Stitz-Zeager_College_Algebra_e-book

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

Ask a homework question - tutors are online