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According to Wolfram Alpha, $(-1)^\pi \approx -0.90 + 0.43i$.

But $\pi$ has proven to be irrational (we can't write $\pi$ in terms of a fraction $a/b$ with $a$ and $b$ integers) and, in the real numbers, a negative number raised to a fraction with even denominator is undefined*.

So why is $(-1)^\pi$ a complex number, if there is no even denominator?

*Of course there are exceptions, like $(-1)^{4/2} = (-1)^2 = 1$ and $(-1)^{6/2} = (-1)^3 = -1$, but consider that the numerator is not a multiple of the denominator.

YuiTo Cheng
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enzo
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    Do you know how $(-1)^\pi$ is defined? –  Mar 09 '19 at 14:13
  • In the complex domain you don't have to have a rational exponent. But if you don't, you'looking as a function with infinitely many possible values. – Oscar Lanzi Mar 09 '19 at 14:15

2 Answers2

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This has to do with complex numbers. $$(-1)^{\pi}=\left(e^{i\pi}\right)^{\pi}=e^{i\pi^2}=\cos{\pi^2}+i\sin{\pi^2}$$ which is the value given by WolframAlpha.

However, since $-1=e^{(2n+1)\pi i}$ for every integer $n,$ there are actually infinitely many values of $(-1)^\pi.$ With complex numbers, exponentiation works differently from the way it does with real numbers.

J. W. Tanner
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saulspatz
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  • You mean $\cdots=\left(e^{i\pi}\right)^{\pi}=\cdots$ I guess. – drhab Mar 09 '19 at 14:23
  • @drhap Of course I do thanks. I struggled with the MathJax, and by the time I got it to accept anything, I wasn't looking too closely any more, I guess. – saulspatz Mar 09 '19 at 14:25
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    You need to justify that $(e^{i\pi})^\pi=e^{i\pi^2}$. In general, we don't have a rule $x^{yz}=(x^y)^z$ for complex numbers. – user1551 Mar 09 '19 at 22:14
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    @user1551 Even when $z$ is real? – saulspatz Mar 09 '19 at 22:47
  • The rule $(x^y)^z=x^{yz}$ fails even when $z$ is real. The problem is that neither $(x^y)^z$ nor $x^{yz}$ are unambigusously defined. In case $x>0$ and $z$ is real, both $x^y$ and $x^{yz}$ are well-defined, but $(x^y)^z$ is still not. In the OP's case, as you said, $(-1)^\pi$ can assume infinitely many values. In fact, $$(-1)^\pi=\exp(\pi\log(-1))=\exp(\pi(i\pi+2in\pi))=e^{i\pi^2}e^{2in\pi^2}$$ but the value of $e^{2in\pi^2}$ varies with $n$. You may see Marc van Leeuwen's answer in another thread about the applicability of the rule $(x^y)^z=x^{yz}$. – user1551 Mar 10 '19 at 05:36
  • @user1551 All I meant was that $e^{i\pi^2}$ is one of the possible values of $(-1)^\pi.$ – saulspatz Mar 10 '19 at 08:33
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For complex numbers, we define $a^b = \exp(b \log a)$. Of course $\log$ is "multivalued", so $a^b$ is also multivalued. Even $(-1)^{1/3}$ has three values, two of them are non-real. In case of $(-1)^\pi$, there are infinitely many values, all non-real.

Wolfram has a system of choosing a "principal value" in these cases, and it uses the logarithm with imaginary part in $(-\pi,\pi]$. So the principal value of $\log(-1)$ is $i\pi$. And the principal value of $(-1)^\pi$ is $\exp(\pi\log(-1)) = \exp(i\pi^2) = \cos(\pi^2)+i\sin(\pi^2)\approx -0.90 - i 0.43$.

GEdgar
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