3
$\begingroup$

If $a,b\in\mathbb{Z}$, and $\frac a b$ is in lowest terms, then $$z^{\frac a b}=1\\\implies z=\exp\left(\frac{2\pi in b}{a}\right)\forall n\in\mathbb{Z}$$ This means that $z$ has exactly $a$ distinct values on the unit circle, as $n\pm a$ gives the same $z$ value as $n$, as the exponent is rational. However, what if the exponent is irrational? If we have $$z^{r}=1,r\notin\mathbb{Q}\\\implies z=\exp\left(\frac{2\pi in }{r}\right)\forall n\in\mathbb{Z}$$ then as this does not loop as with a rational number, does $z$ cover the entire unit circle in the complex plane? Does it cover a dense set of the unit circle? Or is there something wrong with taking irrational roots of unity?

$\endgroup$
  • $\begingroup$ $z^r = \exp(\log(z) r)$. The problem here is that the logarithm of a complex number is a branched function. For different branches you will get different solutions. But the only branch valid for solving the equation is the principal branch. $\endgroup$ – mathreadler Feb 14 '15 at 8:38
  • $\begingroup$ I.e. there's a misconception that $\forall n \in \mathbb{Z}$ is how you solve the original equation. The solutions just happen to overlap in that case. $\endgroup$ – mathreadler Feb 14 '15 at 8:43
  • $\begingroup$ The problem is: what do you mean by $z^{\frac ab}$ and $z^r$? How you define these powers? $\endgroup$ – Fabio Lucchini Feb 14 '15 at 10:53
2
$\begingroup$

Let me bring you an example

$$z^\pi=1$$

I guess an obvious solution which is

$$z_0=e^{\frac{2\pi i}\pi}=e^{2i}=\cos(2)+i \sin(2)$$

In addition, any other number which is natural power of $e^{2i}$ must match the equation. So, $e^{2in}$ where $n\in N$ is a solution. How do you imagine locus of $e^{2i n}$? It covers infinite points of unit circle but not all points of them.

As an example, is $1\angle \frac{\pi}6$ a solution? then you need to find out if there is a $n$ where $$n \times 2 =\frac{\pi}6+2 k\pi$$

So, it has infinite solution on unit circle while they do not cover the whole points of unit circle(Although they are infinite, still they are a very small minority since they are countable).

$\endgroup$
0
$\begingroup$

The complex solutions of $z^r=1$ are $$ \exp((2\pi i n)/r),\qquad n \in \mathbb Z $$ as you said. If $r$ is not rational (for example, if $r$ is not real) then these are all distinct. But there are only countably many of them. If $r$ is real but irrational, these values are all on the unit circle. Countably many of them, dense in the circle, but not the whole circle.

$\endgroup$

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.