$(e^{2\pi i}){}^n \neq e^{2\pi i n}$ where $n\in\mathbb{N}$?

Question

When I type these equations into a calculator I get $({e^{2\pi i}}){}^n = 1$ and something else for $e^{2\pi i n}$. Is that due to the imprecision of the calculator or does the inequality follow logically?

$;e^{2\pi in}=1;$ for any $;n\in\Bbb Z;$ . If your calculator says otherwise change calculator. — DonAntonio, Jan 27 '19 at 10:42
this is for n = 5, asking Google (the url is too long too copy but you can type "(e^(2pi i*5))" into the Google search box). Anyways, thank you for your answers, I guess the Google calculator is not designed for this kind of stuff. — jvdh, Jan 27 '19 at 10:44
What kind of «calculator» are you using? The result should 1 for all $n$ — Thomas Lesgourgues, Jan 27 '19 at 10:44
There are only rounding erros, Google calculator is not meant for these calculation. You should use a formal one — Thomas Lesgourgues, Jan 27 '19 at 10:47

score 3 · Accepted Answer · answered Jan 27 '19 at 10:46

Since $e^{ix}=\cos x+i\sin x$, $e^{ix}=1$ iff $\cos x=1\land\sin x=0$, i.e. iff $2\pi|x$. Thus $(e^{2\pi i})^n=1^n=1$, and $e^{2\pi in}=\cos 2\pi n+i\sin 2\pi n$, which is $1$ if $2\pi |2\pi n$ or equivalently $n\in\Bbb Z$. After a bit of fiddling with your URL, I found Google calculating $e^{2\pi i\times 5}$ as $1-1.2246468\times 10^{-15}i$. Bear in mind computers "think" in rational approximations, and since $\pi$ is irrational it's easy for a multiple of $2\pi$ to seem a little off when its cosine and sine are calculated.

$(e^{2\pi i}){}^n \neq e^{2\pi i n}$ where $n\in\mathbb{N}$?

1 Answers1