833 8.6 De Moivre’s Theorem; Powers and Roots of Complex Numbers Roots of Complex Numbers Every nonzero complex number has exactly n distinct complex nth roots. De Moivre’s theorem can be extended to find all nth roots of a complex number. nth Root For a positive integer n, the complex number a + bi is an nth root of the complex number x + yi if the following holds true. 1 a +bi2n =x +yi To find the three complex cube roots of 81cos 135° + i sin 135°2, for example, look for a complex number, say r 1cos a + i sin a2, that will satisfy 3r 1cos a + i sin a243 = 81cos 135° + i sin 135°2. By De Moivre’s theorem, this equation becomes r31cos 3a + i sin 3a2 = 81cos 135° + i sin 135°2. Set r3 = 8 and cos 3a + i sin 3a = cos 135° + i sin 135° to satisfy this equation. The first of these conditions implies that r = 2, and the second implies that cos 3a = cos 135° and sin 3a = sin 135°. For these equations to be satisfied, 3a must represent an angle that is coterminal with 135°. Therefore, we must have 3a = 135° + 360° # k, k any integer or a = 135° + 360° # k 3 , k any integer. Now, let k take on the integer values 0, 1, and 2. If k = 0, then a = 135° + 360° # 0 3 = 45°. If k = 1, then a = 135° + 360° # 1 3 = 495° 3 = 165°. If k = 2, then a = 135° + 360° # 2 3 = 855° 3 = 285°. In the same way, a = 405° when k = 3. But note that 405° = 45° + 360°, so sin 405° = sin 45° and cos 405° = cos 45°. Similarly, if k = 4, then a = 525°, which has the same sine and cosine values as 165°. Continuing with larger values of k would repeat solutions already found. Therefore, all of the cube roots (three of them) can be found by letting k = 0, 1, and 2, respectively. When k = 0, the root is 21cos 45° + i sin 45°2. When k = 1, the root is 21cos 165° + i sin 165°2. When k = 2, the root is 21cos 285° + i sin 285°2. In summary, we see that 21cos 45° + i sin 45°2, 21cos 165° + i sin 165°2, and 21cos 285° + i sin 285°2 are the three cube roots of 81cos 135° + i sin 135°2.
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