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 '14 at 1:40
  • 5
    $\begingroup$ Indeed, it does not hold in general even for rational numbers. Consider the nonsense expression $-1 = (-1)^{1} = ((-1)^{2})^{1/2} = (1)^{1/2} = 1$ $\endgroup$ – Alex Wertheim Apr 28 '14 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 '14 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 '14 at 3:08
  • $\begingroup$ Nahin does offer a solution in the Appendix D, at least in my 2016 paperback edition. $\endgroup$ – gilbertohasnofb Sep 24 '16 at 23:34

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.