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 $0$.
 A: Hint: Imaginary part of $\frac {a+ib} {c+id}$ equals imaginary part of $\frac {(a+ib) (c-id)} {|c+id|^{2}}$ which is $\frac {bc-ad} {c^{2}+d^{2}}$ and this is $0$ iff $ad=bc$. 
A: We have $\sqrt{3}+i=2e^{i\pi/6}$ and $\sqrt{3}-i=2e^{-i\pi/6}$. So
\begin{eqnarray*}
z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m} = 2^{n-m} e^{ i \pi (n+m) /6}.
\end{eqnarray*}
So we require $ n+m  \equiv 0 \pmod{6}$.
A: Hint:
Use the complex exponential notation and congruences: your final fraction is none other than
$$z=2^{n-m}\frac{\mathrm e^{\tfrac{ni\pi}6}}{\mathrm e^{\tfrac{-mi\pi}6}}=2^{n-m} \mathrm e^{\tfrac{(n+m)i\pi}6},$$
anf it is a real number if and only if
$$\frac{(n+m) \pi}6\equiv 0\mod \pi\iff (n+m)\pi \equiv 0\mod 6\pi\iff n+m\equiv 0\mod 6.$$
A: $\sqrt {3} - i$ is the conjugate to $\sqrt {3} + i$
$\frac {(\sqrt 3 + i)^n}{(\sqrt 3 - i)^m} = \frac {(\sqrt 3 + i)^{n+m}}{(\sqrt 3 - i)^m(\sqrt 3 + i)^m}$
This will rationalize the denominator
$\frac {(\sqrt 3 + i)^n}{(\sqrt 3 - i)^m} = \frac {(\sqrt 3 + i)^{n+m}}{4^{m}}$
We must find where $(\sqrt 3 + i)^{n+m}$ is real
$\sqrt 3 + i = 2(\cos \frac {\pi}{6} + i\sin \frac {\pi}{6})\\
(\sqrt 3 + i)^{n+m} = 2^{n+m}(\cos \frac {\pi}{6} + i\sin \frac {\pi}{6})^{n+m}\\
(\sqrt 3 + i)^{n+m} = 2^{n+m}(\cos \frac {(n+m)\pi}{6} + i\sin \frac {(n+m)\pi}{6})$
if $(\sqrt 3 + i)^{n+m}$ is real $\sin \frac {(n+m)\pi}{6} = 0$
$\frac {(n+m)\pi}{6} = k\pi\\
n+m = 6k$
$6$ divides $n+m$
A: You may continue as follows,
$$z = 2^{n - m} \cdot \dfrac{\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}}
{\cos \dfrac{11 m \pi}{6} + i \sin \dfrac{11 m \pi}{6}}$$
$$=2^{n - m} \cdot \dfrac{\left(\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}\right)
\left(\cos \dfrac{11m \pi}{6} - i \sin \dfrac{11m \pi}{6}\right)}
{\cos^2 \dfrac{11 m \pi}{6} + \sin^2 \dfrac{11 m \pi}{6}}$$
Then, set the the imaginary part of the numerator to zero,
$$I=\sin \dfrac{n \pi}{6}\cos \dfrac{11m \pi}{6} - \cos \dfrac{n \pi}{6}\sin \dfrac{11m \pi}{6} = -\sin\dfrac{(11m -n)\pi }{6}=\sin\dfrac{(m +n)\pi }{6}=0 $$
which leads to $\dfrac{(m+n)\pi }{6}= k\pi$. Thus, the relationship between $m$ and $n$ is
$$m+n=6k$$
with $k=0,1,2,...$
A: 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 $n+m \equiv 0 \bmod 6$.
