Find $$$\sqrt[3]{i}$$$

Find $$$\sqrt[3]{i}$$$.

The polar form of $$$i$$$ is $$$\cos{\left(\frac{\pi}{2} \right)} + i \sin{\left(\frac{\pi}{2} \right)}$$$ (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 $$$r = 1$$$, $$$\theta = \frac{\pi}{2}$$$, and $$$n = 3$$$.

• $$$k = 0$$$: $$$\sqrt[3]{1} \left(\cos{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 0}{3} \right)} + i \sin{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 0}{3} \right)}\right) = \cos{\left(\frac{\pi}{6} \right)} + i \sin{\left(\frac{\pi}{6} \right)} = \frac{\sqrt{3}}{2} + \frac{i}{2}$$$
• $$$k = 1$$$: $$$\sqrt[3]{1} \left(\cos{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 1}{3} \right)} + i \sin{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 1}{3} \right)}\right) = \cos{\left(\frac{5 \pi}{6} \right)} + i \sin{\left(\frac{5 \pi}{6} \right)} = - \frac{\sqrt{3}}{2} + \frac{i}{2}$$$
• $$$k = 2$$$: $$$\sqrt[3]{1} \left(\cos{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 2}{3} \right)} + i \sin{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 2}{3} \right)}\right) = \cos{\left(\frac{3 \pi}{2} \right)} + i \sin{\left(\frac{3 \pi}{2} \right)} = - i$$$

$$$\sqrt[3]{i} = \frac{\sqrt{3}}{2} + \frac{i}{2}\approx 0.866025403784439 + 0.5 i$$$A

$$$\sqrt[3]{i} = - \frac{\sqrt{3}}{2} + \frac{i}{2}\approx -0.866025403784439 + 0.5 i$$$A

$$$\sqrt[3]{i} = - i$$$A