Roots of a complex number and roots of unity

When you reach for it

Over the real numbers, x^3 = 8 has one solution and x^4 = -81 has none. Over the complex numbers both equations have exactly as many solutions as their degree: three cube roots of 8 and four fourth roots of -81. De Moivre’s theorem raises a number to a power by raising its modulus and multiplying its argument; this lesson runs it backwards.

Reach for it whenever an equation has the form z^n = w, or can be brought to that form: finding all square, cube or higher roots of a complex number, solving z^n - w = 0, solving an equation that is quadratic in z^n, and working with the roots of unity, the solutions of z^n = 1, which turn up wherever something repeats in n equal steps around a circle.

The procedure

Throughout, \operatorname{cis}\theta means \cos\theta + i\sin\theta, as in Modulus, argument and polar form, n is a positive integer, and w \ne 0.

Setting up. Write w = R\operatorname{cis}\varphi with R = |w| > 0 and \varphi any argument of w. Look for z in the same form, z = r\operatorname{cis}\theta with r > 0 (z = 0 cannot work, since 0^n = 0 \ne w). By De Moivre’s theorem the equation z^n = w becomes

Two numbers in polar form with positive moduli are equal exactly when their moduli are equal and their angles differ by a whole number of turns. So r^n = R and n\theta = \varphi + 2\pi k for some integer k. The first says \displaystyle r = \sqrt[n]{R}, the one positive real number whose nth power is R. The second says \displaystyle \theta = \frac{\varphi + 2\pi k}{n}.

The nth roots. Every nonzero complex number w = R\operatorname{cis}\varphi has exactly n different nth roots,

Every one of them works: by De Moivre, z_k^n = R\operatorname{cis}(\varphi + 2\pi k) = w, because adding 2\pi k is k whole turns. And the argument above shows there are no others.

Why k stops at n - 1. Each step from k to k + 1 adds \displaystyle \frac{2\pi}{n} to the angle, so n steps add 2\pi, one whole turn. Thus z_n = z_0, z_{n+1} = z_1, and in general z_{k+n} = z_k; a negative k also lands back on one of z_0, \ldots, z_{n-1} after adding a multiple of n. The values k = 0, 1, \ldots, n - 1 give angles from \displaystyle \frac{\varphi}{n} up to, but not reaching, \displaystyle \frac{\varphi}{n} + 2\pi. Any two of these differ by less than a full turn, so they point in different directions and the n roots are all different.

The picture. Every root has the same modulus \displaystyle \sqrt[n]{R}, so all of them lie on the circle of that radius about 0. Neighbors differ in angle by \displaystyle \frac{2\pi}{n}, so the roots split the circle into n equal arcs and sit at the corners of a regular n-gon. This is the polygon of regular polygons and their angles, with its center at 0. Once one root is known, the rest follow by turning it through \displaystyle \frac{2\pi}{n} again and again.

Counting agrees with the fundamental theorem. The nth roots of w are the zeros of the polynomial z^n - w, which has degree n. The The fundamental theorem of algebra says a polynomial of degree n has exactly n zeros counted with repeats. Here there are n different ones, so none repeats.

Roots of unity. The nth roots of 1 are the nth roots of unity. With R = 1 and \varphi = 0 the formula gives \displaystyle \operatorname{cis}\frac{2\pi k}{n}, k = 0, 1, \ldots, n - 1. Write \displaystyle \omega = \operatorname{cis}\frac{2\pi}{n}. By De Moivre, \displaystyle \omega^k = \operatorname{cis}\frac{2\pi k}{n}, so the nth roots of unity are

the corners of a regular n-gon in the unit circle with one corner at 1 (Figure 1). Since \omega^n = 1, any power of \omega can be reduced: \omega^{n + m} = \omega^m.

Their sum is 0 when n \ge 2. Let S = 1 + \omega + \cdots + \omega^{n-1}. Multiplying by \omega shifts every term one place: \omega S = \omega + \omega^2 + \cdots + \omega^n, and the last term \omega^n is 1, so \omega S is the same n numbers in a different order, that is, \omega S = S. Then (\omega - 1)S = 0, and \omega \ne 1 because \displaystyle \frac{2\pi}{n} is not a whole turn when n \ge 2, so S = 0. In the picture: turning the polygon by \displaystyle \frac{2\pi}{n} moves every corner onto the next, so the sum cannot change, and the only number that a turn about 0 leaves in place is 0.

