Trigonometry 7th Edition

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

Chapter 8 - Section 8.4 - Roots of a Complex Number - 8.4 Problem Set - Page 447: 25

Answer

The three cube roots are: \[ \begin{array}{r} \left[\cos 30^{\circ}+i \sin 30^{\circ}\right] 4\\ 4 [\cos 150^{\circ}+i \sin 150^{\circ}] \\ \left[\cos 270^{\circ}+i \sin 270^{\circ}\right]4 \end{array} \]

Work Step by Step

\[ 64 i+0=64 i \] The modulus of a complex number in the standard form $x+y i$ is: \[ \begin{array}{l} \sqrt{y^{2}+x^{2}}=r \\ \sqrt{0^{2}+64^{2}} =r\\ \sqrt{4096}=r \\ 64=r \end{array} \] The $\theta$ angle is the least positive angle for which: \[ 0=\frac{0}{64}=\frac{x}{r}=\cos \theta \] , \[ 1=\frac{64}{64}=\frac{y}{r}=\sin \theta \] Thus, \[ 90^{\circ}=\theta \] Using the definition of the trigonometric form of a complex number: \[ \begin{aligned} &=(\cos \theta+i \sin \theta)r \\ =&\left(\cos 90^{\circ}+i \sin 90^{\circ}\right)64 \end{aligned} \] Using the roots theorem: \[ \left[\cos \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)+i \sin \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)\right]r^{\frac{1}{n}}=u_{k} \] $0=\mathrm{W} \log k$ \[ \left[\cos \left(\frac{90^{\circ}}{3}+\frac{360^{\circ}}{3}(0)\right)+i \sin \left(\frac{90^{\circ}}{3}+\frac{360^{\circ}}{3}(0)\right)\right]64^{\frac{1}{3}}=u_{0} \] Simplify: \[ =\sqrt{4\left[\cos 30^{\circ}+i \sin 30^{\circ}\right]} \] When $1=k$ : \[ \left[\cos \left(\frac{90^{\circ}}{3}+\frac{360^{\circ}}{3}(1)\right)+i \sin \left(\frac{90^{\circ}}{3}+\frac{360^{\circ}}{3}(1)\right)\right]64^{\frac{1}{3}}=w_{1} \] Simplify: \[ \begin{aligned} =\left[\cos \left(30^{\circ}+120^{\circ}\right)+i \sin \left(30^{\circ}+120^{\circ}\right)\right]4 \\ =&\left[4\left[\cos 150^{\circ}+i \sin 150^{\circ}\right]\right. \end{aligned} \] When $2=k$ : \[ \left[\cos \left(\frac{90^{\circ}}{3}+\frac{360^{\circ}}{3}(2)\right)+i \sin \left(\frac{90^{\circ}}{3}+\frac{360^{\circ}}{3}(2)\right)\right]64^{\frac{1}{3}}=w_{2} \] Simplify: \[ \begin{aligned} =\left[\cos \left(30^{\circ}+240^{\circ}\right)+i \sin \left(30^{\circ}+240^{\circ}\right)\right]4 & \\ &=\left[\cos 270^{\circ}+i \sin 270^{\circ}\right]4 \end{aligned} \]
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