3

Given a complex number $e^{ix}$, the nth root can be computed using Euler's formula:

$$e^{i(x + 2k\pi)/n} = \cos((x+2k\pi)/n)+i \sin((x+2k \pi)/n).$$

If $n$ is an irrational number, can $k$ vary from negative infinity to infinity without repeating any solutions?

If so are ALL the points on the complex circle part of the solution?

This might be a really easy question to answer, but I'm a bit stuck. Also I made it up.

Gerry Myerson
  • 179,216
tjent
  • 31
  • 1
    Yes to the first question, and no to the second question. For example, let $x = 2\pi$, so $e^{ix} = 1$. You cannot get $-1$ for any $k$ if $n$ is irrational. – Tunococ Aug 02 '14 at 03:57
  • Thank you, good example. I knew if i could just think of a single counter example then id be good. I'd like to see a bit more about the solutions never over lapping though. – tjent Aug 02 '14 at 04:13
  • 1
    If two points coincide, the difference in their angles must be a multiple of $2\pi$, i.e., $k - k' = 2mn$ where $k$ and $k'$ are the two indices and $m$ is an integer. This contradicts irrationality of $n$. – Tunococ Aug 02 '14 at 04:18
  • A classic (related) exercise is showing that ${\sin(n)}_{n \in \mathbb{N}}$ is a dense subset of $[-1,1]$. – Andrew Maurer Aug 02 '14 at 04:45

0 Answers0