Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 8 - Section 8.3 - Polar Form of Complex Numbers; De Moivre's Theorem - 8.3 Exercises: 32

Answer

$z=2(\displaystyle \cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4})$

Work Step by Step

See p. 604, A complex number $z=a+bi$ has the polar (or trigonometric) form $z=r(\cos\theta+i\sin\theta)$ where $r=|z|=\sqrt{a^{2}+b^{2}}$ and $\tan\theta=b/a$. The number $r$ is the modulus of $z$, and $\theta$ is an argument of $z$. ------- $r=|z|=\sqrt{(-\sqrt{2})^{2}+(-\sqrt{2})^{2}}=\sqrt{2+2}=2$ $\displaystyle \tan\theta=\frac{-\sqrt{2}}{-\sqrt{2}}=1$, In quadrant I, $\displaystyle \tan\frac{\pi}{4}=1$. By symmetry (on the unit circle), $\theta$ can also be $\displaystyle \frac{5\pi}{4}$ (quadrant III) $z=-\sqrt{2}-i\sqrt{2}$ lies in quadrant III, so we select the appropriate argument: $z=2(\displaystyle \cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4})$
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