Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 8 - Application of Trigonometry - Chapter 8 Test Prep - Review Exercises - Page 841: 65

Answer

$$2\sqrt 2 \left( {\cos 135^\circ + i\sin 135^\circ } \right)$$

Work Step by Step

$$\eqalign{ & {\text{Rectangular Form }} - 2 + 2i \cr & {\text{Use }}r = \sqrt {{a^2} + {b^2}} {\text{ and }}\theta = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right),{\text{ so}} \cr & r = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( 2 \right)}^2}} = 2\sqrt 2 \cr & \theta = {\tan ^{ - 1}}\left( {\frac{2}{{ - 2}}} \right) + 180 \cr & \theta = 135^\circ \cr & {\text{write the vector in the trigonometric form }}r\left( {\cos \theta + i\sin \theta } \right) \cr & = 2\sqrt 2 \left( {\cos 135^\circ + i\sin 135^\circ } \right) \cr} $$
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