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Hello i have a function:

$$ e^x = 1 $$

Is it true that this will only hold for $x\!=\! N 2\pi i$ where $N\!=\!1,2,3\dots$. Why is that? Does this has something to do with an Euler identity?

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closed as too localized by 23rd, Raskolnikov, Amzoti, Alex Wertheim, J. M. Jun 6 '13 at 7:36

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Well, $N$ can also be $0$ or a negative integer as well. – Andrew Salmon Jun 6 '13 at 5:57
up vote 1 down vote accepted

Euler's notation states $$e^{it} = \cos (t) + i \sin(t)$$ However, this is only a definition. (A very good definition, since it meshes very well with the real exponential function, but a definition nonetheless.)

But your question asked why the equality held? Simply by plugging in the numbers $$e^{2\pi i N} = \cos(2 \pi i N) + i \sin (2 \pi i N) = 1 + 0$$ (and as someone else pointed out, this holds for any integer $N$, not just positive ones)

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No. Rather, it will hold for $x=N2\pi i$, where $N$ is any integer. There are numerous questions on this site discussing this. (Just search Euler identity.)

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$e^{ix}=\cos(x)+i\sin(x).$ Consider $\cos(x)+i\sin(x)$ as a 2D vector $[\cos(x), \sin(x)]$. Then its magnitude is 1. $e^{ix+y}=e^y(\cos(x)+i\sin(x)).$ If $y$ is real and not 0, the corresponding vector $[e^y\cos(x), e^y\sin(x)]$ has a different magnitude than 1.

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Consider using \sin and \cos: $\sin\ \cos$ to have better typesetting compared to $sin \ cos$. – Lord_Farin Jun 6 '13 at 6:57
@Lord_Farin Thanks. I'll remember that if I can. – Loki Clock Jun 6 '13 at 6:59

$\forall n\in \mathbb Z$: $e^{i2\pi n}= \cos(2\pi n)+i\sin(2\pi n)=1+i0=1$

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As i ve thought an Euler identity :) – 71GA Jun 6 '13 at 6:02
Consider using \sin and \cos: $\sin\ \cos$ to have better typesetting compared to $sin \ cos$. – Lord_Farin Jun 6 '13 at 7:00
Lord_Farin :thanks farin – Maisam Hedyelloo Jun 6 '13 at 7:01

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