Solve for #n#, #(-1/2+ iota sqrt(3)/2)^3 + (-1/2-iota sqrt(3)/2)^(2n)=2# ?
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#n=(3m)/4#, where #m# is an integer
We have to evaluate #(-1/2+isqrt3/2)^3+(-1/2-isqrt3/2)^(2n)#
As #(-1/2+isqrt3/2)=cos((2pi)/3)+isin((2pi)/3)# and
#(-1/2-isqrt3/2)=cos((4pi)/3)+isin((4pi)/3)#
using De Moivre's Theorem
#(-1/2+isqrt3/2)^3=cos((2pi)/3xx3)+isin((2pi)/3xx3)#
= #cos(2pi)+isin(2pi)=1#
and #(-1/2-isqrt3/2)^(2n)=cos((4pi)/3xx2n)+isin((4pi)/3xx2n)#
= #cos((8npi)/3)+isin((8npi)/3)#
Therefore #(-1/2+isqrt3/2)^3+(-1/2-isqrt3/2)^(2n)=2#
#cos((8npi)/3)+isin((8npi)/3)=1=cos(2mpi)+isin(2mpi)#,
where #m# is an integer
i.e. #(8npi)/3=2mpi# or #m=(4n)/3#
and #n=(3m)/4#, where #m# is an integer
#n=3/2 k# with #k = 1,2,3,cdots#
#(-1/2+ iota sqrt(3)/2)^3 + (-1/2-iota sqrt(3)/2)^(2n)=2#
We will using the complex number exponential representation
#x+iy=rho e^(i phi)# where #rho=sqrt(x^2+y^2)# and #phi=arctan(y/x)#
So #rho = sqrt((1/2)^2+(sqrt3 /2)^2) = 1# and #phi=arctan(sqrt(3))=pi/3#
then we have
#e^(i 3 phi)+e^(-i 2n phi)=2#
or
#e^(i 3 phi)+e^(-i3phi((2n)/3))=2#
but #e^(i3phi)= 1# and #e^(-i3phi)= 1# so with #(2n)/3=k, k=1,2,3,cdots#
#1+1^k=2#
then
#n=3/2 k# with #k = 1,2,3,cdots#
