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: 98

Answer

See the explanation below.

Work Step by Step

The product of two complex numbers written in polar form can be found by the use of the product rule of complex numbers. In this method, the modulus and argument of the provided complex numbers are multiplied and added respectively. For example, two complex numbers ${{z}_{1}}={{r}_{1}}\left( \cos \theta +i\sin \theta \right)$ and ${{z}_{2}}={{r}_{2}}\left( \cos \alpha +i\sin \alpha \right)$ can be multiplied by the use of product rule as shown below: ${{z}_{1}}\cdot {{z}_{2}}={{r}_{1}}\cdot {{r}_{2}}\left[ \cos \left( \theta +\alpha \right)+i\sin \left( \theta +\alpha \right) \right]$ Above is the product of two complex numbers provided in polar form by the use of product rule. Example: The above statement can be justified with the help of an example. Let ${{z}_{1}}=5\left( \cos 30{}^\circ +i\sin 30{}^\circ \right)$ and ${{z}_{1}}=10\left( \cos 60{}^\circ +i\sin 60{}^\circ \right)$ be two complex numbers provided in polar form. Then by the use of the multiplication rule of complex numbers, the provided two numbers can be multiplied as shown below: $\begin{align} & {{z}_{1}}\cdot {{z}_{2}}=5\cdot 10\left[ \cos \left( 30{}^\circ +60{}^\circ \right)+i\sin \left( 30{}^\circ +60{}^\circ \right) \right] \\ & =50\left[ \cos 90{}^\circ +i\sin 90{}^\circ \right] \end{align}$ Above is the product of two complex numbers provided in polar form by the use of the product rule.
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