How to find the third root of a complex number by transforming the complex number into a root I'd like to find all 3rd roots of this number z = i - 1. Now I've found formulas on how to do it; First we transform the complex number into this form
$$ \sqrt[n]{r} * e^{i\frac {\phi + 2k\pi}{n}} $$ Where n should be 3 (because of 3rd root) and k should be k = n - 1 (Including 0)
Now what is not really clear to me is what to do with this -1? If I transform this I'd pressume I get this.
$$\sqrt[3]{-1} * e^{i\frac {\pi/2 + 2k\pi}{3}}$$
But than what? If it was a 2nd root I could say that it is simply i (the root of a negative number) but since it is a 3rd root I'm not quite sure what to do next?
Now for the phi (angle) I just assumed this;  $$i = e^{i \frac {\pi}{2}}$$ And I'm not too sure about this
Thanks!
EDIT: With great help from the people in the comments; I think I've solved it.So as already stated in the comments; $$ r = \sqrt 2, \phi = \frac{3\pi}{4} $$ Now If i would to input that into my formula we should  get this;
$$ \sqrt[6]{2} * e^{i\frac{3pi/4 + 2k\pi}{3}} $$
Now as stated before k = 0... n-1;
k = 0 --> $$ \sqrt[6]{2} * e^{i(\pi/4)} $$
k = 1 --> $$ \sqrt[6]{2} * e^{i(11\pi/12)} $$
k = 2 --> $$ \sqrt[6]{2} * e^{i(19\pi/12)} $$
You reckon I did this correctly?
 A: Note that\begin{align}i-1&=-1+i\\&=\sqrt2\left(-\frac1{\sqrt2}+\frac i{\sqrt2}\right)\\&=\sqrt2e^{3\pi i/4}.\end{align}Therefore, the cubic roots of $i-1$ are$$\sqrt[6]2e^{\pi i/4},\ \sqrt[6]2e^{\pi i/4+2\pi i/3}(=\sqrt[6]2e^{11\pi i/12})\text{, and }\sqrt[6]2e^{\pi i/4+4\pi i/3}(=\sqrt[6]2e^{19\pi i/12}).$$
A: I find these much easier to think about visually. The number you start with is $i-1$, or in the more standard notation $-1+i$. That number is located in the second quadrant of the complex plane. Its modulus (i.e., its absolute value, or distance from $0$) is $\sqrt2$, and its argument (the angle between the positive real axis and a line from $0$ to our point) is $\frac{3\pi}{4}$.
You'll find one cube root by taking the cube root of the modulus, and dividing the argument by $3$. This comes from DeMoivre's theorem on powers of complex numbers. Thus, the cube root of $2^{1/2}$ is $2^{1/6}$, and one third of $\frac{3\pi}{4}$ is $\frac{\pi}{4}$.
Therefore, the principal cube root will be $2^{1/6}$ units from $0$, in the direction of $1+i$. In polar form, it's $2^{1/6}\cdot e^{i\pi/4}$ Considering that the real and imaginary parts are equal, and their squares sum to the square of $2^{1/6}$, we obtain the number $a+bi$, where $a=b=2^{-1/3}$. Writing it with radicals, that's $$\frac{1}{\sqrt[3]{2}}+ \frac{1}{\sqrt[3]{2}}i$$
The other two cube roots will have the same modulus, but their arguments are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ greater than the argument of the principal root.
Does this answer your question?
A: Hint: In the formula you used $r$ and $\phi$   are the absolute value and the argument of the complex number $z$.
The absolute value and argument of a complex number $z=x+iy$ can be computed as:
$$
r=\sqrt{x^2+y^2}, \quad \phi=\begin {cases}\arccos\frac xr,&y\ge0\\
2\pi-\arccos\frac xr,&y<0\\
\end {cases}.
$$
Particularly for $z=-1+i$ :
$$r=\sqrt2,\quad \phi=\frac{3\pi}4.
$$
A: $$\sqrt[3]{re^{i\theta+2k\pi}}=\sqrt[3]re^{i\theta/3}e^{i2k\pi/3}.$$
If we let $\omega=e^{i2\pi/3}$, the three roots are
$$\sqrt[3]re^{i\theta/3}, \sqrt[3]re^{i\theta/3}\omega,\sqrt[3]re^{i\theta/3}\omega^2.$$
This is true for any complex number.
Notice that
$$\omega=-\frac12+i\frac{\sqrt3}2,\omega^2=-\frac12-i\frac{\sqrt3}2.$$ The next powers repeat periodically (this is why there are only three distinct roots). The three roots are obtained from any root, multipliying once and twice by $\omega$.
