$(-1)^\pi=(e^{i\pi})^\pi=e^{i\pi^2}=\cos(\pi^2)+i\sin(\pi^2)$. Wolfram Alpha lists this as the only answer. However it started with $e^{i\pi}=1$, although $e^{i\pi(2n+1)}=-1$ also for any integer n. By substituting this new expression for $-1$ and doing the same thing, you get more than one value, like $\cos(3\pi^2)+i\sin(3\pi^2)$. Is Wolfram or my reasoning wrong?

  • $\begingroup$ Anything is 'multivalued' if you work things out like this, e.g $1^{1/n}$ can take n values, namely the nth roots of untiy $\endgroup$ – Displayname Mar 14 at 23:08
  • $\begingroup$ manipulating/using Euler's formula has nothing to do with it really, it's just the fact you're raising a number to the power of something that's not an integer $\endgroup$ – Displayname Mar 14 at 23:15
  • $\begingroup$ @Displayname so it is multivalued then? $\endgroup$ – Benjamin Thoburn Mar 14 at 23:33

$(-1)^x=e^{(1+2n)\pi ix}$ which has infinite possibilities for all irrational $x$.

  • $\begingroup$ Further note. If x is a fraction in lowest terms, it will have a finite number $\gt 1$ possible solutions. $\endgroup$ – herb steinberg Mar 16 at 22:16

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