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?