All roots from one. If z_0 is any one nth root of w, then (z_0\omega^k)^n = z_0^n(\omega^n)^k = w, so z_0, z_0\omega, \ldots, z_0\omega^{n-1} are n different nth roots of w, which is all of them. Multiplying by \omega is the turn by \displaystyle \frac{2\pi}{n} from the picture.

Worked examples

Boundaries

w must be nonzero. 0 has no argument, and its only nth root is 0 (r^n = 0 forces r = 0). In the language of the fundamental theorem, z^n = 0 has the single root 0 repeated n times, not n different roots.

n is a positive integer. The count “exactly n roots” and the regular polygon need it. The sum being 0 needs n \ge 2: the only first root of w is w itself.

\displaystyle \sqrt[n]{R} means the positive real root of the modulus. The modulus of every root is that one positive number; the different roots differ only in angle.

There is no single “the” root of a non-real number. Which root the formula calls z_0 depends on which argument \varphi was chosen. Rules such as \displaystyle \sqrt{a}\sqrt{b} = \sqrt{ab}, true for nonnegative reals, fail for complex roots: \displaystyle \sqrt{-1}\cdot\sqrt{-1} = i \cdot i = -1, while \displaystyle \sqrt{(-1)(-1)} = \sqrt{1} = 1.

Exact rectangular form needs a known angle. The polar form of every root is exact. Writing a root as a + bi exactly needs the cosine and sine of \displaystyle \frac{\varphi + 2\pi k}{n}; otherwise give a decimal.

Where this goes next

Every polynomial of the form z^n - w now splits completely into n linear factors, a concrete case of the The fundamental theorem of algebra. The roots of unity, written as e^{2\pi ik/n} once the exponential form is available, are where much of later mathematics starts: the symmetries of a regular polygon, constructions with ruler and compass, and the discrete Fourier transform of signal processing.

Exercises

Finding all the roots

Give each root as a + bi unless the exercise says otherwise.

1. The two square roots of -16i.
2. The three cube roots of -27.
3. The three cube roots of 64i.
4. The four fourth roots of 625.
5. The four fourth roots of \displaystyle -128 + 128\sqrt{3}\,i.
6. The five fifth roots of \displaystyle 32\operatorname{cis}\frac{5\pi}{6}, in polar form with angles in [0, 2\pi).
7. The two square roots of 7 - 24i, exactly.

Roots of unity

8. List the fourth roots of unity and check that their sum is 0.
9. List the eighth roots of unity as a + bi.
10. Which of the twelfth roots of unity are also cube roots of unity? Which are also fourth roots of unity?
11. Let \displaystyle \omega = \operatorname{cis}\frac{2\pi}{5}. Write \omega^{13} and \omega^{-2} as \omega^m with 0 \le m \le 4, and find 1 + \omega^2 + \omega^4 + \omega^6 + \omega^8.
12. One cube root of a number w is 2 - i. Find w and the other two cube roots.

Solving equations

13. Solve z^4 + 4 = 0, and check by showing z^4 + 4 = (z^2 + 2z + 2)(z^2 - 2z + 2).
14. Solve z^3 + 8i = 0.
15. Solve z^6 + 7z^3 - 8 = 0.
16. The sixth roots of 729 are the corners of a regular hexagon. Find them, and the length of a side of the hexagon.

Reasoning

17. Tiburcio says the cube root of -8 is -2, so the root formula with k = 0 should give -2. Saturnino writes -8 = 8\operatorname{cis}\pi and gets \displaystyle z_0 = 1 + \sqrt{3}\,i. Who is right?
18. Explain why z^n = w (with w \ne 0) never has a solution z = 0, and why z^n = 0 has only one solution.
19. Show that if z_0 and z_1 are both nth roots of the same nonzero w, then \displaystyle \frac{z_1}{z_0} is an nth root of unity.
20. Show that for n \ge 2 the n different nth roots of any nonzero w add to 0.
21. Explain why a positive real number R has exactly one positive real nth root, and why it also has a negative real nth root exactly when n is even.
22. Show that the side of the regular n-gon whose corners are the nth roots of unity has length \displaystyle 2\sin\frac{\pi}{n}. Check the cases n = 4 and n = 6 directly.