James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A58
appendix h Complex Numbers
SOLUTIOn We multiply numerator and denominator by the complex conjugate of
2 1 5i, namely, 2 2 5i, and we take advantage of the result of Example 1:
21 1 3i
2 1 5i
−
21 1 3i
2 1 5i
? 2 2 5i 13 1 11i
− − 13
2 2 5i 2 2 1 5 2 29 1 11
29 i ■
Im
i
0
_i
FIGURE 2
z=a+bi
z=a-bi –
Re
The geometric interpretation of the complex conjugate is shown in Figure 2: z is the
reflection of z in the real axis. We list some of the properties of the complex conjugate
in the following box. The proofs follow from the definition and are requested in Exercise
18.
Properties of Conjugates
z 1 w − z 1 w zw − z w z n − z n
Im
bi
|z|= œ„„„„„„ a@+b@
z=a+bi
b
The modulus, or absolute value, | z | of a complex number z − a 1 bi is its distance
from the origin. From Figure 3 we see that if z − a 1 bi, then
| z | − sa 2 1 b 2
0
a
Re
Notice that
FIGURE 3
zz − sa 1 bidsa 2 bid − a 2 1 abi 2 abi 2 b 2 i 2 − a 2 1 b 2
and so zz − | z | 2
This explains why the division procedure in Example 2 works in general:
z
w − zw
ww −
zw
| w | 2
Since i 2 − 21, we can think of i as a square root of 21. But notice that we also have
s2id 2 − i 2 − 21 and so 2i is also a square root of 21. We say that i is the principal
square root of 21 and write s21 − i. In general, if c is any positive number, we write
s2c − sc i
With this convention, the usual derivation and formula for the roots of the quadratic equation
ax 2 1 bx 1 c − 0 are valid even when b 2 2 4ac , 0:
x − 2b 6 sb 2 2 4ac
2a
EXAMPLE 3 Find the roots of the equation x 2 1 x 1 1 − 0.
SOLUTIOn Using the quadratic formula, we have
x − 21 6 s1 2 2 4 ? 1
2
−
21 6 s23
2
− 21 6 s3 i
2
■
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