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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

 

\begin{align*} (\sqrt{3}/2 - i/2)^{4k + 1} &= (\sqrt{3}/2 - i/2) \cdot (\sqrt{3}/2 - i/2)^{4k} \ &= (\sqrt{3}/2 - i/2) \cdot 1 \ &= \sqrt{3}/2 - i/2. \end{align*}

 

Since this expression has a nonzero imaginary part, it is not real.

 

Case r=2: Then n=4k+2 for some integer k, and

 

\begin{align*} (\sqrt{3}/2 - i/2)^{4k + 2} &= (\sqrt{3}/2 - i/2)^2 \cdot (\sqrt{3}/2 - i/2)^{4k} \ &= \left( \frac{3}{4} - \frac{2 \sqrt{3}}{4} i - \frac{1}{4} \right) \cdot 1 \ &= \frac{3}{4} - \frac{2 \sqrt{3}}{4} i. \end{align*}

 

Since this expression has a nonzero imaginary part, it is not real.

 

Case r=3: Then n=4k+3 for some integer k, and

 

\begin{align*} (\sqrt{3}/2 - i/2)^{4k + 3} &= (\sqrt{3}/2 - i/2) \cdot (\sqrt{3}/2 - i/2)^{4k + 2} \ &= (\sqrt{3}/2 - i/2) \cdot \left( \frac{3}{4} - \frac{2 \sqrt{3}}{4} i \right) \ &= \frac{3 \sqrt{3}}{4} + \frac{1}{4} - \frac{3}{4} i. \end{align*}

 

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​.