so that all the sides have the same length.

  4. zn−1ω=z0ωn−1ω=z0ωn=z0z_{n - 1} \omega = z_0 \resaltar{\omega^{n - 1} \omega} = z_0 \omega^n = \resaltar{z_0}

    The polygon closes up, because ωn=1\omega^n = 1: the rotation carries the last vertex to the first. The nn vertices lie on the circle and the sides are equal. Moreover, the rotation carries the triangle with vertices zkz_k, zk+1z_{k + 1} and zk+2z_{k + 2} to the one with vertices zk+1z_{k + 1}, zk+2z_{k + 2} and zk+3z_{k + 3} while preserving the three sides, and by the law of cosines an angle of a triangle is determined by its three sides: the angles of the polygon, which are those of these triangles at the middle vertex, are equal, and the polygon is regular.

  5. ∣z1−z0∣=∣z0∣∣ω−1∣=2Rnsin⁡πn\abs{z_1 - z_0} = \abs{z_0} \resaltar{\abs{\omega - 1}} = \resaltar{2\sqrt[n]{R} \sin \frac{\pi}{n}}

    The side is computed in the triangle with vertices 00, z0z_0 and z1z_1. By the half-angle formula, ∣ω−1∣2=2−2cos⁡2πn=4sin⁡2πn\abs{\omega - 1}^2 = 2 - 2\cos \frac{2\pi}{n} = 4\sin^2 \frac{\pi}{n}, and the sine of πn\frac{\pi}{n} is positive. For the sixth roots of 64i64i, the side equals 2⋅2⋅12=22 \cdot 2 \cdot \frac{1}{2} = 2, the same as the radius, as in every regular hexagon.

For n=2n = 2, the polygon degenerates into a diameter, whose endpoints are the two opposite roots, and for n=1n = 1 there is a single root, ww itself. That the roots form a regular polygon explains why none of them stands out from the others: the rotation through 2πn\frac{2\pi}{n} permutes them without altering the figure.