1.5 • THE ALGEBRA OF COMPLEX NUMBERS, R EVISITED 35yFigure 1.1 9 The five solutions to t he equation z5 = c.
As before, we get n distinct solutions given by
Zk =p•e .1. i~ n =p• .1. ( COS ¢ +27rk +tS.. IIl ¢+27rk) ,
n n
{1-45)fork= 0, 1, 2, ... , n - 1.Each solution in Equation (1-45) can be considered an nth root of c. Ge-
ometrically, the nth roots of c are equally spaced points that lie on the circle
Gp!; (O) = { z : lzl = p~} and form the vertices ofa regular polygon with n sides.
Figure 1.19 illustrates the case for n = 5.
It is interesting to note that if ( is any particular solution to the equation
z" = c, then all solutions can be generated by multiplying ( by the various nth
roots of unity. That is, the solution set is
(1-46)The reason for this is that if(" = c, then for any j = 0, 1, 2, ... , n - 1,
((w~)" = ( n (w~)" = (" (w;-:)^1 = ("(1) = c, and that multiplying a number
by wn = ei~ increases an argument of that number by^2 ;, so that Expressions
(1- 46 ) contain n distinct values.- EXAMPLE 1.20 Find all cube roots of Si= 8 (cos~+ isin ~).
Solutio n Formula (1- 45 ) gives
(l! + 27rk l! + 27rk)
zk= 2 cos^2
3+isin^2
3, fork=0,1,2.The Cartesian forms of the solutions are zo = v'3+i, z 1 = -J3+i, and z2 = - 2i,
as shown in Figure 1.20.