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 769: 97


See the explanation below.

Work Step by Step

The provided complex number $z=r\left( \cos \theta +i\sin \theta \right)$ is in polar form with $r$ and $\theta $ as its modulus and argument respectively. First find the real and imaginary parts of the complex number, which are represented by $a$ and $b$ respectively. $a=r\cos \theta $ and $b=r\sin \theta $ Substitute the value of $r\cos \theta $ and $r\sin \theta $ in the provided complex number. $\begin{align} & z=a+ib \\ & =\left( r\cos \theta \right)+i\left( r\sin \theta \right) \end{align}$ Above is the rectangular form of the assumed complex number. Example: The above explanation can be justified with the help of an example. Let the provided complex number be $z=5\left( \cos 30{}^\circ +i\sin 30{}^\circ \right)$, in polar form. To convert this complex number into rectangular form, just substitute the value of $\cos 30{}^\circ $ and $\sin 30{}^\circ $ in the provided expression. $z=5\left( \frac{\sqrt{3}}{2}+i\frac{1}{2} \right)$ On further simplification $z=\left( \frac{5\sqrt{3}}{2}+\frac{5}{2}i \right)$ The above expression is the rectangular form of the provided expression.
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