It is well known that \(z^{n}=1\) (where n is an integer) has \(n\) solutions evenly distributed around the unit circle. These are known as the \(n^\text{th}\) roots of unity.

What about when \(n\) is not an integer? What if \(n\) is irrational? Here is a proof that \(z^{r}=1\) has an infinite number of solutions when \(r\) is irrational. There is also a conclusion as to what that means. Can you spot a flaw?

**Step:**

- \(1=e^{2ki\pi}\) for any integer \(k\) so \(z^{r}=e^{2ki\pi}\)
- \(z=e^{2ki\pi/r}\)
- This argument will never be identical to a previous one as \(k_{1}/r\neq k_{2}/r\text{ mod 1}\) for \(k_{1}\neq k_{2}\)
- Since the number \(z\) could have an infinite number of distinct arguments there must be an infinite number of numbers that satisfy the equation.

**Conclusion:**

Every number of magnitude 1 is a solution for \(z\) to the equation \(z^{r}=1\), \(r\) is irrational.

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