To find ( 2 - 2i·sqrt(3) )4
First place the number 2 - 2i·sqrt(3) into "cis form"
-- r = sqrt( (2)2 + [-2·sqrt(3)]2 ) = sqrt( 4 + 12 ) = sqrt( 16 ) = 4
-- theta = tan-1( -2sqrt(3) / 2 ) = 5pi/3
So the number can be written as: 4·cis( 5pi/3 ).
To find the nth power of r·cis( theta) = rn·cis( n·theta)
So: ( 2 - 2i·sqrt(3) )4 = 44·cis( 4·5pi/3 ) = 256·cis( 20pi/3 ).
You'll need to reduce 20pi/3.
Then, to put it back into rectangular form, multiply out the cis form.