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?

  • $\begingroup$ Can you fix your typos please? $\endgroup$ – Dr. Sonnhard Graubner Jan 27 at 10:40
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    $\begingroup$ $\;e^{2\pi in}=1\;$ for any $\;n\in\Bbb Z\;$ . If your calculator says otherwise change calculator. $\endgroup$ – DonAntonio Jan 27 at 10:42
  • $\begingroup$ 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. $\endgroup$ – jvdh Jan 27 at 10:44
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    $\begingroup$ What kind of «calculator» are you using? The result should 1 for all $n$ $\endgroup$ – Thomas Lesgourgues Jan 27 at 10:44
  • $\begingroup$ There are only rounding erros, Google calculator is not meant for these calculation. You should use a formal one $\endgroup$ – Thomas Lesgourgues Jan 27 at 10:47

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.


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