Given: #(sqrt(3) - i)^(-10)# Use DeMoivre's Theorem
dDeMoivre's Theorem: #z^n = r^n[cos theta + i sin theta]^n = r^n [cos(n theta) + i sin (n theta)]#
First put #(sqrt(3) - i) = (a + bi)# into polar form:
#a = sqrt(3); " " b = -1#
Polar form: #z = r(cos theta + i sin theta)#
# " where " r = sqrt(a^2 + b^2) = sqrt(3 +(-1)^2) = sqrt(4) = 2#
#" theta = tan^(-1)(b/a) = tan^(-1)(-1/(sqrt(3))) = -30^@ = (11 pi)/6#
# z = 2[ cos (11 pi)/6 + i sin (11 pi)/6]#
#z^(-10) = 2^(-10) [ cos (-10 * (11 pi)/6) + i sin (-10 * (11 pi)/6)]#
#z^(-10) = 1/1024 [cos ((-110 pi)/6) + i sin ((-110 pi)/6)]#
#(-110 pi/6) * 180/(pi) = -3300^@#
#-3300/360 = -9 *(360^@) + -60^@ = -60^@ = (- pi/3)#
#z^(-10) = 1/1024 [cos (-(pi)/3) + i sin (-(pi)/3)]#
#z^(-10) = 1/1024 [1/2 - (sqrt(3))/2 i]#
#(sqrt(3) - i)^(-10) = z^(-10) = 1/2048 - (sqrt(3))/2048 i#