# $\bigcup_{n=1}^{\infty}\{z\in\Bbb C:z^n=1\}=\{z\in\Bbb C:|z|=1\}$?

Is $$\bigcup_{n=1}^{\infty}\{z\in\Bbb C:z^n=1\}=\{z\in\Bbb C:|z|=1\}$$my argument is that the argument of the elements of the first set are rational multiples of $$\pi$$ whereas the second set also consists of elements with irrational multiples of $$\pi$$. But I am not so sure, there is still some doubt.

Those two sets are certainly not equal.

$$\displaystyle\bigcup_{n=1}^{\infty}\{z\,|\,z^n=1,n\in \mathbb N\}$$ is countable as it is a countable union of finite sets, while $$\{z/|z|=1\}$$ has the cardinality of the continuum.

• What does "power of a continuum" mean? – Karthik Feb 1 at 15:21
• I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $\mathbb R$. – mathcounterexamples.net Feb 1 at 15:23
• "power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I). – Paul Sinclair Feb 1 at 17:22
• @PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste). – mathcounterexamples.net Feb 1 at 17:35
• This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow. – drjpizzle Feb 1 at 18:22

For example, $$e^{\sqrt{2} \pi i}$$ is not in the first set.