Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.5 - Complex Numbers in Polar Form; DeMoivre's Theorem - Exercise Set - Page 769: 111

Answer

See plot and explanations.
1585766955

Work Step by Step

Step 1. Let $z=1=cos0+i\ sin0$; we have the fourth roots as $z_k=\sqrt[4] 1(cos\frac{2k\pi+0}{4}+i\ sin\frac{2k\pi+0}{4} )=cos\frac{k\pi}{2}+i\ sin\frac{k\pi}{2} $, where $k=0,1,2,3$ Step 2. For $k=0$, we have $z_0=1$ Step 3. For $k=1$, we have $z_1=cos\frac{\pi}{2}+i\ sin\frac{\pi}{2}=i $ Step 4. For $k=2$, we have $z_2=cos\frac{2\pi}{2}+i\ sin\frac2{\pi}{2}=-1$ Step 5. For $k=3$, we have $z_3=cos\frac{3\pi}{2}+i\ sin\frac{3\pi}{2}=-i $ Step 6. We can plot the above root as shown in the figure.
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