Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 8 - Review Exercises - Page 410: 43

Answer

$\sqrt[6]{2}cis105^\circ$, $\sqrt[6]{2}cis225^\circ$ and $\sqrt[6]{2}cis345^\circ$ are the three cube roots of $1 - i$.

Work Step by Step

$1 - i$ is at $315^\circ$ with absolute value $\sqrt{1^2 + (-1)^2} = \sqrt{2}$. The equivalent trigonometric form of $1 - i$ is $\sqrt{2}cis315^\circ$. Suppose the cube root of $1 - i$ is represented by $rcis\alpha$, $(rcis\alpha)^3$ = $1 - i$ = $\sqrt{2}cis315^\circ$ By De Moivreā€™s theorem, this equation becomes $r^3cis3\alpha = \sqrt{2}cis315^\circ$ By equating, $r^3 = \sqrt{2}$, $r = \sqrt[6]{2}$, and $cos3\alpha = cos315^\circ, sin3\alpha = sin315^\circ$ For these equations to be satisfied, $3\alpha$ must represent an angle that is coterminal with $315^\circ$. Therefore, we must have $\alpha = \frac{315^\circ + 360^\circ\cdot k}{3}$ and k take on the integer values 0, 1, and 2. When $k = 0, \alpha = 105^\circ$, $k = 1, \alpha = 225^\circ$, and $k = 2, \alpha = 345^\circ$ Hence, when $k = 0,$ the root is $\sqrt[6]{2}cis105^\circ$, $k = 1,$ the root is $\sqrt[6]{2}cis225^\circ$, $k = 2,$ the root is $\sqrt[6]{2}cis345^\circ$, $\sqrt[6]{2}cis105^\circ$, $\sqrt[6]{2}cis225^\circ$ and $\sqrt[6]{2}cis345^\circ$ are the three cube roots of $1 - i$.
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