Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.2 Trigonometric (Polar) Form of Complex Numbers - 8.2 Exercises - Page 365: 35

Answer

$-1-i$

Work Step by Step

$\sqrt 2$ cis $225^{\circ}$=$\sqrt 2(\cos225^{\circ}+i\sin225^{\circ})$ It is known that $\cos225^{\circ}=-\frac{\sqrt 2}{2}$ and $\sin135^{\circ}=-\frac{\sqrt 2}{2}$ Substituting these values in the expression and solving: $\sqrt 2(\cos225^{\circ}+i\sin225^{\circ})=\sqrt 2(-\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i)=(-\frac{\sqrt 2\times\sqrt 2}{2}-\frac{\sqrt 2\times\sqrt 2}{2}i)=(-\frac{2}{2}-\frac{2}{2}i)=-1-i$
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