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?

  • $\begingroup$ $n=0\in\mathbb N_0$ 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 $\lambda^n\in\mathbb R$. $\endgroup$ – gebruiker Apr 17 '16 at 9:37
  • $\begingroup$ Yeah, I meant that n is not 0. I'll edit my question to make that clearer. $\endgroup$ – Martez Apr 17 '16 at 9:39
  • $\begingroup$ That means we need to know what we conclude when no such $n\in\mathbb N$ exists... $\endgroup$ – gebruiker Apr 17 '16 at 9:41
  • $\begingroup$ 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. $\endgroup$ – gebruiker Apr 17 '16 at 9:49
  • $\begingroup$ As gebruiker says, no such $n$ needs to exist, for example you might have $\frac {z_1}{z_2} = \cos(\pi\sqrt 2) + i\sin(\pi\sqrt 2)$. $\endgroup$ – 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 $n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.