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Unformatted text preview: 2kπ + 4 7 + i sin π 2kπ + 4 7 , where k = 0, 1, 2, 3, 4, 5, 6. Example 1.7.2. The case c = 1 is particularly important, as we get the nth roots of 1. Note that R = 1 and α = 0. It follows that the nth roots of unity are given by z = cos
Chapter 1 : The Number System 2kπ 2kπ + i sin , n n where k = 0, 1, . . . , n − 1.
page 13 of 19 First Year Calculus c W W L Chen, 1982, 2005 Example 1.7.3. Consider the polynomial p(z ) = z 3 − z 2 + 2z − 2, and observe that z = 1 is a root. Furthermore, p(z ) = (z − 1)(z 2 + 2), so that two other solutions are given by the roots of the equation z 2 = −2. It is easy to see that −2 = 2(cos π + i sin π ) in polar form. It follows that the two roots of z 2 = −2 are given by z= √ 2 cos π π + i sin 2 2 = √ 2i and z= √ 2 cos 3π 3π + i sin 2 2 √ = − 2i. Example 1.7.4. Consider the polynomial p(z ) = z 6 − 2z 3 +4 = 0. Writing w = z 3 , we have w2 − 2w +4 = 0, with roots √ √ √ 2 ± −12 w= = 1 ± −3 = 1 ± 3i. 2 To ﬁnd the roots of p(z ), we have to ﬁnd all the roots of (16)...
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This note was uploaded on 02/01/2009 for the course MATH 2343124 taught by Professor Staff during the Fall '08 term at UCSD.
 Fall '08
 staff
 Math, Calculus

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