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: 26


$$z=\sqrt{6} \left ( \cos 5.13 +i \sin 5.13 \right )$$

Work Step by Step

We plot the complex number $z=1-\sqrt{5}i$ the same way we plot $(1, -\sqrt{5})$ 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 $z=1-\sqrt{5}i$, we have$$r=\sqrt{1^2+(-\sqrt{5})^2}=\sqrt{6}, \\ \tan \theta =\frac{-\sqrt{5}}{1}=-\sqrt{5}.$$Since $\cos \theta =\frac{a}{r} = \frac{1}{\sqrt{6}} \gt 0$, the argument, $\theta$, must lie in quadrant $IV$. So$$\theta =\tan^{-1} (-\sqrt{5})+\pi \approx 5.13$$Thus, the complex number has the polar form$$z=\sqrt{6} \left ( \cos 5.13 +i \sin 5.13 \right ).$$
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