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Pre-Calc Exam Notes 143

# Pre-Calc Exam Notes 143 - sin θ and integer n ≥ 1 we get...

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Complex Numbers Section 6.3 143 Example6.11 Let z 1 = 6(cos 70 + i sin 70 ) and z 1 = 2(cos 31 + i sin 31 ). Find z 1 z 2 and z 1 z 2 . Solution: By formulas (6.5) and (6.6) we have z 1 z 2 = (6)(2)(cos (70 + 31 ) + i sin (70 + 31 )) z 1 z 2 = 12(cos 101 + i sin 101 ) , and z 1 z 2 = 6 2 (cos (70 31 ) + i sin (70 31 )) z 1 z 2 = 3(cos 39 + i sin 39 ) . For the special case when z 1 = z 2 = z = r (cos θ + i sin θ ) in formula (6.5), we have [ r (cos θ + i sin θ )] 2 = r · r (cos ( θ + θ ) + i sin ( θ + θ )) = r 2 (cos 2 θ + i sin 2 θ ) , and so [ r (cos θ + i sin θ )] 3 = [ r (cos θ + i sin θ )] 2 · r (cos θ + i sin θ ) = r 2 (cos 2 θ + i sin 2 θ ) · r (cos θ + i sin θ ) = r 3 (cos (2 θ + θ ) + i sin (2 θ + θ )) = r 3 (cos 3 θ + i sin 3 θ ) , and continuing like this (i.e. by mathematical induction ), we get: Theorem 6.1. De Moivre’s Theorem: 7 For any integer n 1, [ r (cos θ + i sin θ )] n = r n (cos n θ + i sin n θ ) . (6.7) We define z 0 = 1 and z n = 1/ z n for all integers n 1. So by De Moivre’s Theorem and formula (6.5), for any z = r (cos θ + i sin θ
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Unformatted text preview: sin θ ) and integer n ≥ 1 we get z − n = 1 z n = 1(cos 0 ◦ + i sin 0 ◦ ) r n (cos n θ + i sin n θ ) = 1 r n (cos (0 ◦ − n θ ) + i sin (0 ◦ − n θ )) = r − n (cos ( − n θ ) + i sin ( − n θ )) , and so De Moivre’s Theorem in fact holds for all integers. 8 7 Named after the French statistician and mathematician Abraham de Moivre (1667-1754). 8 There is a way of de±ning z n when n is a real (or complex) number, so that De Moivre’s Theorem holds for any real number n . See pp. 59-60 in R.V. CHURCHILL, Complex Variables and Applications , 2nd ed., New York: McGraw-Hill Book Co., 1960....
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