Dividing by n, we see that O may have any one of the values By substituting, in the assumed expression, rei for the value of the root, we have 2π 4π 6п secutive values we shall be carried all round the circle. The solution which we thus reach may be represented thus: Let the length of each arc be a, so that na = 360° = 2′′. The result can be readily proved by raising each of these quan tities to the sixth power. The roots may also be constructed as in the anexed figure. If angle XOA = AOB etc. = 60°, then, since ОР sin 60° and = cos 60°, ΟΙ B 1. Find and construct the eighth roots of unity, or the roots of the equation a1 = 0. 2. Find the roots of the equation a1 10. 91. Relations between the roots of unity. If we represent by any such quantity as cos ai sin a, we have, by what precedes, a2= cos 2a + i sin 2a; Hence the formation of the powers of x may be represented geometrically by laying off equal arcs around a circle. If x is any nth root of unity, then measuring off its angle a n times will bring us back to the starting-point. If a is itself the nth part of the circumference, then the remaining roots as given in (3) are the first n powers of x. Hence: All the roots of unity are powers of the root corresponding to the smallest arc. From this it follows that if we measure off with a pair of dividers, from 0 to any division-point, the mth, for instance, and repeat the measure n times, the nth measure will end at the zero-point. This is evident of itself, because n measures of m arcs each will measure off mn arcs; and because n of the arcs make up a circumference, the mn arcs will extend around the circle exactly m times. But it does not follow that any such series of n measures will include all the roots. Suppose, for example, that in the preceding figure, where n = 15, we measure arcs of 6a. The 15 successive points reached with the dividers will then be 0, 6, 12, 3, 9, 0, 6, 12, 3, 9, 0, 6, 12, 3, 9, 0. This series includes only 5 of the points of division, each of these 5 being repeated 3 times, while the remaining 10 have not been included at all. If we take the measure 4a in our dividers, the points of division included in the series will be 7 8 . 5 3 13 10 12 14 0, 4, 8, 12, 1, 5, 9, 13, 2, 6, 10, 14, 3, 7, 11, 0, which comprise all n points. Hence in this case all the roots are powers of x or of cos 4a + i sin 4a. 92. We now have the following proposition, where we put If m is prime to n, all the nth roots of unity may be repre sented as powers of cos ma + i sin ma. Proof. Starting from any point of the circle, and measuring off equal arcs, each of length ma, let p be the smallest number of measures which will bring us back to the starting-point. total length of arc measured off will then be pma. The Since we are brought back to the starting-point, we must have measured off an entire number of circumferences. Let q be that entire number. Because each circumference = na, the whole g circumferences measured gna. Therefore Because, by hypothesis, m and n are prime to each other, the n fraction is irreducible, and the smallest values of p and q are n m and m respectively. Therefore any n measures will end at n different points of the circumference, and will therefore include all n points. Def. A root of unity whose powers include all other roots is called a primitive root. Cor. If n is a prime number, all the roots are primitive roots. EXERCISES. 1. The 15th roots of unity being cos 24° + i sin 24°, cos 48° i sin 48°, it is required to find which of these roots are primitive. Prove the following propositions: etc., 2. If n is a prime number, all the nth roots of unity are primitive roots. 3. If a be any primitive nth root of unity, and if p be any number prime to n, then x will also be a primitive root. NOTE. In the preceding theorem this is proved for the case when x is the root corresponding to the smallest angle. The proposition now enunciated extends to the case in which we start from any multiple of this angle prime to n. |
