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 767: 21


$$z=\sqrt{45} \left ( \cos 4.03 +i \sin 4.03 \right )$$

Work Step by Step

We plot the complex number $z=-3\sqrt{2}-3\sqrt{3}i$ the same way we plot $(-3\sqrt{2}, -3\sqrt{3})$ in the rectangular coordinate system. Please note that the complex number $z=a+bi$ is written in polar form as$$z=r(\cos \theta +i \sin \theta ),$$where $a=r\cos \theta$, $b=r\sin \theta$, $r=\sqrt{a^2+b^2}$, and $\tan \theta = \frac{b}{a}$, $0 \le \theta \lt 2\pi $. So, for the complex number $z=-3\sqrt{2}-3\sqrt{3}i$, we have$$r=\sqrt{(-3\sqrt{2})^2+(-3\sqrt{3})^2}=\sqrt{45}, \\ \tan \theta =\frac{-3\sqrt{3}}{-3\sqrt{2}}=\sqrt{1.5}.$$Since $\cos \theta =\frac{a}{r} = \frac{-3\sqrt{2}}{\sqrt{45}} \lt 0$, the argument, $\theta$, must lie in quadrant $III$. So$$\theta =\tan^{-1} (\sqrt{1.5})+\pi \approx 4.03$$Thus, the complex number has the polar form$$z=\sqrt{45} \left ( \cos 4.03 +i \sin 4.03 \right ).$$
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