Pre-Calc Exam Notes 142

# Pre-Calc Exam Notes 142 - 142 Chapter 6 Additional Topics...

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142 Chapter 6 Additional Topics §6.3 Example 6.10 x y 0 r 1 2 z =− 2 i θ Figure 6.3.2 Represent the complex number 2 i in trigonometric form. Solution: Let z = − 2 i = x + yi , so that x = − 2 and y = − 1. Then θ is in QIII, as we see in Figure 6.3.2. So since tan θ = y x = 1 2 = 1 2 , we have θ = 206.6 . Also, r = r x 2 + y 2 = R ( 2) 2 + ( 1) 2 = r 5 . Thus, 2 i = r 5 (cos 206.6 + i sin 206.6 ) , or r 5 cis 206.6 . For complex numbers in trigonometric form, we have the following formulas for multipli- cation and division: Let z 1 = r 1 (cos θ 1 + i sin θ 1 ) and z 2 = r 2 (cos θ 2 + i sin θ 2 ) be complex numbers. Then z 1 z 2 = r 1 r 2 (cos ( θ 1 + θ 2 ) + i sin ( θ 1 + θ 2 )) , and (6.5) z 1 z 2 = r 1 r 2 (cos ( θ 1 θ 2 ) + i sin ( θ 1 θ 2 )) if z 2 n= 0. (6.6) The proofs of these formulas are straightforward: z 1 z 2 = r 1 (cos θ 1 + i sin θ 1 ) · r 2 (cos θ 2 + i sin θ 2 ) = r 1 r 2 [(cos θ 1 cos θ 2 sin θ 1 sin θ 2 ) + i (sin θ 1 cos θ 2 + cos θ 1 sin θ 2 )] = r 1 r 2 (cos ( θ 1 + θ 2 ) + i sin ( θ 1 + θ 2 )) by the addition formulas for sine and cosine. And z 1 z 2 = r 1 (cos θ 1 + i sin θ 1 ) r 2 (cos θ 2 + i sin θ 2 ) = r 1 r 2 · cos θ 1 + i sin θ
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## This note was uploaded on 01/21/2012 for the course MAC 1130 taught by Professor Dr.cheun during the Fall '11 term at FSU.

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