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: 16

Answer

$$z=3 \left ( \cos \frac{3\pi }{2} +i \sin \frac{3\pi }{2} \right )$$
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Work Step by Step

We plot the complex number $z=-3i$ the same way we plot $(0,-3)$ 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 $z=-3i$, we have$$r=\sqrt{0^2+(-3)^2}=\sqrt{9}=3, \\ \sin \theta =\frac{b}{r}=\frac{-3}{3}=-1.$$ So$$\theta =\frac{3\pi }{2} $$Thus, the complex number has the polar form$$z=3 \left ( \cos \frac{3\pi }{2} +i \sin \frac{3\pi }{2} \right ).$$
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