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

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?

• Can you fix your typos please? – Dr. Sonnhard Graubner Jan 27 at 10:40
• $\;e^{2\pi in}=1\;$ for any $\;n\in\Bbb Z\;$ . If your calculator says otherwise change calculator. – DonAntonio Jan 27 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 at 10:44
• What kind of «calculator» are you using? The result should 1 for all $n$ – Thomas Lesgourgues Jan 27 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 at 10:47

## 1 Answer

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.