Chapter 3: Roots of Unity Given a positive integer n, a complex number z is called an nth root of unity if zn = 1. In other words, z is a root of the polynomial Xn 1. Denote by ω , or simply − n by ω if n is understood, the complex number e2πi/n: 2π 2π ω ω = e2πi/n cos + i sin . ≡ n ≡ n n From ωn = (e2πi/n)n = e(2πi/n)n = e2πi = 1 we see that ω is an nth root of unity. The complex numbers 1, ω, ω2, . , ωn−1, (3.1) considered as points in the complex plane, are vertices of a regular n-gon inscribed in the unit circle. For example, when n = 6, they are vertices of a hexagon, as shown in the following figure: Because of this particular geometric arrangement of 1, ω, ω2, . , ωn−1, it is not hard to believe that their sum is zero: 1 + ω + ω2 + ... + ωn−1 = 0. (3.2) We give an algebraic proof of this extremely important identity. From ωn = 1, we have (1 ω)(1 + ω + ω2 + ... + ωn−1) = 1 ωn = 0. − − Clearly ω = 1, that is 1 ω = 0. Hence (2) follows. − Each of ω, ω2, . , ωn−1 is an nth root of unity. Indeed, take z = ωk with 0 k n 1. Then ≤ ≤ − zn = (ωk)n = ωkn = (ωn)k = 1k = 1. 1 Thus 1, ω, ω2, . , ωn−1 are exactly n distinct roots of Xn 1. Each root ωk contributes − to a linear factor X ωk of Xn 1. Hence we have the following factorization of Xn 1: − − − n−1 Xn 1 = (X 1)(X ω)(X ω2) (X ωn−1) = (X ωk). (3.3) − − − − ··· − − k= 0 The last expression is read as the product of X ωk, where k runs from 0 to n 1. − − Exercise 3.1. Prove that n−1 1 ωk = 1. Give a geometric interpretation of this k= 1 | − | identity. Hint: Deduce from (3.3) that n−1(X ωk) = 1 + X + X2 + + Xn−1. k= 1 − ··· Notice that, since ω = 1, we have ωω = ω 2 = 1 and hence ω−1 = ω. On the | | | | other hand, ωn = 1 gives ωn−1ω = 1 and hence ωn−1 = ω−1. Thus we have ωn−1 = ω. Furthermore, from ωn−2ω2 = 1 we have ωn−2 = ω−2 = (ω−1)2 = ω2. we can continue in this manner to get ωn−3 = ω3, etc. In general, we have ωn−k = ωk (3.4) for k = 0, 1, . , n 1. − Example 3.1. Consider the case n = 5. In this case we have 2π 2π ω = e2πi/5 cos + i sin . ≡ 5 5 We are asked to find the value of cos 2π/5. Notice that cos 2π/5 is the real part of ω, that is, (ω + ω)/2. So it is enough to find ω + ω. In the present case, (2) gives 1 + ω + ω2 + ω3 + ω4 = 0. On the other hand, (3.4) above gives ω4 = ω and ω3 = ω2. Thus we have 1 + ω + ω2 + ω2 + ω = 0, or 1 + (ω + ω) + (ω2 + ω2) = 0. Now (ω + ω)2 = ω2 + ω2 + 2ωω = ω2 + ω2 + 2. (Recall that ωω = ω 2 = 1.). So | | ω2 + ω2 = (ω + ω)2 2. − 2 Hence 1 + (ω + ω) + (ω2 + ω2) = 0 becomes (ω + ω)2 + (ω + ω) 1 = 0. We have shown − that ω+ω 2 cos 2π/5 is a root of X2 +X 1. The roots of X2 +X 1 can be obtained ≡ − − by the usual formula for solving quadratic equations. The result is X = ( 1 √5)/2. − ± Since cos 2π/5 is a positive number, necessarily 2 cos 2π/5 = ( 1 + √5)/2. Thus − 2π √5 1 cos = − 5 4 which is the answer we are looking for. Example 3.2. Now we try to factorize the polynomial X5 1 into a product of real − polynomials. In the present case, identity (3.3) becomes X5 1 = (X 1)(X ω)(X ω2)(X ω3)(X ω4). (3.5) − − − − − − This is not the factorization we are seeking for, since X ωk (1 k 4) are not real − ≤ ≤ polynomials. Using ω4 = ω and ω3 = ω2, we rewrite this identity as X5 1 = (X 1) (X ω)(X ω) (X ω2)(X ω2) . − − { − − }{ − − } Now (X ω)(X ω) = X2 (ω + ω)X + ωω. From the previous example we see that − − − ω + ω = (√5 1)/2 and ωω = 1. So we have − √5 1 (X ω)(X ω) = X2 − X + 1 − − − 2 which is a real polynomial. Next, we have (X ω2)(X ω2) = X2 (ω2 + ω2)X + ω2ω2. − − − Here, ω2ω2 = (ωω)2 = 1 and 2 √5 1 √5 + 1 ω2 + ω2 = (ω + ω)2 2 = − 2 = . − 2 − − 2 Substituting the last expressions into (5), we have 1 √5 1 + √5 x5 1 = (x 1) x2 + − x + 1 x2 + x + 1 , − − 2 2 which is the required factorization. Example 3.3. We would like to find the values of cos π/12 and sin π/12. As we know, it is wise to consider eiπ/12 instead. The following computation is inspired by a 1 1 1 simple observation: = . 