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: 31

Answer

$z=2\displaystyle \sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})$

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 $z$. ------- $r=|z|=\sqrt{(-2)^{2}+(2)^{2}}=\sqrt{8}=2\sqrt{2}$ $\displaystyle \tan\theta=\frac{2}{-2}=-1$, In quadrant I, $\displaystyle \tan\frac{\pi}{4}=1$ (reference), so, by symmetry (on the unit circle), $\theta$ is either $\displaystyle \frac{3\pi}{4}$ (quadrant II) or $\displaystyle \frac{7\pi}{4}$ (quadrant IV) z=-2+2i lies in quadrant II, so we select the appropriate argument: $z=2\displaystyle \sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})$
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