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We know that $(-1)^{\frac 12}$, $(-1)^{\frac 14}$, ... $(-1)^{\frac 1{even}}$ all result in complex answers, whereas $(-1)^{\frac 13}$, $(-1)^{\frac 15}$, ... $(-1)^{\frac 1{odd}}$ all result in real answers. I believe this comes from the definitions that $(-1)^{\frac 12} = \sqrt{-1} = i$ and $(-1)^{\frac 13}=\sqrt[3]{-1}=-1$. Moreover, of the set that provides real answers, if the numerator of the exponent is even, the result then becomes $+1$ instead of $-1$. This explains why the graph $y = (-1)^x$ results in an infinite amount of discontinuous points on $y=1$ and $y=-1$, only existing where the denominator of the exponent is odd:

graph of negative one to the x

But the sets I provided only explain $(-1)^{rational}$; in other words, this assumes that we are only raising $-1$ to a fraction. When I plug in $(-1)^e$, $(-1)^{\pi}$, $(-1)^{\sqrt{2}}$, or any other irrational number, the answer comes back undefined/imaginary.

negative one to irrational powers resulting in undefined

This cannot come from the explanation that I provided earlier because, even though it is the fundamental explanation for where imaginary results come from, it assumes the exponent is a fraction. My intuition is that there would be some other rule that would sometimes result in an imaginary answer and sometimes result in a real number - similar to how rational exponents work - but it seems to always be imaginary. What, then, is the rule for this, and can we prove that this is the case for all irrational numbers?

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    $\begingroup$ $x^n$ is defined as $\exp(n \log (x))$ for $x$ real valued. When $x$ is negative, and $n$ is non-integral, the (complex) logarithm of $x$ is multiply valued, hence why $-1^{1/4}$ has $4$ values, $\pm i, \pm -1$. $\endgroup$ – rubikscube09 Mar 11 at 2:46
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    $\begingroup$ I believe you'll find the answers to a similar MSE question at Non-integer powers of negative numbers to be useful. $\endgroup$ – John Omielan Mar 11 at 3:06
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This is because the most general definition of $x^\alpha$ for complex $x,\alpha$ is $x^\alpha:=\exp(\alpha\log x)$, where $\exp$ is the exponential function and $\log$ is the (principal) natural logarithm. Pictorially, this means that taking $x$ to the power of $\alpha$ means multiplying the argument of $x$ by $\alpha$; that is, rotating $x$ with respect to the origin by an angle of $\alpha$ times the seperation between $x$ and the positive real axis. So if $x$ is a negative real number and $\alpha$ is an irrational real number, then we must rotate $x$ by some irrational multiple of $\pi$ (i.e. $180^\circ$). Obviously, irrational numbers are never integers, so the result after the rotation never ends up on the real axis; that is, $x^\alpha$ is not real.

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  • $\begingroup$ I like this answer! I can't edit it at the moment, but I'd appreciate a reformatting of some things, like changing natural log to ln, etc. $\endgroup$ – Starkle Apr 2 at 1:25
  • $\begingroup$ @Starkle It's quite rare for the national logarithm to be denoted $\ln$ instead of $\log$, especially in complex analysis. $\endgroup$ – YiFan Apr 2 at 1:28
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When we write $x^{\frac{1}{n}}$, this is just a notation. We mean to say that there should be a real number $y$ such that $y^n = x$. This definition is valid only when $n \in \mathbb{Z}$. But, when we want to do the same for irrationals, we do not have any definition. The usual intuition of exponents, which is multiplying the number itself as many times, will not work in case of irrationals. This is why we have to go into complex numbers where $w^z$ is defined for any complex numbers $w$ and $z$.

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Every $z\in \mathbb{C}\setminus\{0\}$ can be written as $z=r\cdot e^{i\theta}$, with $r>0$ and $\theta\in\mathbb{R}$.

If $z$ is a negative real number, $r=-z$ and $\theta=(2n+1)\pi$, $n\in\mathbb{Z}$.

So, if $z^x=r^x\cdot e^{ix\theta}=(-z)^x\cdot e^{ix(2n+1)\pi}$ is real, then

$$(-z)^x\cdot e^{ix(2n+1)\pi} = r_0\cdot e^{ik\pi}$$ for some $r_0>0$ and $k\in\mathbb{Z}$. (Remember: $e^{ik\pi}=\pm1$, if $k\in\mathbb{Z}$)

So $ix(2n+1)\pi=ik\pi$ and then $x=\frac{k}{2n+1}\in\mathbb{Q}$.

In particular, if $x$ is irrational, so $z^x$ is not a real number

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When considering $-1=e^{i\pi}$, the function $$f:\mathbb{R} \to \mathbb{C} \,\, ; \,\, x \mapsto (-1)^x=e^{i\pi x \, (\text{mod }2\pi)}=\cos(\pi x)+i\cdot \sin(\pi x)$$ becomes continuous. The expression $f(x)$ is real only if $x$ is an integer (you would need $\pi x$ to be equal to $\pi k$ for some integer $k$). Now this differs from the $n^{\text{th}}$-root concept you presented, for $(-1)^\frac{1}{3}=f(\frac{1}{3})=\frac{1}{2}+\frac{\sqrt{3}}{2}\cdot i \neq -1$. Also, with this you get $$f(\pi) \approx -0.9-0.43 \cdot i \,\,\,\, , \,\,\,\, f(e) \approx -0.633 +0.774 \cdot i$$

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