We need to find a certain complex number #z# such that #z^4=-8+8isqrt(3)#.
First of all, all complex numbers can be written in polar form. Thus, we can try to rewrite the right-hand side from Cartesian form to polar form #a+bi=r(cos(theta)+i sin(theta))#, where #r=sqrt(a^2+b^2)# and #theta=arctan(b/a)#. Note that, since #-8+8isqrt(3)# is in the third quadrant of the Argand plane, #theta# must be in the third quadrant of the unit circle.
Then, #r=sqrt((-8)^2+(8sqrt(3))^2)=16# and #theta=arctan((8sqrt(3))/-8)=(2pi)/3+2kpi,k in ZZ# (the reason for #k# will be clear soon, but realize that adding an integer multiple of #2pi# to an angle does not change its value).
So we have #-8+8isqrt(3)=16(cos((2pi)/3+2kpi)+i sin((2pi)/3+2kpi))#.
Then, #z^4=16(cos((2pi)/3+2kpi)+i sin((2pi)/3+2kpi)#, or #z=(16(cos((2pi)/3+2kpi)+i sin((2pi)/3+2kpi))^(1/4)#
Using de Moivre's formula, we know that #(r(cos(theta)+i sin(theta)))^n=r^n(cos(ntheta)+i sin(ntheta))#.
Thus, #z=(2(cos(((2pi)/3+2kpi)/4)+i sin(((2pi)/3+2kpi)/4))),k in ZZ#.
Since #cos(theta)# and #sin(theta)# are periodic every #2pi#, we only need to substitute #0≤k<4# (substituting other integers will reveal the same solution with one in this range).
The solutions are the following:
#z=2(cos(pi/6)+i sin(pi/6))=sqrt(3)+1/2i#
#z=2(cos((2pi)/3)+i sin((2pi)/3))=-1+isqrt(3)#
#z=2(cos((7pi)/6)+i sin((7pi)/6))=-sqrt(3)-1/2i#
#z=2(cos((5pi)/3)+i sin((5pi)/3))=1-isqrt(3)#