12 3 − 4 π π π π eπi/12 = e(πi/3)−(πi/4) = eπi/3e−πi/4 = cos + i sin cos i sin 3 3 4 − 4 1 √3 √2 √2 √6 + √2 √6 √2 = + i i = + i − . 2 2 2 − 4 2 4 3 π √6 + √2 π √6 √2 Hence we have cos = and sin = − . 12 4 12 4 Alternatively, we let ω = a + bi = eπi/12. Then √3 1 ω2 = eπi/6 = + i. 2 2 On the other hand, ω2 = (a2 b2) + 2ab i. Comparing the real and the imaginary parts of − these two expressions of ω, we obtain 1 √3 2ab = and a2 b2 = . 2 − 2 Add the last identity to a2 + b2 = ω 2 = 1, we get | | 1 + √3 2a2 = . 2 Dividing both sides by 2 and then taking the square root, we arrive at π 1 cos = a = 1 + √3. 12 2 The reader should check that this is the same answer as the previous one for cos π/12, even though they look different. √6 + √2 1 Exercise 3.2. Verify = 1 + √3 directly. 4 2 Exercise 3.3. Plot the points of the set Z[ω] Z + Zω ≡ consisting of all those complex numbers of the form a+bω, where a, b are integers, for each of the following values of ω: (a) ω = i; (b) ω = eπi/6; (c) ω = eπi/3. Exercise 3.4. Let ω = e2πi/3 and consider the ring R = Z[ω] Z+Zω; for the notation, ≡ see the last exercise). Prove: (a) for each Z R, z 2 is an integer; and (b)* if an element ∈ | | z in R is invertible (in the sense that there is an element w in R such that zw = 1), then z is one of the following: 1, i, (1 + ω). (The “*” sign means that it is optional.) ± ± ± Take any positive integer and let ω = e2πi/n. Consider the set C = 1, ω, ω2, ω3, . , ωn−1 n { } 4 of all nth roots of unity. This set forms a group with respect to multiplication in the sense that any product of two elements in Cn is also in Cn and any element in Cn has an inverse 7 4 6 10 7+ 3 3 in Cn. For example, when n = 7, we have ω = 1 and hence ω ω = ω = ω = ω , ω−3 = ω7ω−3 = ω4, etc. The group Cn is a cyclic group in the sense that there is an element called a gen- erator, such that its powers fill the whole group. Clearly, ω is a generator. But usually this is not the only one. More Exercises 3.5. Write out all nth roots of unity, for the following values of n: n = 2, 3, 4, 6, 8, 12. 3.6. Let ω = eπi/3. Check: (a) 1+ω3 = 0, 1+ω2 +ω4 = 0 and 1+ω+ω2 +ω3 +ω4 +ω5 = 0; (b) ω + ω5 = 1, ω ω5 = √3 i and ω4 = ω2. − 3.7. Check that if ω = eπi/m (m is a positive integer), then, for any integer k, ωk = ωm+ k. − 3.8. Verify that if ω = e2πi/5, then a = ω + ω4 and b = ω2 + ω3 are roots of t2 + t 1. − 3.9. Prove that if n 3 and if ω = 2πi/n, then 1 + ω2 + ω4 + ω6 + + ω2n−2 = 0. ≥ ··· 3.10 Factorize the polynomials X6 1 and X8 1 into real (irreducible) polynomials as in − − Example 3.2. 3.11 Verify that if ω = e2πi/7, then a = ω + ω2 + ω4 and b = ω3 + ω5 + ω6 are roots of t2 + t + 2. The material of the rest of this chapter is optional. We continue with the discussion of the cyclic group Cn. Suppose that m is a positive integer divisible by n, say n = mk, where k is another positive integer. Then Cm is generated by e2πi/m. By the substitution m = n/k, e2πi/m becomes e2πik/n ωk; (here, ≡ we still use the symbol ω for e2πi/n). Clearly a power of ωk is a power of ω. This shows that Cm is contained in Cn. Certainly Cm is itself a group under multiplication. So l Cm is a subgroup of Cn. To give a simple example to illustrate this, let us take n = 6 so that ω = e2πi/6 ωpii/3. Then ω6 = 1 and C = 1, ω, ω2, ω3, ω4, ω5 . Both C and C ≡ 6 { } 2 3 are subgroups of C ; in fact, C = 1, ω3 1, 1 and C = 1, ω2, ω4 1, η, η2 , 6 2 { } ≡ { − } 3 { } ≡ { } where η = ω2 e2πi/3.