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Section 98 Powers and Roots of Complex Numbers

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WRITE IN RECT. = 16(1 i(0)) = 16 (x yi)1/p = r1/p (cos ?/p isin ?/p) Example: 3i = i1/3 ... RECT= 3/2 1/2 i (1 i) 1/3. r = (12 12) = 2. T = tan -1 (1 ... – PowerPoint PPT presentation

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Title: Section 98 Powers and Roots of Complex Numbers


1
Section 9-8Powers and Roots ofComplex Numbers
2
DeMoivres Thm.
  • r(cos?isin ?)n rn (cos n? isin n?)
  • n is a pos. integer
  • Find (1-i)8 .this would take forever to
    expandso
  • r ?(12 (-1)2) ?2
  • ? tan-1 (-1/1) -?/4 OR 7?/4
  • ?2(cos 7?/4 isin 7?/4)8
  • (?2)8 (cos 8 7?/4 isin 8 7?/4)
  • POLAR 16(cos 14? isin 14?)
  • WRITE IN RECT. 16(1 i(0)) 16

3
(xyi)1/p r1/p (cos ?/p isin ?/p)
  • Example
  • 3vi i1/3
  • (01i)1/3
  • r v(02 12) v11
  • T tan-1 (1/0) ?/2
  • POLAR 1(cos ?/2 isin ?/2)1/3
  • 1(cos ?/6 isin ?/6)
  • 1(v3/2 i 1/2)
  • RECT v3/2 1/2 i

4
(1 i) 1/3
  • r v(12 12) v2
  • T tan -1 (1/1) ?/4
  • v2(cos ?/4 isin ?/4) 1/3
  • (21/2)1/3 (cos ?/12 isin ?/12)
  • 21/6 (cos ?/12 isin ?/12)
  • 6v2 (cos ?/12 isin ?/12 )
  • 1.08 .29i

5
Any complex has
  • 2, square roots
  • 3, cube roots
  • 4, 4th roots
  • 5, 5th roots
  • P, Pth roots

6
The p distinct pth roots of xyi
  • (xyi)1/p r(cos(?2n?) isin(?2n?)1/p
  • For n0,1,2,,p-1

7
Find the 5,5th roots of -243
  • (xyi)1/p r(cos(?2n?) isin(?2n?)1/p
  • (-2430i)1/5
  • r 243 and ? tan-1 (0/-243) ?
  • 243(cos(?2n?)isin(?2n
    ?)1/5
  • n0 3(cos(?/5) isin(?/5)) 2.43 1.76i
  • n1 3(cos(3?/5) isin(3?/5)) -.93 2.85i
  • n2 3(cos(5?/5) isin(5?/5)) -3 0i -3
  • n3 3(cos(7?/5) isin(7?/5)) -.93 2.85i
  • n4 3(cos(9?/5) isin(9?/5)) 2.43 1.76i

8
Solve x3270
  • X3 -27
  • Find the 3 cube roots -27 0i
  • r 27 , ? ?
  • 27(cos (? 2n?) isin (? 2n?) 1/3
  • n0 3(cos (?/3) isin (?/3)) (3/2)(3v3i)

    2
  • n1 3(cos (3?/3) isin (3?/3)) (-3 0i)
  • n2 3(cos (5?/3) isin (5?/3))
    (3/2)-(3v3i)
    2

9
Homework
  • P 516
  • 11-21 odd, 22,24,35-39
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