## Precalculus (6th Edition)

$[2{cis45^{\circ},2cis 135^{\circ},2cis 225^{\circ},2cis 315^{\circ}}]$
Given: $x^{4}+16=0$ or $x^{4}=-16$ $-16$ can be written in trigonometric form as: $-16=16(\cos 180^{\circ}+\sin 180^{\circ})$ $x^{4}=16(\cos 180^{\circ}+\sin 180^{\circ})$ $x^{4}=2^{4}[\cos (180^{\circ}+2k\pi)+i \sin (180^{\circ}+2k\pi)]$ $x=2^{4\times \frac{1}{4}}[(\cos (180^{\circ}+2k\pi)+i \sin (180^{\circ}+2k\pi))]^{\frac{1}{4}}$ $x=2[(\cos (180^{\circ}+2k\pi)+i \sin (180^{\circ}+2k\pi))]^{\frac{1}{4}}$ Apply De-Moivre's Theorem $x=2[(\cos \frac{(180^{\circ}+2k\pi)}{4}+i \sin\frac{(180^{\circ}+2k\pi)}{4})]$ Now, the arguments can be written as: $\frac{180^{\circ}+2k\pi}{4}$ for $k=0,1,2,3$ Arguments are: $45^{\circ}$ for $k=0$ $135^{\circ}$ for $k=1$ $225^{\circ}$ for $k=2$ $315^{\circ}$ for $k=3$ Solution set of the equation can be written as: $[2{cis45^{\circ},2cis 135^{\circ},2cis 225^{\circ},2cis 315^{\circ}}]$