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7.6 Powers and Roots of Complex Numbers • Powers of Complex numbers [r (cos i sin )]2 [r (cos i sin )r (cos i sin )] r r[cos( ) i sin( )] r 2 (cos 2 i sin 2 ) In the same way, [r (cos i sin )]3 r 3 (cos 3 i sin 3 ) Copyright © 2007 Pearson Education, Inc. Slide 10-1 7.6 De Moivre’s Theorem De Moivre’s Theorem If r(cos + i sin ) is a complex number, and n is any real number, then [r (cos i sin )]n r n (cos n i sin n ). In compact form, this is written [r cis ]n r n (cis n ). Copyright © 2007 Pearson Education, Inc. Slide 10-2 7.6 Finding a Power of a Complex Number Example Find (1 i 3) and express the result in rectangular form. 8 Solution (1 i 3)8 [2(cos 60 i sin 60 )]8 2 [cos(8 60 ) i sin(8 60 )] 8 Convert to trigonometric form. 256(cos 480 i sin 480 ) 256(cos 120 i sin 120 ) 3 1 256 i 2 2 128 128i 3 Copyright © 2007 Pearson Education, Inc. 480º and 120º are coterminal. cos120º = -1/2; sin120º = 3 / 2 Rectangular form Slide 10-3 7.6 Roots of Complex Numbers nth Root For a positive integer n, the complex number a+bi is the nth of the complex number x + yi if (a + bi)n = x + yi. • To find three complex cube roots of 8(cos 135º + i sin 135º), for example, look for a complex number, say r(cos + sin ), that will satisfy [r (cos i sin )]3 8(cos135 i sin 135 ). Copyright © 2007 Pearson Education, Inc. Slide 10-4 7.6 Roots of Complex Numbers By De Moivre’s Theorem, [r (cos i sin )]3 8(cos135 i sin 135 ). becomes r 3 (cos 3 i sin 3 ) 8(cos135 i sin 135 ). Therefore, we must have r3 = 8, or r = 2, and 3 135 360 k , k any integer 135 360 k , k any integer. 3 Copyright © 2007 Pearson Education, Inc. Slide 10-5 7.6 Roots of Complex Numbers Let k take on integer values 0, 1, and 2. 135 0 k 0, 45 3 135 360 k 1, 165 3 135 720 k 2, 285 3 It can be shown that for integers k = 3, 4, and 5, these values have repeating solutions. Therefore, all of the cube roots (three of them) can be found by letting k = 0, 1, and 2. Copyright © 2007 Pearson Education, Inc. Slide 10-6 7.6 Roots of Complex Numbers When k = 0, the root is When k = 1, the root is When k = 2, the root is 2(cos 45º + i sin 45º). 2(cos 165º + i sin 165º). 2(cos 285º + i sin 285º). nth Root Theorem If n is any positive integer, r is a positive real number, and is in degrees, then the nonzero complex number r(cos + i sin ) has exactly n distinct nth roots, given by n r (cos i sin ), where 360 k n Copyright © 2007 Pearson Education, Inc. 360 k or , k 0,1, 2, , n 1 n n Slide 10-7 7.6 Finding Complex Roots Example Find the two square roots of 4i. Write the roots in rectangular form, and check your results directly with a calculator. Solution First write 4i in trigonometric form as 4i 4 cos i sin . 2 2 Here, r = 4 and = /2. The square roots have modulus 4 2 and arguments as follows. 2 k k 2 2 4 Copyright © 2007 Pearson Education, Inc. 2 Slide 10-8 7.6 Finding Complex Roots Since there are two roots, let k = 0 and 1. If k = 0, then If k = 1, then 0 . 4 4 5 1 . 4 4 Using these values for , the square roots are 2 cis 4 and 2 cis 54 , which can be written in rectangular form as 2 i 2 and 2 i 2. Copyright © 2007 Pearson Education, Inc. Slide 10-9 7.6 Finding Complex Roots Example Find all fourth roots of 8 8i 3. Write the roots in rectangular form. Solution 8 8i 3 16 cis 120 r 16 and 120 Modulus 4 16 2 120 360 k Arguments 30 90 k 4 4 If k = 0, then If k = 1, then If k = 2, then If k = 3, then Copyright © 2007 Pearson Education, Inc. = 30º + 90º·0 = 30º. = 30º + 90º·1 = 120º. = 30º + 90º·2 = 210º. = 30º + 90º·3 = 300º. Slide 10-10 7.6 Finding Complex Roots Using these angles on the previous slide, the fourth roots are 2 cis 30º, 2 cis 120º, 2 cis 210º, and 2 cis 300º. These four roots can be written in rectangular form as 3 i, 1 i 3, 3 i, 1 i 3. Copyright © 2007 Pearson Education, Inc. Slide 10-11 7.6 Solving an Equation by Finding Complex Roots Example Find all complex number solutions of x5 – 1 = 0. Graph them as vectors in the complex plane. Solution Write the equation as x5 1 0 x 1. To find the five complex number solutions, write 1 in polar form as 1 1 0i 1(cos 0 i sin 0 ). 5 The modulus of the fifth roots is 5 1 1. Copyright © 2007 Pearson Education, Inc. Slide 10-12 7.6 Solving an Equation by Finding Complex Roots The arguments are given by 0 72 k , k 0, 1, 2, 3, or 4. Using these arguments, the fifth roots are 1(cos 0º + i sin 0º), 1(cos 72º + i sin 72º), 1(cos 144º + i sin 144º), 1(cos 216º + i sin 216º), 1(cos 288º + i sin 288º), Copyright © 2007 Pearson Education, Inc. k=0 k=1 k=2 k=3 k=4 Slide 10-13