you are looking for:
#x = (1-i)^(1/3)#
as a rule, this complex number will have 3 cube roots, same absolute value but spaced equally about the Argand diagram.
using #cis (theta) equiv e^(i theta)# in the notation, we can express this in polar coordinates as follows:
#1-i = \sqrt 2 \ cis(- pi /4 color(red)(+ 2n pi))# where #n in Z#
So
#(1-i)^(1/3) = (\sqrt 2 \ cis(- pi /4 + 2n pi))^(1/3)#
# = 2^(1/6) \ cis(- pi /12 + (2n pi)/3 )#
working through values of n
#n = 0 to 2^(1/6) cis(- pi /12) = 2^(1/6) exp(- (i pi) /12)#
#n = 1 to 2^(1/6) cis( (7pi) /12) = 2^(1/6) exp( (7i pi) /12)#
#n = 2 to 2^(1/6) cis( (5pi)/4 ) = 2^(1/6) exp( (5 i pi)/4 )#
and then the roots start repeating as follows
#n = 3 to 2^(1/6) cis(- pi /12 + 2pi ) = 2^(1/6) cis(- pi /12 )#
