Stitz-Zeager_College_Algebra_e-book

# Hence we may parametrize the path as x f t y g t

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Unformatted text preview: D then we get D = [cos(β ) + i sin(β )] [cos(β ) − i sin(β )] = cos2 (β ) − i cos(β ) sin(β ) + i cos(β ) sin(β ) − i2 sin2 (β ) Expand = cos2 (β ) − i2 sin2 (β ) Simplify = cos2 (β ) + sin2 (β ) Again, i2 = −1 =1 Pythagorean Identity Putting it all together, we get z w = = = |z | cos(α) + i sin(α) cos(β ) − i sin(β ) · |w| cos(β ) + i sin(β ) cos(β ) − i sin(β ) |z | cis(α − β ) |w| 1 |z | cis(α − β ) |w| and we are done. The next example makes good use of Theorem 11.16. 850 Applications of Trigonometry √ √ Example 11.7.3. Let z = 2 3 + 2i and w = −1 + i 3. Use Theorem 11.16 to ﬁnd the following. 2. w5 1. zw 3. z w Write your ﬁnal answers in rectangular form. √ Solution. In order to use Theorem 11.16, we need to write z and w in polar form. For z = 2 3+2i, √ √ √ z) 2 we ﬁnd |z | = (2 3)2 + (2)2 = 16 = 4. If θ ∈ arg(z ), we know tan(θ) = Im(z ) = 2√3 = 33 . Since Re( √ z lies in Quadrant I, we have θ = π + 2πk for integers k ....
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## This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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