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David Terr
Ph.D. Math, UC Berkeley

 

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6.6. De Moivre's Theorem

In the last section, we looked at the polar form of complex numers and proved a beautiful theorem regarding them. In this section, we prove another beautiful result, known as De Moivre's Theorem, which allows us to easily compute powers and roots of complex numbers given in polar form. We will also apply this theorem to many examples.

 

Theorem 6.6.1 (De Moivre's Theorem): For every real number θ and every positive integer n, we have

  • (6.6.2) (cos θ + i sin θ)n = cos nθ + i sin nθ.

Proof: We prove this theorem by induction, i.e. first we prove it for n=1 and then we prove that if (6.6.2) holds for a particular value of n, then it holds for n+1 as well. This suffices to prove the theorem for every positive integer n. (Induction is a commonly-used method for proving mathematical results.) The case n=1 is trivial. Assume (6.6.2) holds for n. Then we have

(cos θ + i sin θ)n+1 = (cos θ + i sin θ)n (cos θ + i sin θ) = (cos nθ + i sin nθ) (cos θ + i sin θ),

where in the last step we used the induction hypothesis, i.e. the assumption that (6.6.2) holds for n. Computing the product on the right yiedls

(cos θ + i sin θ)n+1 = (cos nθ cos θ - sin nθ sin θ) + i (sin nθ cos θ + cos nθ sin θ) = cos (n+1)θ + i sin (n+1)θ,

where we used the addition formulas for sine and cosine in the last step. (Altermatively, we could have applied Theorem 6.5.4 to the two factors on the right side of the previous equation.) Thus, (6.6.2) holds for n+1 as well, whence it holds for every positive integer n.

QED

 

One reason De Moivre's Theorem is useful is that it allows us to compute large powers of complex numbers expressed in polar form without having to carry out every multiplication explicitly. The following examples show this.

 

Example 1: Use De Moivre's Theorem to compute (1 + i)12.

Solution: As we have seen in the previous section, the polar form of 1 + i is √2 (cos π/4 + i sin π/4). Thus, by De Moivre's Theorem, we have

(1 + i)12 = [√2 (cos π/4 + i sin π/4)]12

= (√2)12(cos π/4 + i sin π/4)12

= 26 (cos 3π + i sin 3π)

= 64(cos π + i sin π) = 64(-1) = -64.

 

Example 2: Use De Moivre's Theorem to compute (√3 + i)5.

Solution: It is straightforward to show that the polar form of √3 + i is 2(cos π/6 + i sin π/6). Thus we have

(√3 + i)5 = [2(cos π/6 + i sin π/6)]5

= 25(cos π/6 + i sin π/6)5

= 32(cos 5π/6 + i sin 5π/6)

= 32(-√3/2 + 1/2 i)

= -16√3 + 16 i.

 

A very important application of De Moivre's Theorem is computing nth roots of complex numbers, where n is a posiive integer. First we look at nth roots of 1, also known as nth roots of unity. First we note that 1 may be written in polar form as 1 = cos 2πm + i sin 2πm for every integer m, since cosine and sine are periodic with period 2π. Now consider the complex number z = cos (2πm/n) + i sin (2πm/n), where n is an arbitrary positive integer. By De Moivre's Theorem we have

zn = [cos (2πm/n) + i sin (2πm/n)]n = cos 2πm + i sin 2πm = 1.

Thus we have shown that cos (2πm/n) + i sin (2πm/n) is an nth root of unity. In fact, all the nth roots of unity are obtained this way by plugging in all integer values of m from 0 to n-1. (Every other integer m yields a root of unity identical to one of these.) Thus we now know how to find all n nth roots of unity for every positive integer n.

 

Example 3: Compute the three cube roots of unity.

Solution: From our above discussion, we see that the three cube roots of unity have the form cos (2πm/3) + i sin (2πm/3) for m=0, 1, or 2. Plugging in m=0 yields the root cos 0 + i sin 0 = 1. (It is easy to see that 1 is an nth root of unity for every integer n, since 1n = 1. Plugging in m=0 will always yield this root.) Plugging in m=1 yields the root cos (2π/3) + i sin (2π/3) = -1/2 - √3/2 i and plugging in m=2 yields the root cos (4π/3) + i sin (4π/3) = -1/2 - √3/2 i.

 

Example 4: Compute the four fourth roots of unity.

Solution: The four fourth roots of unity have the form cos (2πm/4) + i sin (2πm/4) for m=0, 1, 2, or 3. As usual, plugging in m=0 yields the root 1. Plugging in m=1 yields the root cos π/2 + i sin π/2 = i. Plugging in m=2 yields cos π + i sin π = -1. Finally, plugging in m=3 yields cos 3π/2 + i sin 3π/2 = -i.

 

The nth roots of unity have a nice geometric interpretation in terms of where they lie in the complex plane. They form a regular n-gon on the unit circle with one vertex at 1. The case n=6 is shown in the following figure.

sixth roots of unity

Figure 6.6.3: The Six Sixth Roots of Unity

So much for nth roots of unity - how about nth roots of general complex numbers? The following lemma, which is an immediate consequence of De Moivre's Theorem, tells us how to compute the n nth roots of an arbitrary complex number, given in polar form.

 

Lemma 6.6.4: Let z = r (cos θ + i sin θ) be an arbitrary complex number. Then the n nth roots w of z are each of the form

  • (6.6.5) w = nr {cos [(θ + 2πm)/n] + i sin [(θ + 2πm)/n]}

where m is an integer ranging from 0 to n-1.

 

Proof: If we apply De Moivre's Theorem to each of the alleged nth roots of z, we find

wn = (nr {cos [(θ + 2πm)/n] + i sin [(θ + 2πm)/n]})n

= r {cos [(θ + 2πm)/n] + i sin [(θ + 2πm)/n]}n

= r [cos (θ + 2πm) + i sin (θ + 2πm)]

= r (cos θ + i sin θ) = z.

Thus, each of these is indeed an nth root of z. Since we have found n distinct roots, this must be all of them.

QED

 

It should be noted that in Lemma 6.6.4 as well as our derivation of the formula for the nth roots of unity, we have implicitly used a very powerful result, known as the Fundamental Theorem of Algebra. This theorem says that every polynomial of degree n factors completely into a product of n linear factors over the complex numbers. Among other things, this theorem implies that every polynomial of degree n has at most n complex roots. Thus, there can be at most n nth roots of a fixed complex number a, since these roots satisfy the degree-n polynomial equation xn - a = 0. We will not prove the Fundamental Theorem of Algebra in this module.

 

We close this section with a few examples.

 

Example 5: Compute the two square roots of i.

Solution: It is easy to see that i has the polar form cos π/2 + i sin π/2. Thus, by Lemma 6.6.4, its square roots are cos π/4 + i sin π/4 = √2/2 + √2/2 i and cos 3π/4 + i sin 3π/4 = -√2/2 - √2/2 i.

 

Example 6: Compute the three cube roots of -8.

Solution: Since -8 has the polar form 8 (cos π + i sin π), its three cube roots have the form 38 {cos[(π + 2πm)/3] + i sin[(π + 2πm)/3]} for m=0, 1, and 2. Thus the roots are 2 (cos π/3 + i sin π/3) = 1 + √3 i, 2 (cos π + i sin π) = -2, and 2 (cos 5π/3 + i sin 5π/3) = 1 - √3 i.

 

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