Find $\sqrt[3]{-8}$

Find $\sqrt[3]{-8}$.

The polar form of $-8$ is $8 \left(\cos{\left(\pi \right)} + i \sin{\left(\pi \right)}\right)$ (for steps, see polar form calculator).

According to the De Moivre's Formula, all $n$-th roots of a complex number $r \left(\cos{\left(\theta \right)} + i \sin{\left(\theta \right)}\right)$ are given by $r^{\frac{1}{n}} \left(\cos{\left(\frac{\theta + 2 \pi k}{n} \right)} + i \sin{\left(\frac{\theta + 2 \pi k}{n} \right)}\right)$, $k=\overline{0..n-1}$.

We have that $r = 8$, $\theta = \pi$, and $n = 3$.

• $k = 0$: $\sqrt[3]{8} \left(\cos{\left(\frac{\pi + 2\cdot \pi\cdot 0}{3} \right)} + i \sin{\left(\frac{\pi + 2\cdot \pi\cdot 0}{3} \right)}\right) = 2 \left(\cos{\left(\frac{\pi}{3} \right)} + i \sin{\left(\frac{\pi}{3} \right)}\right) = 1 + \sqrt{3} i$
• $k = 1$: $\sqrt[3]{8} \left(\cos{\left(\frac{\pi + 2\cdot \pi\cdot 1}{3} \right)} + i \sin{\left(\frac{\pi + 2\cdot \pi\cdot 1}{3} \right)}\right) = 2 \left(\cos{\left(\pi \right)} + i \sin{\left(\pi \right)}\right) = -2$
• $k = 2$: $\sqrt[3]{8} \left(\cos{\left(\frac{\pi + 2\cdot \pi\cdot 2}{3} \right)} + i \sin{\left(\frac{\pi + 2\cdot \pi\cdot 2}{3} \right)}\right) = 2 \left(\cos{\left(\frac{5 \pi}{3} \right)} + i \sin{\left(\frac{5 \pi}{3} \right)}\right) = 1 - \sqrt{3} i$

$\sqrt[3]{-8} = 1 + \sqrt{3} i\approx 1 + 1.732050807568877 i$A

$\sqrt[3]{-8} = -2$A

$\sqrt[3]{-8} = 1 - \sqrt{3} i\approx 1 - 1.732050807568877 i$A