# Problematic Proof? (II)

Algebra Level 5

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. $$1=e^{2ki\pi}$$ for any integer $$k$$ so $$z^{r}=e^{2ki\pi}$$
2. $$z=e^{2ki\pi/r}$$
3. This argument will never be identical to a previous one as $$k_{1}/r\neq k_{2}/r\text{ mod 1}$$ for $$k_{1}\neq k_{2}$$
4. 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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