1

If $\operatorname{cis}(2\pi) = \operatorname{cis}(4\pi)$, then don't we have $$\big(\operatorname{cis}(2\pi)\big)^{1/5} = \big(\operatorname{cis}(4\pi)\big)^{1/5}?$$

This isn't yielding the same answer, even though I'm raising two equal complex numbers to the power $1/5$ (or for that matter any number?).

gebruiker
  • 6,154
Dieblitzen
  • 129
  • 4
    $z^w$ is not single-valued unless $w$ is an integer. – mrf May 19 '16 at 12:58
  • 2
    The five fifth-roots lie at the vertices of a regular pentagon. As you successively raise cis $ \frac{2 \pi}{5} \ $ or cis $ \frac{4 \pi}{5} \ $ to higher (integer) powers, on the left-hand side of your "equation", you encounter each root one after another, while on the right-hand side, you skip every other root. So in this "cycling", you only "meet up" at $ \ z \ = \ 1 \ $ again after raising each of these roots to the fifth power. (Something similar would happen for any prime-number root.) – colormegone May 19 '16 at 13:12
  • "This isn't yielding the same answer" Meaning, when using some kind of program? Which one? – Did May 19 '16 at 13:33
  • 1
    Related (actually, probably duplicate): http://math.stackexchange.com/q/415807/ – Did May 19 '16 at 13:41

1 Answers1

1

The concept of "raising to the power $1/n$" is not well defined in the complex numbers: generally there are $n$ candidates for what "$z^{1/n}\,$" might mean. The five possible values of the left-hand side of your equation are just the same as the five possible values of the right-hand side, namely the five complex roots of $z^5=1$. In the real numbers, it is possible to assign a unique real value to $x^{1/5}$, and the notation is therefore not problematic for real $x$.

John Bentin
  • 18,454
  • What I'm doing is I'm equating two complex numbers, but when I raise them to 1/n, the complex numbers aren't turning out to be the same anymore The equation in my question was just an example. Am I missing some law regarding raising complex numbers in polar form to some power? Do the laws of indices still work? – Dieblitzen May 19 '16 at 13:52
  • @Dieblitzen: I hope that my expanded explanation makes this clear now. – John Bentin May 19 '16 at 14:14
  • Oh right I realise it now. The 5 values are equal, I just compared the immediate answer using De Moivre's. Thanks – Dieblitzen May 19 '16 at 14:57