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 8.3 The Product and Quotient Theorems - 8.3 Exercises - Page 370: 23

Answer

$\frac{2\sqrt{6} - 2i\sqrt{2}}{\sqrt{2} - i\sqrt{6}}$ = $\sqrt{3} + i$

Work Step by Step

Now, $2\sqrt{6} - 2i\sqrt{2} = 2\sqrt{2}(\sqrt{3} - i)$ where $(\sqrt{3} - i)$ is at $330^\circ$ with absolute value $\sqrt{(\sqrt{3})^2 + (-1)^2} = 2$, and $\sqrt{2} - i\sqrt{6} = \sqrt{2}(1 - i\sqrt{3})$ where $(1 - i\sqrt{3})$ is at $300^\circ$ with absolute value $\sqrt{1^2 + (-\sqrt{3})^2} = 2$ Therefore, $\frac{2\sqrt{6} - 2i\sqrt{2}}{\sqrt{2} - i\sqrt{6}}$ = $\frac{2\sqrt{2}(\sqrt{3} - i)}{\sqrt{2}(1 - i\sqrt{3})}$ = $2\cdot \frac{2cis330^\circ}{2cis300^\circ}$ = $2cis(330^\circ - 300^\circ)$ (Quotient Theorem) = $2cis30^\circ$ = $2(\frac{\sqrt{3}}{2} + \frac{1}{2} i)$ = $\sqrt{3} + i$
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