1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

1.5 • THE ALOEBRA OF COMPLEX NUMBERS, REVISITED 31

y

~Zl

~) _
Z1
-+------- --x -+-------+-x

Figure 1.16 Figure 1.17

1.5 The Algebra of Complex Numbers, Revisited

The real numbers are deficient in the sense that not a.JI algebraic operations on
them produce real numbers. Thus, for .;=I to make sense, we must consider the
domain of complex numbers. Do complex numbers have this same deficiency?
That is, if we are to make sense of expressions such as JI+ i, must we appeal
to yet another new number system? The answer to this question is no. In other
words, any reasonable algebraic operation performed on complex numbers gives
complex numbers. Later we show how to evaluate intriguing expressions such as
ii. For now we only look at integral powers and roots of complex numbers.
The important players in this regard are the exponential and polar forms of
a nonzero complex number z = rei^9 = r( cos 8 + i sin 8). By the laws of exponents
(which, you recall, we have promised to prove in Chapter 5) we have

zn = (rei^9 t = rneine = rn [cos (n8) + isin (nil)) , and

z-n = (rei^9 ) - n = r-ne- ine = r-n [cos (-n8) + isin (-n8)).

• EXAMPLE 1. 15 Show that (-J3 - i)^3 = -8i in two ways.

(1-39)

Solution (Method 1): The binomial formula (Exercise 14 of Section 1.2) gives

(Method 2): Using Identity (1-39) and Example 1.12 yields

( - v'3- i)

3

= (2e^1 (~))

3

= (2^3 ei(~)) =8(cos -~

5

7T +isin -~

5

71")

= -8i.

Which method would you use if you were asked to compute (-J3 - i)^30?