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We have $e^{2\pi i n}=1$

So we have $e^{2\pi in+1}=e$

which implies $(e^{2\pi in+1})^{2\pi in+1}=e^{2\pi in+1}=e$ Thus we have $e^{-4\pi^{2}n^{2}+4\pi in+1}=e$

This implies $e^{-4\pi^{2}n^{2}}=1$

Taking the limit when $n\rightarrow \infty$ gives $0=1$.

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    This puzzle is due to [Thomas Clausen](http://en.wikipedia.org/wiki/Thomas_Clausen_(mathematician)). Have you perhaps found it in Nahin's "Story of $\sqrt{-1}$"?2012-08-04

3 Answers 3

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Your error is (as in most of those fake-proofs) in the step where you use the power law $(a^b)^c=a^{bc}$ without the conditions of that power law being fulfilled.

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    But such conditions are not easy to produce, except for some special cases: For example, having $a$ and $b$ real (with $a$ nonnegative) is sufficient, unless you base your powers on a highly unusual branch of the natural log function.2012-08-04
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The proof is wrong because an expression of the form $x^y$ is actually ambiguous, when $x$ is a complex number: Rewrite it as $e^{y\ln x}$ and note the multivalued nature of the natural logarithm as used on complex numbers. For your proof to be correct, you would need $\ln e^{2\pi in+1}=2\pi in+1$, but that is not consistent with $\ln e=1$.

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    @Auke: No. That would lead to a paraphrase of half a chapter's worth (or more!) of some complex analysis book, talking about branch points, Riemann surfaces, and I don't know what. This is not the purpose of this site. Anyway, the point of all this is that the union of complex analysis and the blind application of certain rules learned from real analysis *is* an inconsistent theory.2012-08-04
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From wikipedia on Euler identity

The identity is a special case of Euler's formula from complex analysis, which states that $e^{i x}=\cos x+i\sin x$. for any real number $x$.

Note $x$ should be real number.

$e e^{i x} \neq e^{i x + 1} = e^{i(x-i)} = \text{undefined} $

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    @Auke I know that. But never got a chance to write equations into TeX system. Only used for resume n final report purpose :)2012-08-04