Bridge to Abstract Mathematics: Mathematical Proof and Structures

9.4 PROPERTIES OF THE COMPLEX NUMBER FIELD 325

The primary use of the polar form is in connection with multiplication of
complex numbers. The reason this form is of value in complex multiplica-
tion is that we may calculate polar form of a product of two complex num-
bers, expressed in polar form, by adding the arguments (and multiplying
the moduli). We approach this important fact and some of its consequences
by using Definition 4.

DEFINITION 4

If z is any complex number x + yi, we define the complex exponential eZ by

the rule eZ = eX(cos y + i sin y).

If z is purely imaginary, so that x = 0, ez equals eiy which has the value
cos y + i sin y. Hence ex+" = eX(cos y + i sin y) = exeiy for any x, y E R, a
result consistent with a familiar property of the ordinary (real) exponential
function. The complex-valued function f (z) = ez is an extension of the real
exponential function, in that if the complex number z is real, then
ez = ex(cos 0 + i sin 0) = e" (recall Exercise 13, Article 8.1). Other proper-
ties of the real exponential shared by ez are listed in the next theorem.

THEOREM 4

The complex exponential eZ has these properties:

(a) If z = 0, then e" = 1 (i.e., e0 = 1)

(b) @+"=eZew foranyw,z~C

(c) (eZ)" = en" for any n E N and z E C

(d) e-z = 1 I@ for any z E C

The proof of Theorem 4, which should not be difficult, given the following
result, is left as an exercise (Exercise 10).

LEMMA

If X, y E R, then ei(x+y) = eixeiy.

The special case eiY, y E R, of the complex exponential, is the case of

primary interest for our purposes, and this lemma focuses on its most

important property. Note that any nonzero complex number can be ex-

pressed z = reio, where r = 1.~1 > 0 and leiBI = 1.

Proof of lemma -ei(X + Y) = cos (x + y) + i sin (x + y) = (cos x cos y -

sin x sin y) + i(sin x cos y + cos x sin y) = (cos x + i sin x)(cos y +

i sin y) = eixeiy.

THEOREM 5

(a) If z, = r,eiel and z2 = r2eie2, then z1z2 = rlr,ei(e1+e2)

(b) If z = and n E N, then 2' = reine

Proof (a) z, z2 = (r, eiB1)(r2eio2) = rl r2eiB1eie2 - - rrr2ei(B1 +02), where the last

step follows from the lemma.