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: 23

Answer

$$z=5(\cos 2.21+i \sin 2.21 )$$
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Work Step by Step

We plot the complex number $z=-3+4i$ the same way we plot $(-3,4)$ 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 $z=-3+4i$, we have$$r=\sqrt{(-3)^2+4^2}=\sqrt{25}=5, \\ \tan \theta =\frac{4}{-3}=-\frac{4}{3}.$$Since $\cos \theta =\frac{a}{r} = \frac{-3}{\sqrt{25}} \lt 0$, the argument, $\theta$, must lie in quadrant $II$. So$$\theta =\tan^{-1} (-\frac{4}{3}) \approx 2.21$$Thus, the complex number has the polar form$$z=5 \left ( \cos 2.21 +i \sin 2.21 \right ).$$
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