how to find cube roots of complex number

Question

$c^3 = -1 + i$

How does one calculate the complex number(s) that satisfy above. Would like a general method, if there is one?

Answer 1

Using the definition of atan2, $$\displaystyle c^3=\sqrt2e^{\dfrac{3\pi i}4+2n\pi i}=\sqrt2e^{\dfrac{(3+8n)\pi i}4}$$ where $n$ is any integer

$$\displaystyle \implies c=2^{\dfrac16}e^{\dfrac{(3+8n)\pi i}{12}}$$ where $0\le n\le2$

Answer 2

Let's call call c^3 = z^3.

We know that

$z = cos(\theta) + isin(\theta)$.

By De Moivre ...

$z^3 = cos(3 \theta) + isin(3 \theta)$

We know that $cos(3 \theta) = -1$ and $sin(3 \theta) = 1$.

$3 \theta = cos^{-1}(-1)$ and $3 \theta = sin^{-1}(1)$

So $3 \theta = \pi + 2k \pi$ and $3 \theta = \frac{\pi}{2} + 2k \pi$ for integers k.

I'm also learning complex analysis, I'm pretty sure that's how you do it. I hope that helps.