How to solve equation ${z^3 = i}$ in the complex plane using polar form?

Question

I'm just wondering how can I solve equation

$${z^3 = i}$$

in polar form of complex numbers?

Answer 1

You almost had it in the comment section; the mistake was

$${3\phi = k\cdot \frac{\pi}{2}}$$

Instead, it should be

$${3\phi = \frac{\pi}{2} + 2\pi k, k = \{0,1,2\}}$$

Since ${\sin(x)}$ and ${\cos(x)}$ both have period ${2\pi}$. This will mean that

$${\cos\left(\frac{\pi}{2}\right)=\cos\left(\frac{\pi}{2} + 2\pi k\right), k \in \mathbb{Z}}$$

and

$${\sin\left(\frac{\pi}{2}\right)=\sin\left(\frac{\pi}{2} + 2\pi k\right), k \in \mathbb{Z}}$$

This gives you

$${\phi = \frac{\pi}{6} + \frac{2\pi k}{3}, k=\{0,1,2\}}$$

Can you take it from here?