There are a natural number $n$ that verify $(1/5 + i (3^{1/2})/5)^n = k$; where $k$ is a real number

Question

What I did was calculate the complex in trigonometric form and then try to plot it in the complex plane to see if the nth root of some real number could be obtained from this complex but I do not conclude anything and I would not know of another way to study this statement.

Answer 1

Let $z = \dfrac{1+i \sqrt3}{5} = \dfrac{2}{5}\omega$, where $\omega = \dfrac{1+i \sqrt3}{2}$.

Then $z^n$ is real iff $\dfrac{5}{2}z^n$ is real iff $\omega^n$ is real.

Now, $\omega^2 = \dfrac{-1+i \sqrt3}{2}$ and $\omega^3 = -1$, and so $\omega^{3m} = (-1)^m$, $\omega^{3m+1} = (-1)^m \omega$, $\omega^{3m+2} = (-1)^m \omega^2$.

Thus, $\omega^n$ is real iff $n$ is a multiple of $3$.

Answer 2

$$z=\frac {1}{5}+i\frac {\sqrt {3}}{5}=$$

$$\frac {2}{5}(\frac {1}{2}+i\frac {\sqrt {3}}{2})=\frac {2}{5}e^{i\frac{\pi}{3}} .$$

$$z^n=\frac {2^n}{5^n}e^{i\frac {n\pi}{3}}\in \mathbb R $$ $$\iff (\exists m\in\mathbb Z) \;:\;\frac{n\pi}{3}=m\pi$$

$$\iff (\exists m\in \mathbb Z)\;:\; n=3m .$$