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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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    I took the liberty of doing an edit that actually changed the meaning of your question, in the middle formula. It made no sense as it stood, but it does now, I think.2012-08-04
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    well, the reason for the joke is this does not make sense.2012-08-04
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    Do you want to undo my edit? You certainly may do so, if you insist.2012-08-04
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    You actually show that 1 is equal to infinitely many other numbers (if the steps of the argument are correct)2012-08-04
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    well, we are here to please everyone.2012-08-04
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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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    It could be useful to recall these conditions, or give a reference.2012-08-04
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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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    Would be useful to clarify why this doesn't mean complex analysis is an inconsistent theory.2012-08-04
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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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    Euler's formula also holds for complex $x$. In fact, it is a great trick to *define* cos and sin for arbitrary arguments.2012-08-04
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    Oh! so $e*e^{ix} = e^{ix + 1}$ is true?2012-08-04
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    @Ankush: Yes. But please, please don't use the asterisk $*$ for multiplication! In mathematics, it usually means convolution. It's an abomination coming from programming languages, where it arose due to the lack of a more appropriate symbol in ASCII (or EBCDIC, or …). Use `\cdot` instead: $e\cdot e^{ix}$.2012-08-04
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    ok :) I'm new to writing equations in $2012-08-04
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    @Ankush: it is called LaTeX.2012-08-04
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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