Mathematics, 22.11.2019 05:31 capo9972
Find the cube roots of -8. write the answer in a + bi form.
Answers: 1
![\bf \textit{ roots of complex numbers, DeMoivre's theorem} \\\\ \sqrt[{{ n}}]{z}=\sqrt[{{ n}}]{r}\left[ cos\left( \cfrac{\theta+2\pi k}{{{ n}}} \right) +i\ sin\left( \cfrac{\theta+2\pi k}{{{ n}}} \right)\right]\quad \begin{array}{llll} k\ roots\\ 0,1,2,3,... \end{array}\\\\ -------------------------------\\\\ z=-8+0i\qquad \qquad \sqrt[3]{z}=\sqrt[{{ 3}}]{8}\left[ cos\left( \cfrac{\pi+2\pi k}{{{ 3}}} \right) +i\ sin\left( \cfrac{\pi+2\pi k}{{{ 3}}} \right)\right]\\\\](/tpl/images/0385/9549/6b38e.png)
![\bf -------------------------------\\\\ \sqrt[{{ 3}}]{8}\left[ cos\left( \cfrac{\pi+2\pi (0)}{{{ 3}}} \right) +i\ sin\left( \cfrac{\pi+2\pi (0)}{{{ 3}}} \right)\right]\impliedby \begin{array}{llll} first\ root\\ k=0 \end{array} \\\\\\ 2\left[ cos\left( \frac{\pi }{3} \right)+i\ sin\left( \frac{\pi }{3} \right)\right] \implies 2\cdot \cfrac{1}{2}+2\cdot \cfrac{\sqrt{3}}{2}\ i\implies 1+\sqrt{3}\ i](/tpl/images/0385/9549/123e6.png)
![\bf -------------------------------\\\\ \sqrt[{{ 3}}]{8}\left[ cos\left( \cfrac{\pi+2\pi (1)}{{{ 3}}} \right) +i\ sin\left( \cfrac{\pi+2\pi (1)}{{{ 3}}} \right)\right]\impliedby \begin{array}{llll} second\ root\\ k=1 \end{array} \\\\\\ 2\left[ cos\left( \pi \right)+i\ sin\left( \pi \right)\right]\implies 2\cdot -1+2\cdot 0\ i\implies -2+0i](/tpl/images/0385/9549/5d726.png)
![\bf -------------------------------\\\\ \sqrt[{{ 3}}]{8}\left[ cos\left( \cfrac{\pi+2\pi (2)}{{{ 3}}} \right) +i\ sin\left( \cfrac{\pi+2\pi (2)}{{{ 3}}} \right)\right]\impliedby \begin{array}{llll} third\ root\\ k=2 \end{array} \\\\\\ 2\left[ cos\left( \frac{5\pi }{3} \right)+i\ sin\left( \frac{5\pi }{3} \right)\right]\implies 2\cdot \cfrac{1}{2}+2\cdot -\cfrac{\sqrt{3}}{2}\ i\implies 1-\sqrt{3}\ i](/tpl/images/0385/9549/4a41f.png)
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Find the cube roots of -8. write the answer in a + bi form....
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