The following false "proof" is attributed to Thomas Clausen in 1827, and was stated on page 79 of Nahin's An Imaginary Tale.

$e^{i2\pi n}=1$ for all integers $n$. So

\begin{align*} ee^{i2\pi n}=e&=e^{1+i2\pi n}\\ &=\left(e^{1+i2\pi n}\right)^{1+i2\pi n}\\ &=e^{(1+i2\pi n)^2}=e^{1+i4\pi n-4\pi^2n^2}\\ &=e^{1+i4\pi n}e^{-4\pi^2n^2} \end{align*} But $e^{1+i4\pi n}=e$, therefore $e^{-4\pi^2n^2}=1$. But that last equation is only true for $n=0$. We started with a statement true for all integers $n$, and through a series of (apparently) valid steps ended with a statement true only for $0$. Therefore all integers are $0$. Where is the mistake?

  • 11
    $\begingroup$ The identity $(e^a)^b=e^{ab}$ does not hold in general for complex numbers. Not that this explains anything, but that is the error in the proof. $\endgroup$
    – user142299
    Apr 28, 2014 at 1:40
  • 6
    $\begingroup$ Indeed, it does not hold in general even for rational numbers. Consider the nonsense expression $-1 = (-1)^{1} = ((-1)^{2})^{1/2} = (1)^{1/2} = 1$ $\endgroup$ Apr 28, 2014 at 1:41
  • $\begingroup$ Now that we have an answer in a comment, tell us: What did Clausen say about it? $\endgroup$
    – GEdgar
    Apr 28, 2014 at 2:58
  • $\begingroup$ As far as I know, Clausen himself never offered a resolution, and Nahin doesn't either. You can find a (German) publication of his work containing this in the following citation: Thomas Clausen, Aufgabe 53, J. Reine Angew. Math. 2 (1827), 286–287. $\endgroup$
    – ant11
    Apr 28, 2014 at 3:08
  • $\begingroup$ Nahin does offer a solution in the Appendix D, at least in my 2016 paperback edition. $\endgroup$ Sep 24, 2016 at 23:34

1 Answer 1


See Failure of power and logarithm identities and Complex exponents with positive real bases 2 which explains this exact problem. Also consider the related problem of $e^{-4\pi^2n^2}=\left(e^{i2\pi n}\right)^{i2\pi n}=1^{i2\pi n}=1$, which forgets that, in this scheme, the power of 1 is involves a multivalued log of 1 in an exponent.


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