If $z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m}$, find the relation between $m$ and $n$ such that $z$ is a real number. I am given the following number $z$:
$$z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m}$$
with $n, m \in \mathbb{N}$. I have to find a relation between the natural numbers $n$ and $m$ such that the number $z$ is real. I know that for a complex number to be real, its imaginary part must equal $0$, but I can't isolate the imaginary part. This is as far as I got:
$$\sqrt{3} + i =
2 \bigg (\frac{\sqrt{3}}{2} + i\frac{1}{2} \bigg) =
2 \bigg( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) $$
$$\sqrt{3} - 1 =
2 \bigg ( \dfrac{\sqrt{3}}{2} - i \dfrac{1}{2} \bigg) = 
2 \bigg ( \cos \dfrac{\pi}{6} - i \sin \dfrac{\pi}{6} \bigg ) =  
2 \bigg ( \cos \dfrac{11\pi}{6} + i \sin \dfrac{11\pi}{6} \bigg )$$
So I got the numerator and the denominator in a form that I can use DeMoivre's formula on. So, next I'd have:
$$z = \dfrac{\bigg [2 \bigg ( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) \bigg ]^n}
{\bigg [2 \bigg( \cos \dfrac{11 \pi}{6} + i \sin \dfrac{11 \pi}{6} \bigg ) \bigg ]^m }$$
$$z = 2^{n - m} \cdot \dfrac{\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}}
{\cos \dfrac{11 m \pi}{6} + i \sin \dfrac{11 m \pi}{6}}$$
But this is where I got stuck. I still can't isolate the imaginary part of $z$ in order to equal it to $0$.

The key insight is to recognize roots of unity in the expression.
We have
$$
z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m} = \frac{(2\omega)^n}{(2\omega^5)^m} = 2^{n-5m}\omega^{n-5m}
$$
where $\omega^6=-1$. Therefore, we need $n-5m \equiv 0 \bmod 6$, or $n+m \equiv 0 \bmod 6$.
