Stitz-Zeager_College_Algebra_e-book

The reader can verify that x cost y sint for 32

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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...
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