835 8.6 De Moivre’s Theorem; Powers and Roots of Complex Numbers Because there are four fourth roots, let k = 0, 1, 2, and 3. If k = 0, then a = 30° + 90° # 0 = 30°. If k = 1, then a = 30° + 90° # 1 = 120°. If k = 2, then a = 30° + 90° # 2 = 210°. If k = 3, then a = 30° + 90° # 3 = 300°. Using these angles, the fourth roots are 2 cis 30°, 2 cis 120°, 2 cis 210°, and 2 cis 300°. These four roots can be written in rectangular form as 2 3 + i, -1 + i23, -23 - i, and 1 - i23. The graphs of these roots lie on a circle with center at the origin and radius 2. See Figure 54. The roots are equally spaced about the circle, 90° apart. (For convenience, we label the real axis x and the imaginary axis y.) y –2 x 2 2 –2 2 cis 308 308 2 cis 3008 2 cis 1208 2 cis 2108 0 Figure 54 S Now Try Exercise 29. EXAMPLE 4 Solving an Equation (Complex Roots) 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, or x5 = 1. Because 15 = 1, there is a real number solution, 1, and it is the only one. There are a total of five complex number solutions. To find these solutions, first write 1 in trigonometric form. 1 = 1 + 0i = 11cos 0° + i sin 0°2 Trigonometric form The absolute value of the fifth roots is 25 1 = 1. The arguments are given by 0° + 72° # k, k = 0, 1, 2, 3, and 4. By using these arguments, we find that the fifth roots are as follows. 11cos 0° + i sin 0°2, k = 0 11cos 72° + i sin 72°2, k = 1 11cos 144° + i sin 144°2, k = 2 11cos 216° + i sin 216°2, k = 3 11cos 288° + i sin 288°2 k = 4 Real solution
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