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

1 Answer
Feb 25, 2017

Answer:

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#