How this $ \exp(i 10 \pi)^{\frac56}=\exp(\frac{i \pi}{3})$ true in the below paper ? And how is de Moivre's formula applied for non integer exponent?

Question

I discussed with my friend the following problem which is related to this paper which is written by Curtis D. Bennett, A. M. W. Glass and Gábor J. Székely under this title Fermat's Last Theorem for Rational Exponents in Monthly American journal , Really in the introduction Authors wrote for the identity :$1^{5/6}+1^{5/6}=1^{5/6}$ , They wrote :$ \exp(i 10 \pi)^{\frac56}=\exp(\frac{i \pi}{3})$ , How this true and how $10$ comes ? and how is de Moivre's formula applied here for rational number for non integer exponent?? I did many attempt to let de Moivre's formula applied for integers but I failed ?

Answer 1

The De Moivre formula only works with integers, otherwise you would have $$ \forall\theta\in\mathbb{R},e^{i\theta}=(e^{2i\pi})^{\frac{\theta}{2\pi}}=1^{\frac{\theta}{2\pi}}=1 $$ Moreover, the definition of $1^{5/6}$ is (for me) $1^{5/6}:=e^{\frac{5}{6}\ln 1}$ so that $1^{5/6}=1$ and thus $1^{5/6}+1^{5/6}=2\neq 1^{5/6}$.

Answer 2

$\exp(i 10 \pi)^{5/6} = \exp(i \pi 50/6) = \exp(i \pi 8) . \exp (i \pi /3) = \exp (i \pi /3)$

Indeed $1^{5/6}$ shall refer a to real root of 1. But I think this is why the authors write "new" solution (in quotation mark).

Answer 3

In $\mathbb C$, if we say $\zeta=1^{1/6}$ means $\zeta^6=1$, there are $6$ possibilities for $\zeta$:

$\exp(2\pi i/6), \exp(4\pi i/6), \exp(6\pi i/6)=-1, \exp(8\pi i/6), \exp(10\pi i/6),$ and $ \exp(12\pi i/6)=1$,

and thus $6$ possibilities for $\zeta^5$:

$\exp(10\pi i/6)=\color{green}{\exp(5\pi i/3)}, \exp(20\pi i/6), \exp(30\pi i/6)=-1, \exp(40\pi i/6), $

$\exp(50\pi i/6)=\color{brown}{\exp(\pi i/3)}$, and $ \exp(60\pi i/6)=\color{blue}1$.

Now $\color{green}{\exp(5\pi i/3)}+\color{brown}{\exp(\pi i/3)}=\color{blue}1,$

so we can have $\color{green}{1^{5/6}}+\color{brown}{1^{5/6}}=\color{blue}{1^{5/6}}$ by making different choices for $1^{5/6}$.

Answer 4

we know that $\exp(2πi)=\exp(10πi)$. so if Moivre formula works with rational number we get: $ \exp(2πi)^{5/6}=\exp(5πi/3)=\exp(-iπ/3)$ and: $\exp(10πi)^{5/6}=\exp(50πi/6)=\exp(iπ/3)$ contradiction.