Roots of Unity - $x^3 = -i$

Question

I need to find the roots of unity for the following equation:

$$x^3 + i = 0$$

Thus, $x^3 = -i$. I know that $-i = \exp[i(\frac{3\pi}{2} + 2n \pi)]$ however I do not know how to get all roots.

Note: I apologize for the terrible formatting, I just joined the website today and am not sure how to add pre-formatted equations yet.

TIA!

Answer 1

You can do this "trick"

All the roots of $x^3=1$ are $1,\omega,\omega^2$ where $\omega=e^{2i\pi/6}$.

So, all roots of $x^3=-i$ would be roots of $1$ times $i$ (since $i^3=-i$). Thus the roots are $i,i\omega,i\omega^2$

Answer 2

$x^3 + i = 0 \to x^3 - i^3 = 0 \to (x-i)(x^2 + xi - 1) = 0$. Can you use quadratic formula to complete the answer.

Answer 3

In the case of equation like $$ x^n-\alpha=0 $$ where $\alpha\in\mathbb C$ the best trick is to write $\alpha$ in polar form: $\alpha=re^{i\theta}$. But $e^{i\theta}=e^{i(\theta+2k\pi)},\;k\in\mathbb Z$. Then you have $$ x^n=re^{i\theta+2k\pi i} $$ from which immediately $$ x=\sqrt[n]re^{i(\frac{\theta+2k\pi }{n})} $$ and as $k=0,1,\dots,n-1$ you get all and only the $n$ roots of the equation.

In your case $n=3$ and $\alpha=-i=e^{i\frac32\pi+2k\pi i}$ hence $$ x_k=e^{i(\frac{\pi}{2}+\frac23k\pi)},\;\;k=0,1,2 $$ are the 3 roots of your equation.