Question

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

Answer 1

`n=(3m)/4`, where `m` is an integer

Explanation:

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

`n=3/2 k` with `k = 1,2,3,cdots`

Explanation:

`(-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`