$\sqrt[3]{- 1} =$ $root(3)(-1) =$ ?

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Answer 1 · 2017-02-25T13:57:24

See below.

Putting $x^{3} = - 1 = e^{- i \frac{π}{2}} = - cos (π) - i sin (π)$ $x^3=-1 = e^(-i pi/2) = -cos(pi)-isin(pi)$ (de Moivre's identity)

we have:

$x^{3} = e^{- i (π + 2 k π)}$ $x^3=e^(-i (pi + 2k pi))$ so $x = e^{- i (\frac{π}{3} + \frac{2}{3} k π)}$ $x = e^(-i(pi/3+2/3 k pi)$

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

$\frac{1}{2} - i \frac{\sqrt{3}}{2}, - 1, \frac{1}{2} + i \frac{\sqrt{3}}{2}$ $1/2 - i sqrt[3]/2,-1,1/2 + i sqrt[3]/2$

3√−1=root(3)(-1) = ?