Hence the product of any number of complex numbers is a complex number which has for its modulus the product of the moduli, and for its amplitude the sum of the amplitudes of the factors. This is known as DeMoivre's Theorem, for n a positive integer. Now suppose n a negative integer = -m, then = rm [cos moj sin me] [r (cos + j sin 0)]" Rationalizing the denominator, 1 [r (cos + j sin 0)]-m m = p―m [cos (m) 0+j sin (−m) 0] = rm [cos (−m) 0 + j sin (−m) 0] This proves DeMoivre's theorem for n = -m. The following exercises are to be solved by means of DeMoivre's Theorem. 1. Show that sin 20 = 2 sin cos 0 and cos 20 =cos2 sin2 0. cos2 + 2j sin cos 0 + j2 sin2 = cos 20+j sin 2 0. By Art. 91, the real parts are equal and the coefficients of j are equal. That is, cos 20 = cos2 - sin20 and sin 20 = 2 sin 0 cos 0. 2. Prove that [cos 45° + j sin 45°]2 = j. Solution. With 0 (cos 45° + j sin 45°)2 = 45° and n = = 2, by (3) = cos 90° + j sin 90° = j, since cos 90° = 0 and sin 90° = 1. Hence the quotient of two complex numbers is a complex number having an amplitude equal to the amplitude of the dividend minus the amplitude of the divisor, and a modulus equal to the modulus of the dividend divided by the modulus of the divisor. For example, 8 (cos 60°+j sin 60°) 4 (cos 30°+j sin 30°) = 2 [cos (60° - 30°) + j sin (60° – 30°)] = 2 (cos 30° + j sin 30°). 22 96. Square roots of a number. Let z2 = r2 (cos + j sin 0) be the number. Now the square root of a number is one of the two equal factors of the number. By Art. 94, when a number is squared the modulus is squared and the amplitude is doubled Therefore, in extracting the square root, we extract the positive square root of the modulus and divide the amplitude by 2. = cos (0 + 2 kя) and sin sin (0 + 2 km). = Now cos 0 Then the most general form in which the original number may be written, is These are the only square roots, for if k = 2, the amplitude Ө becomes + 2 π, the functions of which are the same as those of 2 0 2 0 3, we have 3 π, the functions of which are the same as 2 +, etc. Any other values of k will give either zo or 21. Example. Extract the square root of the number 1. Solution. The complex number 1 is plotted on the x-axis, one unit to the right of the origin. The modulus is 1 and the amplitude is 0 + 2km. 97. Cube roots of a number. The cube root of a number is one of the three equal factors of the number. The number is the cube of this factor. Since in raising a number to the third power, we cube the modulus, and multiply the amplitude by 3, then in extracting the cube root, we extract the cube root of the modulus and divide the amplitude by 3. Suppose 23 = r (cos 0 + j sin 0) is the number. Writing the number in the most general form we have 0 But the functions of and 3 3 3+2 are equal and therefore the roots corresponding to k = 0 and k = 3 are the same. Any other values of k will give an amplitude with the functions equal to those Therefore there are but three cube roots of 0, 1, or 2. for k = Solution. Let 23 = 1 = cos (0 + 2 kπ) + j sin (0 + 2 kπ). These three roots may be represented graphically as follows: Draw a circle of radius 1, Fig. 112. Since the moduli of 20, 21, and 2 are each 1, their terminal points will lie on the circumference of the circle. The amplitudes of zo, 21, and 22 are 0, 3 π, and respectively. Therefore zo OA, z1 = OB, and z2 = OC. Example. Find the three cube roots of z3 = −8. = 21 z2 = π Solution. Plot -8 OA, Fig. 113. The amplitude is and the modulus is 8. Then Let k = 2, then z2 = 2 [cos § π + j sin § π] = 1 − j √3. = Now the moduli of 20, 21, and 22 are each equal to 2. The points representing the cube roots are therefore on the circumference of a circle of radius 2, Fig. 113. Constructing the amplitudes 60°, 180°, and 300°, we have zo OB, 21 = OC, and Z2 = OD. 98. To find the nth roots of a number. An nth root of a number is one of its n equal factors. By DeMoivre's Theorem, when a number is raised to the nth power, the modulus is raised to the nth power, while the amplitude is multiplied by n. Then to extract the nth root, we extract the positive nth root of the modulus, and divide the amplitude by n. * * The student is already familiar with the fact that there is one positive square root of a positive number, and one positive cube root of a positive number. It is true, in general, that there is one and only one positive nth root of any positive number. |