Find the cube roots of $ -8 i $ and plot them on a plane.

Question

I can’t figure out the angle of this equation.

I set it up like this: $$ z^{3} = 0 - 8 i. $$ I find that the $ r $-value is $ 2 $, but when I try to find the angle, I’m stuck. I can’t divide by $ 0 $, so where did I go wrong? Or what am I missing to solve this?

Answer 1

$$\text{You want to use Demoivre. So you have,} \ \ z^3=-8i \Rightarrow 8(\cos(\frac{-\pi}{2})+i\sin(\frac{-\pi}{2}))$$

$$\text{By Demoivre}: z = 2\{\cos(\frac{\frac{-\pi}{2}+2\pi k}{3})+i\sin(\frac{\frac{-\pi}{2}+2\pi k}{3})\}, \ \text{for $k=0,1,2$}$$

Answer 2

Hint: $-8i$ lies on the negative imaginary axis, so you will have trouble computing the angle as an arctangent. But, thinking about the geometry, isn't the angle just $-\pi/2$?

Answer 3

You can do this without DeMoivre and trigonometry if you’ve observed that $i^3=-i$ and that therefore $(2i)^3=-8i$. Thus $X^3+8i$, as the difference of two cubes, factors in the way that I hope you already know, $a^3-b^3=(a-b)(a^2+ab+b^2)$, giving $$ X^3+8i=(X-2i)(X^2+2iX-4)\,, $$ and the quadratic part you can run the Quadratic Formula on.