5.8 Dr. Trig Learns Complex Numbers
195
Multiplying by 2z3 and rearranging the terms, we get z6 + z5 + z4 + z3 + z2 + z + 1 = 0, or z7 − 1 = 0, z−1 which is obvious. Problem 2.75 Prove the equality √ n π 2π (n − 1)π sin sin · · · sin = n−1 . 2n 2n 2n 2 Solution Consider the polynomial P (X) = X 2n − 1. Its roots are the numbers xk = kπ cos kπ n + i sin n , with k = 0, 1, . . . , 2n − 1. It follows that P (X) = (X − x0 ) × (X − x1 ) · · · (X − x2n−1 ). We observe that, except x0 = 1 and xn = −1, all the other roots are non-real complex numbers and that, for 1 ≤ k ≤ n − 1, x k = x2n−k . Thus, we can write n−1 P (X) = X 2 − 1 (X − xk )(X − x k ) k=1
n−1 x 2 − (xk + x k )x + xk x k . = X2 − 1 k=1 2 Now, xk + x k = 2 cos kπ n and xk x k = 1. Dividing both sides by x − 1, we obtain
x 2n−2 + x 2n−4 + · · · + x 4 + x 2 + 1 =
n−1
x 2 − 2x cos
k=1
kπ +1 . n
For x = 1, the above equality becomes n=2
n−1
n−1 k=1
= 2n−1
kπ 1 − cos n
n−1
sin
k=1
kπ 2n
= 2n−1
2 .
Since sin kπ 2n > 0 for k = 1, . . . , n − 1, we obtain n−1 k=1
sin
√ n kπ = n−1 . 2n 2
n−1 k=1
2 sin2
kπ 2n