827 8.5 Trigonometric (Polar) Form of Complex Numbers; Products and Quotients Writing 1 + i23, -223 + 2i, and - 1 2 i in trigonometric form gives 1 + i23 = 21cos 60° + i sin 60°2, -223 + 2i = 41cos 150° + i sin 150°2, Use r =2x2 +y2 and tan u = y x . and - 1 2 i = 1 2 3cos1-90°2 + i sin1-90°24. -90° can be replaced by 270°. Quotient Theorem If r11cos u1 + i sin u12 and r21cos u2 + i sin u22 are any two complex numbers, where r21cos u2 + i sin u22 ≠0, then the following holds true. r11cos U1 +i sin U12 r21cos U2 +i sin U22 = r1 r2 3 cos1U1 −U22 +i sin1U1 −U22 4 In compact form, this is written r1 cis U1 r2 cis U2 = r1 r2 cis1U1 −U22. That is, to divide complex numbers in trigonometric form, divide their absolute values and subtract their arguments. EXAMPLE 7 Using the QuotientTheorem Find the quotient 10 cis1-60°2 5 cis 150° . Write the answer in rectangular form. SOLUTION 10 cis1-60°2 5 cis 150° = 10 5 cis1-60° - 150°2 Quotient theorem = 2 cis1-210°2 Divide and subtract. = 2 3cos1-210°2 + i sin1-210°24 Rewrite. = 2 c - 23 2 + i a 1 2b d = -23 + i Distributive property S Now Try Exercise 87. cos1-210°2 = - 23 2 ; sin1-210°2 = 1 2 Here, the absolute value of the quotient, 1 2 , is the quotient of the two absolute values, 2 4 = 1 2 . The argument of the quotient, -90°, is the difference of the two arguments, 60° - 150° = -90°. That is, 21cos 60° + i sin 60°2 41cos 150° + i sin 150°2 Trigonometric form of 1 + i 23 -223 + 2i = 1 23cos1-90°2 + i sin1-90°24 = 1 21cos 270° + i sin 270°2. Write so that 0° … u 6360°. Divide absolute values and subtract arguments.
RkJQdWJsaXNoZXIy NjM5ODQ=