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: 35

Answer

$z=4(\displaystyle \cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6})$

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 $z$. ------- $z=2\sqrt{3}-2i$ $r=|z|=\sqrt{(2\sqrt{3})^{2}+(-2)^{2}}=\sqrt{4\cdot 3+4}=4$ $\displaystyle \tan\theta=\frac{-2}{2\sqrt{3}}=-\frac{\sqrt{3}}{3}$, In quadrant I, $\displaystyle \tan\frac{\pi}{6}=+\frac{\sqrt{3}}{3}$. By symmetry (on the unit circle), $\theta$ can be $\displaystyle \frac{5\pi}{6}$ (quadrant II) or $\displaystyle \frac{11\pi}{6}$ (quadrant IV) $z=2\sqrt{3}-2i$ lies in quadrant IV, so we select the appropriate argument: $z=4(\displaystyle \cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6})$
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