What is the smallest positive integer value of n such that (√3/2-i/2)^n is real?
Consider the expression (3/2−i/2)n. Let's analyze the powers of this complex number according to the four cases for the remainder r of n when divided by 4:
Case r=0: Then n=4k for some integer k, and
[(\sqrt{3}/2 - i/2)^{4k} = \left( (\sqrt{3}/2)^2 + (i/2)^2 \right)^{2k} = \left( \frac{3}{4} + \frac{1}{4} \right)^{2k} = 1^{2k} = 1.]
Since 1 is real, this case produces a real number.
Case r=1: Then n=4k+1 for some integer k, and
(√3/2−i/2)4k+1=(√3/2−i/2)⋅(√3/2−i/2)4k =(√3/2−i/2)⋅1 =√3/2−i/2.
Since this expression has a nonzero imaginary part, it is not real.
Case r=2: Then n=4k+2 for some integer k, and
(√3/2−i/2)4k+2=(√3/2−i/2)2⋅(√3/2−i/2)4k =(34−2√34i−14)⋅1 =34−2√34i.
Since this expression has a nonzero imaginary part, it is not real.
Case r=3: Then n=4k+3 for some integer k, and
(√3/2−i/2)4k+3=(√3/2−i/2)⋅(√3/2−i/2)4k+2 =(√3/2−i/2)⋅(34−2√34i) =3√34+14−34i.
Since this expression has a nonzero imaginary part, it is not real.
In conclusion, the smallest positive integer value of n such that (3/2−i/2)^n is real is 4.