$$ (-1+i)^{\frac{1}{3}} $$
Below is what I've attempted, but I'm not 100% positive if it's right. Also, and more importantly, how do I know if I've computed ALL of the roots of a complex number?
$$|(-1+i)|=\sqrt{2}$$ $$\theta=tan^{-1}(\frac{1}{1})=\frac{\pi}{4}$$ $$z=\sqrt{2}(e^{i(2\pi k+\frac{\pi}{4})})^{\frac{1}{3}}$$
Then if $k=0,1,2$
$$z_0=\sqrt{2}e^{i\frac{\pi}{12}}$$ $$z_1=\sqrt{2}e^{i\frac{3\pi}{4}}$$ $$z_2=\sqrt{2}e^{i\frac{17\pi}{12}}$$
On
If $z = re^{i\theta} \ne 0$ and $n \in \mathbb N; n > 0$, then $z^{1/n}$ can have precisely $n$ possible values and those values are: $r^{1/n}e^{i*\frac{\theta + k*2\pi}{n}}; k = 0, 1, ... ,n-1$.
This is a fundamental theorem.
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Proof: If $w = te^{i\rho}$ is such that $w^n = z$ then $w^n = t^n e^{i*n*\rho} = r e^{i\theta}$.
So $t^n = r$ and $\cos n\rho = \cos \theta$ and $\sin n\rho = \sin \theta$
As $t, r \in \mathbb R$ and $t,r > 0$ then $t = r^{1/n}$ is the unique solution that always exists.
If $\rho = (\theta + k*2\pi)/n$ then $n \rho = \theta + k*2\pi$$;k \in \mathbb Z$, will obviously always be solutions. But are they the only ones.
Yes, because for any given $\cos \eta = x; \sin \eta = y; \eta \in [0, 2\pi]$ has one precise solution for $\eta$.
So no other solutions for $\rho$ exist. (Because $n \rho$ must equal $\theta + 2k*\pi$ for both $\sin n\rho = \sin \theta$ and $\cos n\rho = \cos \theta$.)
As $e^i{(\theta + 2*(k + n*m)\pi)/n}= e^i{(\theta + 2*k\pi)/n}$, then $\rho = (\theta + 2*k*\pi)/n; k = 0,1,..., n-1$ are the precise distinct possible values for $\rho$.
You have $z=-1+i$, so $|z|=\sqrt{2}$. Note $\tan^{-1}(\tfrac{1}{-1})=-\frac{\pi}{4}$, so you need $\pi-\frac{\pi}{4}=\tfrac{3\pi}{4}$ as $z$ is in the second quadrant. Now using $z=\sqrt{2}e^{i\left(2k\pi+\tfrac{3\pi}{4}\right)}$ you need the cube roots of this. To this end let $w^3=z$, then
\begin{align} w_k=\left(\sqrt{2}e^{i \big(2k\pi+\tfrac{3\pi}{4}\big)}\right)^{\frac{1}{3}}. \end{align} Now using Euler's formula, $$ e^{i \theta}=\cos \theta + i \sin \theta, $$ gives the following: $$ w_k= 2^{1/6}\left(\cos\left(\frac{(8k+3)\pi}{4}\right)+i\sin\left(\frac{(8k+3)\pi}{4}\right)\right)^{\frac{1}{3}}. $$ Further using de Moivre's formula, $$ (\cos \theta + i \sin \theta)^n = \cos \,n\theta + i \sin \,n\theta, $$ gives the following: $$ w_k=2^{1/6}\left(\cos\left(\frac{(8k+3)\pi}{12}\right)+i\sin\left(\frac{(8k+3)\pi}{12}\right)\right), \qquad \text{for $k=0,1,2,\dotsc.$} $$ This gives the three cube roots of $z$ as $$ w_0=2^{1/6}\left(\cos\left(\frac{3\pi}{12}\right)+i\sin\left(\frac{3\pi}{12}\right)\right)=2^{1/6}e^{\frac{i\pi}{4}}; $$
$$ w_1=2^{1/6}\left(\cos\left(\frac{11\pi}{12}\right)+i\sin\left(\frac{11\pi}{12}\right)\right)=2^{1/6}e^{\frac{11i\pi}{12}}; $$
$$ w_2=2^{1/6}\left(\cos\left(\frac{19\pi}{12}\right)+i\sin\left(\frac{19\pi}{12}\right)\right)=2^{1/6}e^{\frac{-5i\pi}{12}}. $$ Where the last root, $w_2$, has been simplified in the exponential form since $19\pi/12-2\pi=-5\pi/12$, and we usually take the principal value to be in the interval $(-\pi\text{ rad}, \pi\text{ rad}]$, i.e, the one where $-\pi$ is not allowed. Another way to think of this is if we take $k=-1,0,1$ in the above formula for $w_k$ we obtain the roots having principal value without further simplification, but this time in the order $w_2$, $w_0$, $w_1$.
Note that in our formula for $w_k$ we could have immediately worked out $$ w_k=2^{1/6}e^{i\left(\tfrac{(8k+3)\pi}{12}\right)} $$ by laws of exponents, and then moved to Euler's formula without the need for de Moivre's formula to deal with the $\frac{1}{3}$rd power, but I thought it best to cover both methods as they are useful when flitting between the exponential and trigonometrical forms.
As to the question of how many roots there are: You only ever get $n$ $n$th roots of a complex number by the Fundamental Theorem of Algebra. In this question then you only need $k=0,1,2$ in the last formula. If plotted in the complex plane the $n$th root of a complex number $s$ will give $n$ complex numbers equally spaced $\frac{2\pi}{n}$ radians apart on a circle of radius $|s|$ about the origin. Moreover if these $n$ complex numbers were connected by straight lines in sequence, $s_0$ to $s_1$, $s_1$ to $s_2$, and so on, a regular $n$-gon is constructed with the roots as vertices.