#z=2sqrt3 -2i#
#|z|=sqrt((2sqrt3)^2 +(2)^2)=sqrt(4*3+4)=4#
DeMoivre's formula:
#[|z|(cosvarphi+isinvarphi)]^n=|z|^n(cos(nvarphi)+isin(nvarphi))#
Let: #a=2sqrt3# and #b=-2#
In order to find #varphi# the following statement must be true:
#cosvarphi=a/|z|# #^^# #sinvarphi=b/|z|#
#cosvarphi=(2sqrt3)/4=sqrt3/2 =>#
#=> varphi_1=pi/6# #vv# #varphi_2=(11pi)/6#
#sinvarphi=-2/4=-1/2 =>#
#=> varphi_1=(5pi)/6# #vv# #varphi_2=(11pi)/6#
The right #varphi# is the one that matches which is #varphi=(11pi)/6#
YOU CAN SIMPLY SKIP THIS BY WRITING A CIRCLE WITH RADIUS EQUAL TO 1 AND POINTING TREIR VALUES AND FINDING COMMON ANGLE
#(2sqrt3 -2i)^5=#
#=[|4|(cos((11pi)/6)+isin((11pi)/6))]^5=#
#=4^5(cos(5(11pi)/6)+isin(5(11pi)/6))=#
#=1024(cos((55pi)/6)+isin((55pi)/6))=#
#55pi=12*2*2pi+7pi#
#=1024(cos((7pi)/6)+isin((7pi)/6))=#
#=1024(-cos(pi/6)+isin(-pi/6))=#
#=1024(-sqrt3/2-i1/2)=-512sqrt3-i512#
(Hi guys, do you use the same process?)
