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The complex numbers. The set of real numbers. A function f : AmapsB is said to be surjective (or onto) if for every y element of B there exists x element of A such that f(x) = y. Let R and S be rings. A ring homomorphism from R to S is a function f :RmapsS such that for a,b element of R, we have (a) f(a + b) = f(a) + f(b),, (b) f(a · b) = f(a) · f(b). If <f is injective, f is a ring isomorphism A group is considered cyclic if it is generated by a single element. phi:nmapsphi(n) = the number of integers a, (1 < a < n) such that (a,n) = 1. If E is a field, then a subfield of E is a subset of E which is also a field with respect to the operations of E.

The Complex Numbers

The complex numbers C are introduced in high school as the set of all quantities of the form a + bi, where a and b are real numbers and i = square root of -1. This last condition means simply that i2 = -1. We say that two complex numbers a + bi and c + di are equal if a = c and b = d. Further, we define addition and multiplication of complex numbers by

(1)
(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 0 + 0i and the multiplicative identity is 1 + 0i. The additive inverse of a + bi is (-a) + (-b)i, while if a + bi is nonzero, then either a not equal 0 or bnot equal 0 and the multiplicative inverse of a + bi 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 a element of R with the complex number a + 0i. More precisely, the mapping

amapsa + 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 i = square root of -1. And it is not at all clear what i is or where it comes from. This gives complex numbers a somewhat mysterious quality and accounts for the terminology which calls i an "imaginary number." The way out of this logical morass was discovered by Gauss in his dissertation. Instead of considering quantities of the form a + bi, let us consider ordered pairs (a,b). We say that two ordered pairs (a,b) and (c,d) are equal if a = c and b = d. We define addition and multiplication of the ordered pairs via

(1')
(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 {(a,0) | a element of R} of C. Thus we have explicitly constructed a field C containing R. Consider the element (0,1) of C. From the definition of multiplication in C,

(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 = (0,1). Then every complex number can be uniquely written in the form a + b · i, where a,b element of R. The field C which we have constructed is precisely what we are accustomed to think of as the field of complex numbers.

Let us proceed to study the properties of the complex numbers in somewhat more detail. If alpha = a + ib element of C, we define the complex conjugate alpha conjugate of alpha by

alpha conjugate = a - ib.

The following are easily verified properties of complex conjugation: Let alpha,beta element of C.

I. conjugate sum = alpha conjugate + beta conjugate.

II. conjugate product = alpha conjugate · beta conjugate.

III. conjugate of conjugate = alpha.

IV. alpha conjugate = alpha if and only if alpha is real.

V. The mapping alphamapsalpha conjugate of C onto C is a surjective isomorphism which is the identity isomorphism on R.(This follows from I, II and IV.)

VI. alpha · alpha conjugate is a real number and alpha · alpha conjugate > 0. (In fact, if x = a + ib, then alpha · alpha conjugate = a2 + b2 > 0.)

We define the norm of the complex number a + ib to be a2 - b2.

Let us define the absolute value of alpha, denoted |alpha|, by

|alpha| = conjugate square root = real square root    (alpha = a + ib).

By VI, alpha · alpha conjugate is a nonnegative real number, so |alpha| makes sense and is a nonnegative real number. It is possible to give the absolute value of a complex number a geometric interpretation as follows: Let us associate to the complex number alpha = a + ib the point on the Cartesian plane with coordinates (a,b). Then this sets up a one-to-one correspondence between complex numbers and points of the Cartesian plane. Moreover, it is clear from Figure 1 that |a + ib| is just the distance of the point (a,b) from the origin.

complex plane
Figure 1: One-to-One Correspondence Between R2 and C.

Let rho = |a + ib| = real square root and let theta be the angle which the line connecting the origin and (a,b) makes with the positive half of the x-axis. By the definitions of sintheta and costheta, we know that

(3)
a = rhocostheta,   b = rhosintheta.

Therefore, we may represent the complex number a + ib in the form

(4)
a + ib = rho(costheta + i sintheta).

This representation is called the polar form of a + ib and theta is called the argument of a + ib.

Let us illustrate geometrically the meaning of addition and multiplication of complex numbers. Let a + ib, c + id element of C. Then

(a + ib) + (c + id) = (a + c) + i(b + d)

corresponds geometrically to the point (a + c, b + d). Thus, complex numbers add according to the usual "parallelogram law" of elementary physics (see Figure 2).

complex addition
Figure 2: Addition in C

In order to describe multiplication, it is best to use the polar form. Let

a + ib = rho1(costheta1 + i sintheta1),
c + id = rho2(costheta2 + i sintheta2),

Then

(a + ib) · (c + id) = rho1rho2([costheta1costheta2 - sintheta1sintheta2]
(5)
+ i[costheta1sintheta2 + costheta2sintheta1])
= rho1rho2[cos(theta1 + theta2) + i sin(theta1 + theta2)].

Thus, (a + ib) · (c + id) corresponds to the point at distance rho1rho2 from the origin and has argument theta1 + theta2. Note that we have used the addition formula for sin and cos, which should be familiar from trigonometry. From (5), an induction argument immediately shows that for n > 0,

(6)
(a + ib)n = rhon[cos(ntheta) + i sin(ntheta)].

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 alpha = rho(costheta + i sintheta) be a complex number. Let us inquire as to whether alpha has any nth roots in C, where n is a positive integer. We say that beta is an nth root of alpha if betan = alpha. If alpha = 0, then beta = 0 is an nth root, and is in fact the only nth root. Thus, let us assume that alphanot equal 0. Then we have rho > 0. First observe that alpha has at most n nth roots in C, since any nth root of alpha is a zero of the polynomial Xn - alpha, and this polynomial has at most n zeros. Next, consider the following n distinct complex numbers:

(7)
betak = rho1/n[cos(theta/n + 360/n · k) + i sin(theta/n + 360/n · k)]
(k = 0,1,...,n - 1),

where rho1/n is the positive nth root of rho. By de Moivre's theorem,

betakn = (rho1/n)n[cos(theta + 360 · k) + i sin(theta + 360 · k)]
= rho[costheta + i sintheta]
= alpha.

Therefore, betak is and nth root of alpha 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

zetan = cos(360/n) + i sin(360/n).
Then from (8), zetan is an nth root of 1, and in fact, corresponds to k = 1. Moreover, from de Moivre's theorem, we see that

zetank = cos(360k/n) + i sin(360k/n).

Therefore, from (8), we see that all the nth roots of unity are powers of zetan. 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 zetan.

We will denote the group of nth roots of unity by Xn. A generator of Xn is called a primitive nth root of unity. Since the order of zetakn is n/(k,n), we see that the order of zetakn is n only if (k,n) = 1. Therefore, the primitive nth roots of unity are given by

zetank    [0 < k < n - 1, (k,n) = 1].

There are exactly phi(n) such roots of unity.

We found above that the n nth roots of a nonzero complex number alpha = rho(costheta + i sintheta) are given by betak (k = 0,...,n - 1). where betak is defined by (7). Note however, that by (5), we have

betak = (rho1/n[cos(theta/n) + i sin(theta/n)])[cos(360k/n) + i sin(360k/n)]
= beta0zetank.

Thus we obtain the n nth roots of alpha by multiplying the fixed nth root beta0 by all possible nth roots of unity. Therefore, we have

Theorem 3: Let alpha be a nonzero complex number. Let beta be an nth root of alpha. Then the nth roots of alpha are all of the form beta · zeta, where zeta is an nth root of unity. Conversely, every such quantity is an nth root of alpha.