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The Complex NumbersThe complex numbers C are introduced in high school as the set of all quantities of the form
(a + bi) + (c + di) = (a + c) + (b + d)i,
(2)
(a + bi) · (c + di) = (ac  bd) + (ad + bc)i.
With respect to these operations, C becomes a field. The additive identity is
We will leave the verification of these facts, as well as the associativity, commutativity, and distributivity laws for the reader to check. Note that we may regard the real numbers R to be a subfield of C by defining
aa + 0i
is an isomorphism. In what follows, we will always regard R as a subfield of C. All of the above is probably more or less familiar from high school algebra. However, there is a basic defect with the above definition of C. In our above explanation, we have introduced
(a,b) + (c,d) = (a + c, b + d),
(2')
(a,b) · (c,d) = (ac  bd, ad + bc).
We define C to be the set of all these ordered pairs. With respect to the addition (1') and multiplication (2'), C becomes a field. Moreover, R may be identified with the subfield
(0,1) · (0,1) = (1,0).
That is, (0,1)^{2} equals 1. Thus, our field C contains a square root of 1. Let us set i = Let us proceed to study the properties of the complex numbers in somewhat more detail. If
= a  ib.
The following are easily verified properties of complex conjugation: Let II. III. IV. V. The mapping VI. We define the norm of the complex number Let us define the absolute value of , denoted
 = = ( = a + ib).
By VI, Let
a = cos, b = sin.
Therefore, we may represent the complex number
a + ib = (cos + i sin).
This representation is called the polar form of Let us illustrate geometrically the meaning of addition and multiplication of complex numbers. Let
(a + ib) + (c + id) = (a + c) + i(b + d)
corresponds geometrically to the point
Figure 2: Addition in C
In order to describe multiplication, it is best to use the polar form. Let
a + ib = _{1}(cos_{1} + i sin_{1}),
c + id = _{2}(cos_{2} + i sin_{2}),
Then
(a + ib) · (c + id) = _{1}_{2}([cos_{1}cos_{2}  sin_{1}sin_{2}]
(5)
+ i[cos_{1}sin_{2} + cos_{2}sin_{1}])
= _{1}_{2}[cos(_{1} + _{2}) + i sin(_{1} + _{2})].
Thus,
(a + ib)^{n} = ^{n}[cos(n) + i sin(n)].
This last formula is usually known as de Moivre's theorem. Let us use de Moivre's theorem to get some information about the nth roots of complex numbers. Let
_{k} = ^{1/n}[cos(/n + 360/n · k) + i sin(/n + 360/n · k)]
(k = 0,1,...,n  1),
where ^{1/n} is the positive nth root of . By de Moivre's theorem,
_{k}^{n} = (^{1/n})^{n}[cos( + 360 · k) + i sin( + 360 · k)]
= [cos + i sin]
= .
Therefore, _{k} is and nth root of and we have proved the following theorem. Theorem 1: Every nonzero complex number has n nth roots in C. The formula (7) can be used to compute nth roots. For example let us look at the formula for the nth roots of 1. Since
1 = 1 · (cos 0 + i sin 0),
we see that the nth roots of 1 are given by (8)
cos(360k/n) + i sin(360k/n) (k = 0,1,...,n  1).
Let
_{n} = cos(360/n) + i sin(360/n).
Then from (8), _{n} is an nth root of 1, and in fact, corresponds to
_{n}^{k} = cos(360k/n) + i sin(360k/n).
Therefore, from (8), we see that all the nth roots of unity are powers of _{n}. Thus, we have the following theorem. Theorem 2: The nth roots of 1 in C form a cyclic group of order n. A generator of this group is _{n}. We will denote the group of nth roots of unity by X_{n}. A generator of X_{n} is called a primitive nth root of unity. Since the order of _{k}^{n} is
_{n}^{k} [0 < k < n  1, (k,n) = 1].
There are exactly (n) such roots of unity. We found above that the n nth roots of a nonzero complex number
_{k} = (^{1/n}[cos(/n) + i sin(/n)])[cos(360k/n) + i sin(360k/n)]
= _{0}_{n}^{k}.
Thus we obtain the n nth roots of by multiplying the fixed nth root _{0} by all possible nth roots of unity. Therefore, we have Theorem 3: Let be a nonzero complex number. Let be an nth root of . Then the nth roots of are all of the form · , where is an nth root of unity. Conversely, every such quantity is an nth root of . 
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