**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