## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 799: 61

#### Answer

The rectangular form of $z=8\left( \cos 60{}^\circ +i\sin 60{}^\circ \right)$ is $z=4+4i\sqrt{3}$.

#### 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 $r=8$ and $\theta =60{}^\circ$. Since, $\cos 60{}^\circ =\frac{1}{2}\ \ \text{ and }\ \sin 60{}^\circ =\frac{\sqrt{3}}{2}\ \$ Therefore, \begin{align} & z=8\left( \cos 60{}^\circ +i\sin 60{}^\circ \right) \\ & =8\left( \frac{1}{2}+i\frac{\sqrt{3}}{2} \right) \\ & =\left( \frac{8}{2}+i\frac{\sqrt{3}\times 8}{2} \right) \\ & =\left( 4+4i\sqrt{3} \right) \end{align}

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