= ( = = = = cos(2π) + 0.355 + 1.092i isin(2π) ≈ Observe that for the angles 2kπ, where is an integer and 5, the values of are repeated as the k k ≥ same as its values when = 0, 1, 2, 3, and 4. In conclusion, all complex solutions of = 2 are k z5 1.149, 0.355 + 1.092i, -0.929 + 0.675i, -0.929 - 0.675i, and 0.355 - 1.092i. 67. If = 1 then = . In the complex plane, 1 would sit on the horizontal axis at an angle of 0, z6 z giving the polar form . Then = e0 = cos (0) + = 1. isin(0) Since the angles 2π, 4π, 6π, 8π, 10π, and 12π are coterminal with the angle of 0, can be represented by turns as , , , , , and . = = cos() + = + isin() i = = cos() + = + isin() i = = cos(π) + = -1 isin(π) = = cos() + = - isin() i = = cos() + = - isin() i = = cos(2π) + = 1 isin(2π) Observe that for the angles 2kπ, where is an integer and 6, the values of are repeated as the k k ≥ same as its values when = 0, 1, 2, 3, 4, and 5. In conclusion, all complex solutions of = 1 are k z6 1, + + -1, - and - i, i, i, i. 8.4 Solutions to Exercises 1. Initial point (4,0); terminal point (0,2). The vector component form is .
= ( = = = = cos(2π) + isin(2π) ≈ 0.355 + 1.092i Observe that for the angles 2kπ, where k is an integer and k ≥ 5, the values of are repeated as the same as its values when k = 0, 1, 2, 3, and 4. In conclusion, all complex solutions of z5 = 2 are 1.149, 0.355 + 1.092i, -0.929 + 0.675i, -0.929 - 0.675i, and 0.355 - 1.092i. 67. If z6 = 1 then z = . In the complex plane, 1 would sit on the horizontal axis at an angle of 0, giving the polar form . Then = e0 = cos (0) + isin(0) = 1. Since the angles 2π, 4π, 6π, 8π, 10π, and 12π are coterminal with the angle of 0, can be represented by turns as , , , , , and . = = cos() + isin() = + i = = cos() + isin() = + i = = cos(π) + isin(π) = -1 = = cos() + isin() = - i = = cos() + isin() = - i = = cos(2π) + isin(2π) = 1 Observe that for the angles 2kπ, where k is an integer and k ≥ 6, the values of are repeated as the same as its values when k = 0, 1, 2, 3, 4, and 5. In conclusion, all complex solutions of z6 = 1 are 1, + i, + i, -1, - i, and - i. 8.4 Solutions to Exercises 1. Initial point (4,0); terminal point (0,2). The vector component form is .
Chapter%208%20Solutions.docx