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

A62
appendix h Complex Numbers
From the fact that sine and cosine have period 2, it follows that
Thus
n − 1 2k or − 1 2k
n
w − r
1ynFcosS
1 2k
n
D 1 i sinS 1 2k
n
Since this expression gives a different value of w for k − 0, 1, 2, . . . , n 2 1, we have
the following.
DG
3 Roots of a Complex Number Let z − rscos 1 i sin d and let n be a positive
integer. Then z has the n distinct nth roots
1 2k
w k − r 1ynFcosS D 1 i sinS DG
1 2k
n
n
where k − 0, 1, 2, . . . , n 2 1.
Notice that each of the nth roots of z has modulus | w k | − r 1yn . Thus all the nth roots of
z lie on the circle of radius r 1yn in the complex plane. Also, since the argument of each
suc cessive nth root exceeds the argument of the previous root by 2yn, we see that the
nth roots of z are equally spaced on this circle.
EXAMPLE 7 Find the six sixth roots of z − 28 and graph these roots in the complex
plane.
SOLUTIOn In trigonometric form, z − 8scos 1 i sin d. Applying Equation 3 with
n − 6, we get
1 2k
w k − 81y6Scos 1 i sin D
1 2k
6
6
We get the six sixth roots of 28 by taking k − 0, 1, 2, 3, 4, 5 in this formula:
w 0 − 8 1y6Scos 6 6D 1 i sin − s2 S D
s3
2 1 1 2 i
w
Im
œ„2i
w¡
w¸
w 1 − 8 1y6Scos 2 1 i sin 2D − s2 i
w 2 − 8 1y6Scos 5 6
w 3 − 8 1y6Scos 7 6
D 1 i sin 5 − s2 S2 D
s3
6 2 1 1 2 i
D 1 i sin 7 − s2 S2 D
s3
6 2 2 1 2 i
_œ„2
0
œ„2
w£
_œ„2i
w¢
w∞
FIGURE 9
The six sixth roots of z − 28
Re
w 4 − 8 1y6Scos 3 2
1 i sin 3 2
D − 2s2 i
11
w 5 − 81y6Scos 1 i sin D 11 − s2 S D
s3
6 6 2 2 1 2 i
All these points lie on the circle of radius s2 as shown in Figure 9.
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