How do you find the three cube roots of #-2-2i \sqrt{3}#?

1 Answer
Oct 24, 2014

Let #z=re^{i theta}# be the three roots.

#z^3=-2-2isqrt{3}=4(-1/2-sqrt{3}/2i)#

by rewriting in exponential form,

#Rightarrow r^3e^{i(3theta)}=4e^{i({4pi}/3+2npi)}#,

where #n# is any integer,

#Rightarrow {(r^3=4 Rightarrow r=root(3){4}),(3theta={4pi}/3+2npi Rightarrow theta={4pi}/9+{2npi}/3):}#

Hence, the three roots are

#z=\{ root(3){4}e^{i{4pi}/9},root(3){4}e^{i{10pi}/9},root(3){4}e^{i{16pi}/9}}#.


I hope that this was helpful.