Solve for #n#, #(-1/2+ iota sqrt(3)/2)^3 + (-1/2-iota sqrt(3)/2)^(2n)=2# ?

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Answer 1 · 2017-03-12T16:29:54

#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

Answer 2 · 2017-03-12T17:14:15

#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#