To use DeMoivre's Theorem, we need to convert
#z=-sqrt3+i=x+iy# into Polar Form, i.e.,
#r(costheta+isintheta), where, r>0, &, theta in (-pi,pi]#.
#z=x+iy=r(costheta+isintheta)=rcistheta#
#rArr x=rcostheta, y=rsintheta, r=sqrt(x^2+y^2)#
#:. r=sqrt{(-sqrt3)^2+1^2}=2#
Hence, from #x=rcostheta, costheta=-sqrt3/2, "&, similarly,"# #sintheta=1/2#
#"clearly", theta=(5pi)/6#
Thus, #-sqrt3+i=2cis((5pi)/6)#.
Now, by DeMoivre's Theorem, #(rcistheta)^n=r^ncis(ntheta)#
In our case, #n=5, r=2,theta=(5pi)/6#
#:.(-sqrt3+i)^5=(2cis(5pi)/6)^5#
#=(2^5)(cis(5*(5pi)/6))#
#=32(cis(25pi)/6)#
#=32(cis(4pi+pi/6))#
#=32(cos(4pi+pi/6)+isin(4pi+pi/6))#
#=32(cos(pi/6)+isin(pi/6))#
#=32(sqrt3/2+1/2i)#
#=16(sqrt3+i)#.
Enjoy Maths.!
