Solving $z^3=-1+i$

Question

First, is there a better way than using x+iy and solving the system? I tried letting $z=e^{i3\theta}$ and using the cosine and i*sine way but I don't see how that can equal -1 and i at the same time, either cosine is -1 or sine is 1 not both?

Anyway I've arrived at the real and imaginary parts by plugging in $x+iy$ to get the system

$$x^3-3y^2x=-1$$ $$3yx^2 -y^3=1$$

We can conclude y and x are nonzero but how do we solve?

Answer 1

You can't put $z^3=-1+i=e^{3i\theta}$ because $\lvert -1+i\rvert = \sqrt{2}$ but $\lvert e^{3i\theta}\rvert = 1$, you need to normalize $z$, i.e. $z=re^{i\theta}$, $r\in\mathbb{R}$. Then $r^3 = \sqrt{2}$ and after that, solve for $\theta$.

Answer 2

You can use polar form. Notice that $-1+i = \sqrt{2}e^{i\left(\frac{3\pi}{4} + 2k\pi\right)}$, with $k \in \Bbb Z$. Write $z = re^{i\theta}$. So: $$r^3e^{i(3\theta)} = \sqrt{2}e^{i\left(\frac{3\pi}{4} + 2k\pi\right)} \implies r = \sqrt[6]{2} \quad \mbox{and} \quad \theta = \frac{\pi}{4} + \frac{2k\pi}{3}.$$ Make $k = 0,1$ and $2$ to get your roots.

Answer 3

Let $-1+i= z^3 = r^3 e^{3i\theta}$. Then we have $r^3= \sqrt{(-1)^2+1^2} = \sqrt{2}$ and $\theta = 3\pi/4$. Then we get $\theta=\pi/4$ or you can use $\pi/4\pm2\pi/3$.