Question

`root(3)(-1) =` ?

Answer 1

See below.

Explanation:

Putting `x^3=-1 = e^(-i pi/2) = -cos(pi)-isin(pi)` (de Moivre's identity)

we have:

`x^3=e^(-i (pi + 2k pi))` so `x = e^(-i(pi/3+2/3 k pi)`

Now, making `k=0,1,2` we have the tree roots of `-1`

`1/2 - i sqrt[3]/2,-1,1/2 + i sqrt[3]/2`