# Number of roots of unity on the unit circle

I have to find the number of roots of unity on the unit circle |z|=1 in the argand plane.

I know that there are n roots of the the equation $$z^n=1$$ and all of them lie on the given circle. Does that mean that there are infinite roots of unity on this circle (as the sum 1+2+3..... diverges to infinity) or am I missing something?

• I think you're right. This is the torsion subgroup of the circle group.
– user403337
Commented Nov 5, 2018 at 18:35
• @ChrisCuster Sorry, I am a high school student and have no idea of groups. Can you dumb it down a bit for me? Thanks Commented Nov 5, 2018 at 18:44
• A group is one of the objects, together with rings and fields, studied in Abstract Algebra. You needn't worry too much about them at this point.
– user403337
Commented Nov 5, 2018 at 18:59

If $$m$$ is a multiple of $$n$$, then all the roots of $$z^n = 1$$ are also roots of $$z^m = 1$$. So counting the roots is not a simple matter of adding up $$1 + 2 + 3 + ...$$, as you are counting the repeats.

However, for any $$m$$, the first root of $$z^m = 1$$ encountered while moving counter-clockwise around the circle from $$1$$ is a "primitive" root. That is, it is not a root of $$z^k = 1$$ for any $$k < m$$. Thus the mapping $$m \mapsto \cos \frac {2\pi}m + i\sin \frac{2\pi}m$$ assigns a unique root of unity to each positive integer $$m$$.

Thus, there are at least as many roots of unity as there are positive integers. That is, an infinite number.

Another argument, similar to your accounting, shows that there are at least as many positive integers as there are roots of unity. Putting the two results together*, there must be exactly as many roots of unity as there are positive integers, which means that the set of all roots of unity is countably infinite.

(There are infinite sets that cannot be put in such a correspondence with the positive integers. Such sets are called uncountable. The set of real numbers is one example, as is the set of all points on the unit circle.)

$$^*$$ By applying the Schröder-Bernstein theorem.

• Why the need to apply Schröder-Bernstein theorem? Isn't $\{z: z^n=1\mbox{ for some } n\ge 1\}$ a countable union of finite sets of the form $\{z: z^n=1\mbox{ for fixed } n\ge 1\}$ & so countable ?
– user10575
Commented Oct 20, 2021 at 12:27
• @Shahab - I was talking to someone with no concept of countability and its properties. Thus I used an intuitively obvious argument instead of one depending on things my audience was unaware of. But since I knew that the "intuitively obvious" statement was actually quite difficult to prove, I chose to acknowledge it by explicitly mentioning the theorem where it was demonstrated in a footnote. Commented Oct 20, 2021 at 12:42