## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 799: 64

#### Answer

The rectangular form of $z=0.6\left( \cos 100{}^\circ +i\sin 100{}^\circ \right)$ is $z\approx \left( -0.1+0.6i \right)$.

#### Work Step by Step

The rectangular form of a complex number is $z=a+ib$, where $a$ and $b$ are rectangular coordinates. In the polar form the complex number $z=a+ib$ is represented as $z=r\left( \cos \theta +i\sin \theta \right)$, where r is the distance of the complex number from the origin and $\theta$ is the respective angle. Here, the given complex number is in the polar form with $r=0.6$ and $\theta =100{}^\circ$. Since, $\cos 100{}^\circ \approx -\ 0.17\ \text{ and }\ \sin 100{}^\circ =0.98$ Therefore, \begin{align} & z=0.6\left( \cos 100{}^\circ +i\sin 100{}^\circ \right) \\ & \approx 0.6\left( -0.17+i0.98 \right) \\ & \approx \left( \left( -0.17 \right)\left( 0.6 \right)+i\left( 0.98 \right)\left( 0.6 \right) \right) \\ & \approx \left( -0.102+0.588i \right) \end{align} $\approx \left( -0.1+0.6i \right)$

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