+118658 find the complex fourth of -16
\(-16 = -16+0i = 16e^{i\theta}=16(cos\theta+isin\theta)\\ cos\theta =-1 \quad and \quad sin\theta =0\\ so\\ \theta = \pi+2n\pi=\pi(1+2n)\qquad \\ -16=16e^{(1+2n)\pi i }\\ (-16)^{0.25}=16^{0.25}e^{0.25(1+2n)\pi i }\qquad n=0,1,2,3\\ \sqrt[4]{-16}=2e^{0.25(1+2n)\pi i }\\ \sqrt[4]{-16}=2e^{0.25\pi i },\qquad 2e^{0.25(3)\pi i },\qquad 2e^{0.25(5)\pi i },\qquad 2e^{0.25(7)\pi i }\\~\\ \sqrt[4]{-16}\\=2e^{\frac{\pi i}{4} },\qquad 2e^{\frac{3\pi i}{4} },\qquad 2e^{\frac{5\pi i}{4} },\qquad 2e^{\frac{7\pi i}{4} } \\ =2(cos(\frac{\pi}{4})+isin(\frac{\pi}{4})),\qquad 2(cos(\frac{3\pi}{4})+isin(\frac{3\pi}{4})),\qquad 2(cos(\frac{5\pi}{4})+isin(\frac{5\pi}{4})),\qquad 2(cos(\frac{7\pi}{4})+isin(\frac{7\pi}{4}))\\ =2(cos(45)+isin(45)),\qquad 2(-cos(45)+isin(45)),\qquad 2(-cos(45)-isin(45)),\qquad 2(cos(45)-isin(45))\\ =2(\frac{1+i}{\sqrt{2}}),\qquad 2(\frac{-1+i}{\sqrt{2}}),\qquad 2((\frac{-1-i}{\sqrt{2}}),\qquad 2(\frac{1-i}{\sqrt{2}})\\~\\ \sqrt[4]{-16}=\sqrt2 +\sqrt2\;i, \qquad -\sqrt2 +\sqrt2\;i, \qquad -\sqrt2 -\sqrt2\;i, \qquad \sqrt2 -\sqrt2\;i, \qquad \)
+118658 I used this video to help me to get started. (for revision)
https://www.youtube.com/watch?v=WpoBmnt0Re4
ans
This has a grph to help you visualize it.
