Question

Find the cube roots of -8. Write the answer in a + bi form.

Answer 1

So... let's see, so is -8, or -8 + 0i then

$\bf \begin{array}{rllll} -8&,&0i\\ a&&b \end{array}\qquad r=\sqrt{a^2+b^2}\implies r=\sqrt{(-8)^2+0^2}\implies r=8 \\\\\\ \textit{and notice the picture below}\qquad \theta=\pi$

$\bf \textit{ roots of complex numbers, DeMoivre's theorem} \\\\ \sqrt[{{ n}}]{z}=\sqrt[{{ n}}]{r}\left[ cos\left( \cfrac{\theta+2\pi k}{{{ n}}} \right) +i\ sin\left( \cfrac{\theta+2\pi k}{{{ n}}} \right)\right]\quad \begin{array}{llll} k\ roots\\ 0,1,2,3,... \end{array}\\\\ -------------------------------\\\\ z=-8+0i\qquad \qquad \sqrt[3]{z}=\sqrt[{{ 3}}]{8}\left[ cos\left( \cfrac{\pi+2\pi k}{{{ 3}}} \right) +i\ sin\left( \cfrac{\pi+2\pi k}{{{ 3}}} \right)\right]$

$\bf -------------------------------\\\\ \sqrt[{{ 3}}]{8}\left[ cos\left( \cfrac{\pi+2\pi (0)}{{{ 3}}} \right) +i\ sin\left( \cfrac{\pi+2\pi (0)}{{{ 3}}} \right)\right]\impliedby \begin{array}{llll} first\ root\\ k=0 \end{array} \\\\\\ 2\left[ cos\left( \frac{\pi }{3} \right)+i\ sin\left( \frac{\pi }{3} \right)\right] \implies 2\cdot \cfrac{1}{2}+2\cdot \cfrac{\sqrt{3}}{2}\ i\implies 1+\sqrt{3}\ i$

$\bf -------------------------------\\\\ \sqrt[{{ 3}}]{8}\left[ cos\left( \cfrac{\pi+2\pi (1)}{{{ 3}}} \right) +i\ sin\left( \cfrac{\pi+2\pi (1)}{{{ 3}}} \right)\right]\impliedby \begin{array}{llll} second\ root\\ k=1 \end{array} \\\\\\ 2\left[ cos\left( \pi \right)+i\ sin\left( \pi \right)\right]\implies 2\cdot -1+2\cdot 0\ i\implies -2+0i$

$\bf -------------------------------\\\\ \sqrt[{{ 3}}]{8}\left[ cos\left( \cfrac{\pi+2\pi (2)}{{{ 3}}} \right) +i\ sin\left( \cfrac{\pi+2\pi (2)}{{{ 3}}} \right)\right]\impliedby \begin{array}{llll} third\ root\\ k=2 \end{array} \\\\\\ 2\left[ cos\left( \frac{5\pi }{3} \right)+i\ sin\left( \frac{5\pi }{3} \right)\right]\implies 2\cdot \cfrac{1}{2}+2\cdot -\cfrac{\sqrt{3}}{2}\ i\implies 1-\sqrt{3}\ i$

Answer 2

Answer:

D

Step-by-step explanation:

just took it on ed-geniuty, hope it helps :)