-2

We know from Euler's identity that $$e^{\pi i} = -1.$$ So that means $$(e^{\pi i})^{-i} = e^{\pi} = (-1)^{-i} = ((-1)^{-1})^{i} = (-1)^{i}.$$ But also we have $$ (-1)^{i} =( e^{\pi i})^{i} = e^{-\pi}$$ But how can this be, since $e^{\pi} \neq e^{-\pi}$?

Blue
  • 75,673
Ruby
  • 311
  • 1
    Well, this is exactly the reason why $z\to z^i$ (which first of all has to be defined of course) is not a function in general. Indeed, it returns infinitely many different values. This is why we have to define branches of logarithm to turn this into a function. – Mark Jul 21 '19 at 22:00
  • You can check the top answer here, it addresses the same issue – Paulo Mourão Jul 21 '19 at 22:01
  • Cf. this answer – J. W. Tanner Jul 21 '19 at 22:03
  • 1
    So many errors in your fake proof. I suppose you actually don't realize that you haven't been taught properly what exponentiation really is. Please read this very carefully, and then this. If you have started learning basic real analysis, then you should also read through this, but if you are only at high-school level then just make sure you know that every law applies only under certain conditions, and you must know exactly what those are. – user21820 Jul 22 '19 at 15:41
  • By the way, I myself was taught improperly too, and also did such nonsense with exponentiation when I was young. It should not be surprising that every mathematics student who is taught improperly is likely to do the same. – user21820 Jul 22 '19 at 15:43

1 Answers1

1

You are right about $(-1)^{-i} = (-1)^i$, but it doesn't mean $e^{\pi} = e^{-\pi}$. Please consider the following case as an analogy: $$ (-2)^2 = 4 = (+2)^2, $$ but it does not mean $-2 = +2$. In essence, your equations state $$ (e^{-\pi})^i = (e^{\pi})^i, $$ but it does not imply $e^{-\pi} = e^{\pi}$.

guest
  • 1,764