1 .5 • THE ALGEBRA OF COMPLEX NUMBERS, REVISITED 33Theorem 1.4 implies that if we can find n distinct solutions to the equationzn = c (or zn - c = 0), we will have found all the solutions. We begin our
search for these solutions by looking at the simpler equation zn = 1. Solving
this equation will enable us to handle the more general one quite easily.
To solve zn = 1 we first note that, from Identities (1-29) and (1-37), we
can deduce an important condition that determines when two nonzero complex
numbers are equal. If we let z 1 = r 1 ei^91 and Z2 = r2eill>, then
(1-41)where k is an integer. That is, two nonzero complex numbers are equal iff their
moduli agree and an argument of one equals an argument of the other to within
an integral multiple of 27r.
We now find all solutions to z n = 1 in two stages, with each stage corre-
sponding to one direction in the iff part of Relation (1- 41 ). First, we show that
if we have a solution to zn = 1, then the solution must have a certain form.
Second, we show that any quantity with that form is indeed a solution.
For the first stage, suppose that z = r e^19 is a solution to z" = 1. P utting
the latter equation in exponential form gives r neinll = l · e^1 ·^0 , so Relation (1-41)
implies that r n = 1 and nO = 0 + 27rk. In other words,
21rk
r = 1 and () = - ,
n
where k is an integer.
(1-42)So, if z = rei^9 is a solution to zn = 1, then Relation (1-42) must be true.
This observation completes the first stage of our solution strategy. For the second
stage, we note that if r = 1, and () =^2 ~k, then z = rei^9 = e^12 ~· is indeed a
solution to zn = 1 because zn = ( e'" »w•)" = e' ·2 .. k = 1. For example, if n = 7
and k = 3, then z = eillf is a solution to z^7 = 1 because ( ei~)
7
= e'^6 " = 1.Furthermore, it is easy to verify that we get n distinct solutions to zn = 1
(and, therefore, all solutions , by Theorem 1.4) by setting k = 0, 1, 2, ... , n - 1.
The solutions for k = n , n + 1, ... merely repeat those for k = 0, 1,... , because
the arguments so generated agree to within an integral multiple of 2 7r. As we
stated in Section 1.1, then solutions can be expressed as
-2h 2 7rk.. 27rk
Z k = e'....., = cos --+ism - , for k = 0, 1, 2,... , n - 1.
n n
(1-43)T hey are called the n t h roots o f un ity.
·2n·O 0
When k = 0 in Equat ion (1-43), we get zo = e•-.-= e = 1, which is
a rather trivial result. T he first interesting root of unity occurs when k = 1,
giving z 1 = e^1 1f. This particular value shows up so often that mathematicians
have given it a special symbol.