1 Objectives Use De Moivres Theorem to find

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Objectives Use De. Moivre’s Theorem to find powers of complex numbers. Find nth roots of complex numbers. 2

Powers of Complex Numbers 3

Powers of Complex Numbers The trigonometric form of a complex number is used to raise a complex number to a power. To accomplish this, consider repeated use of the multiplication rule. z = r (cos + i sin ) z 2 = r (cos + i sin ) = r 2(cos 2 + i sin 2 ) z 3 = r 2(cos 2 + i sin 2 )r (cos + i sin ) = r 3(cos 3 + i sin 3 ) 4

Powers of Complex Numbers z 4 = r 4(cos 4 + i sin 4 ). . . z 5 = r 5(cos 5 + i sin 5 ) This pattern leads to De. Moivre’s Theorem, which is named after the French mathematician Abraham De. Moivre. 5

Example 7 – Finding Powers of a Complex Number Use De. Moivre’s Theorem to find (– 1 + Solution: The absolute value of )12. is and the argument is given by Because lies in Quadrant II, 6

Example 7 – Solution cont’d So, the trigonometric form is Then, by De. Moivre’s Theorem, you have 7

Example 7 – Solution cont’d 8

Roots of Complex Numbers 9

Roots of Complex Numbers We know that a consequence of the Fundamental Theorem of Algebra is that a polynomial equation of degree n has n solutions in the complex number system. So, the equation x 6 = 1 has six solutions, and in this particular case you can find the six solutions by factoring and using the Quadratic Formula. x 6 – 1 = 0 (x 3 – 1)(x 3 + 1) = 0 (x – 1)(x 2 + x + 1)(x 2 – x + 1) = 0 10

Roots of Complex Numbers Consequently, the solutions are and Each of these numbers is a sixth root of 1. In general, an nth root of a complex number is defined as follows. 11

Roots of Complex Numbers To find a formula for an nth root of a complex number, let u be an nth root of z, where u = s(cos + i sin ) and z = r (cos + i sin ). By De. Moivre’s Theorem and the fact that un = z, you have sn(cos n + i sin n ) = r (cos + i sin ). 12

Roots of Complex Numbers Taking the absolute value of each side of this equation, it follows that sn = r. Substituting back into the previous equation and dividing by r, you get cos n + i sin n = cos + i sin . So, it follows that cos n = cos and sin n = sin . Because both sine and cosine have a period of 2 , these last two equations have solutions if and only if the angles differ by a multiple of 2. 13

Roots of Complex Numbers Consequently, there must exist an integer k such that By substituting this value of into the trigonometric form of u, you get the result stated below. 14

Roots of Complex Numbers When k > n – 1, the roots begin to repeat. For instance, if k = n, the angle is coterminal with n, which is also obtained when k = 0. The formula for the nth roots of a complex number z has a nice geometrical interpretation, as shown in Figure 6. 38 15

Roots of Complex Numbers Note that because the nth roots of z all have the same magnitude , they all lie on a circle of radius with center at the origin. Furthermore, because successive nth roots have arguments that differ by 2 n, the n roots are equally spaced around the circle. You have already found the sixth roots of 1 by factoring and using the Quadratic Formula. Example 9 shows how you can solve the same problem with the formula for nth roots. 16

Example 9 – Finding the nth Roots of a Complex Number Find the three cube roots of z = – 2 + 2 i. Solution: The absolute value of z is and the argument is given by 17

Example 9 – Solution cont’d Because z lies in Quadrant II, the trigonometric form of z is = + arctan(– 1) = 3 / 4 = 135 By the formula for nth roots, the cube roots have the form 18

Example 9 – Solution cont’d Finally, for k = 0, 1, and 2 you obtain the roots 19

Example 9 – Solution cont’d See Figure 6. 40 20