Stitz-Zeager_College_Algebra_e-book

# The reader can verify that x cost y sint for 32

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Hence, z = 4cis π . For w = −1 + i 3, 6 6 √ √ √ 3 we have |w| = (−1)2 + ( 3)2 = 2. For an argument θ of w, we have tan(θ) = −1 = − 3. Since π π w lies in Quadrant II, θ = 23 + 2πk for integers k and w = 2cis 23 . We can now proceed. π 1. We get zw = 4cis π 2cis 23 = 8cis 6 √ After simplifying, we get zw = −4 3 + 4. π 6 + 2π 3 5π 6 = 8cis 5 = 8 cos 5π 6 + i sin π π 2. We use DeMoivre’s Theorem which yields w5 = 2cis 23 = 25 cis 5 · 23 = 32cis √ π π π Since 10π is coterminal with 43 , we get w5 = 32 cos 43 + i sin 43 = −16 − 16i 3. 3 5π 6 . 10π 3 . z 4cis( π ) π = 2cis 26 = 4 cis π − 23 = 2cis − π . Since − π is a π 2 6 2 2 (3) w quadrantal angle, we can ‘see’ the rectangular form by moving out 2 units along the positive real axis, then rotating π radians clockwise to arrive at the point 2 units below 0 on the 2 z imaginary axis. The long and short of it is that w = −2i. 3. Last, but not least, we have Some remarks are in order. First, the reader may not be sold on using the polar form of complex numbers to multiply complex numbers – especially if they aren’t...
View Full Document

## 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.

Ask a homework question - tutors are online