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: 33

Answer

$z=2(\displaystyle \cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6})$

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 $z$. ------- $z=-\sqrt{3}-i$ $r=|z|=\sqrt{(-\sqrt{3})^{2}+(-1)^{2}}=\sqrt{3+1}=2$ $\displaystyle \tan\theta=\frac{-1}{-\sqrt{3}}=\frac{\sqrt{3}}{3}$, In quadrant I, $\displaystyle \tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$. By symmetry (on the unit circle), $\theta$ can also be $\displaystyle \frac{7\pi}{6}$ (quadrant III) $z=-\sqrt{3}-i$ lies in quadrant III, so we select the appropriate argument: $z=2(\displaystyle \cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6})$
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