# Is $(-1)^x$ multivalued?

$$(-1)^\pi=(e^{i\pi})^\pi=e^{i\pi^2}=\cos(\pi^2)+i\sin(\pi^2)$$. 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?

• 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 – Displayname Mar 14 at 23:08
• 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 – Displayname Mar 14 at 23:15
• @Displayname so it is multivalued then? – Benjamin Thoburn Mar 14 at 23:33

$$(-1)^x=e^{(1+2n)\pi ix}$$ which has infinite possibilities for all irrational $$x$$.
• Further note. If x is a fraction in lowest terms, it will have a finite number $\gt 1$ possible solutions. – herb steinberg Mar 16 at 22:16