# Find the smallest $n\in\mathbb N_{\geq1}$, such that $\left(\frac {z_1}{z_2}\right)^n\in\mathbb R$

I've been trying to solve this problem in multiple ways but can't seem to find a reasonable solution.

We've got two complex numbers: $z_1 = a+bi$ and $z_2 = c+di$. Say $\lambda = \frac{z_1}{z_2}$. What's the smallest value of $n \in\mathbb N_{\geq 1}$ such that $\lambda^n \in\mathbb R$?

Anyone got an idea?

• $n=0\in\mathbb N_0$ is always the smallest such $n$. If we don't want $n=0$, then we need more conditions on $z_1$ and $z_2$, because in the most general case it is not always possible to find a natural $n$ such that $\lambda^n\in\mathbb R$. – gebruiker Apr 17 '16 at 9:37
• Yeah, I meant that n is not 0. I'll edit my question to make that clearer. – Martez Apr 17 '16 at 9:39
• That means we need to know what we conclude when no such $n\in\mathbb N$ exists... – gebruiker Apr 17 '16 at 9:41
• Before I forget: Welcome to MSE. As you can see I made an edit to your question. You already seem to be familiar with MatJaX, but there were still a few things that could be improved. If you are ever unsure how to typeset your maths on this site, you can check out this quick reference. Also, I made your title more informative. More on this can be found here. – gebruiker Apr 17 '16 at 9:49
• As gebruiker says, no such $n$ needs to exist, for example you might have $\frac {z_1}{z_2} = \cos(\pi\sqrt 2) + i\sin(\pi\sqrt 2)$. – Ennar Apr 17 '16 at 9:50

Let $z = \frac{z_1}{z_2} = \cos\alpha + i\sin\alpha$ (norm is irrelevant, so we will assume $|z| = 1$). Then, there exists $n\in\mathbb N$ such that $z^n = \cos n\alpha + i\sin n\alpha$ is in $\mathbb R$ if and only if there exists $n\in\mathbb N$ such that $n\alpha\in \pi\mathbb Z$, i.e. if and only if $\alpha\in\pi\mathbb Q$.
Thus, find $\frac ab\in\mathbb Q$ such that $a$ and $b$ are relatively prime and $\alpha =\frac ab\pi$ and set $n = b$ for minimal $n$.