Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 8 - Section 8.3 - Products and Quotients in Trigonometric Form - 8.3 Problem Set - Page 440: 53

Answer

$-4-4i$

Work Step by Step

In trigonometric form, $1+i=\sqrt {2}(\cos 45^{\circ}+i\sin 45^{\circ})$ $2i=2(\cos 90^{\circ}+i\sin90^{\circ})$ $-2+2i=2\sqrt 2(\cos135^{\circ}+i\sin135^{\circ})$ Using de Moivre's theorem ($z^{n}=r^{n} cis \,n\theta$ where $z=r\, cis\,\theta$ and $n$ is an integer), we get $(1+i)^{4}=[\sqrt {2}(\cos 45^{\circ}+i\sin 45^{\circ})]^{4}$ $=(\sqrt {2})^{4}[\cos(4\cdot45^{\circ})+i\sin(4\cdot45^{\circ})]$ $=4(\cos180^{\circ}+i\sin180^{\circ})$ $(2i)^{2}=[2(\cos 90^{\circ}+i\sin90^{\circ})]^{2}$ $=(2)^{2}[\cos(2\cdot90^{\circ})+i\sin(2\cdot90^{\circ})]$ $=4(\cos 180^{\circ}+i\sin 180^{\circ})$ Now, $(1+i)^{4}(2i)^{2}=$ $[4(\cos 180^{\circ}+i\sin 180^{\circ})][4(\cos 180^{\circ}+i\sin 180^{\circ})]=$ $4\cdot4[\cos(180^{\circ}+180^{\circ})+i\sin(180^{\circ}+180^{\circ})]$ $=16(\cos 360^{\circ}+i\sin360^{\circ})$ $\frac{(1+i)^{4}(2i)^{2}}{-2+2i}=\frac{16(\cos 360^{\circ}+i\sin360^{\circ})}{2\sqrt 2(\cos135^{\circ}+i\sin135^{\circ})}$ $=\frac{16}{2\sqrt 2}[\cos(360^{\circ}-135^{\circ})+i\sin(360^{\circ}-135^{\circ})]$ $=4\sqrt {2}(\cos225^{\circ}+i\sin225^{\circ})$ which in standard form is $=4\sqrt {2}(-\frac{1}{\sqrt {2}}+i\cdot-\frac{1}{\sqrt {2}})$ $=-4-4i$
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