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 $0$.
Best Answer
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 $ad=bc$.