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Is $\large\sqrt[\pi]{-1}$ a real, imaginary, or complex number? Or maybe, is it something else entirely? Is there even a way to evaluate this?

3 Answers3

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It is not one number, since exponentiation of complex numbers is generally multi-valued.

There are four numbers $z$ where the principal value of $z^{\pi}$ is $-1$, namely $$ e^{-3i}, e^{-i}, e^{i}, e^{3i} $$ whose explicit $a+bi$ forms are $$ \cos(3) \pm \sin(3)i\quad\text{and}\quad \cos(1) \pm \sin(1)i $$

In general, the numbers such that $-1$ is a possible value of $z^{\pi}$ are $e^{ki}$ for any odd (positive or negative) integer $k$. These numbers lie densely on the unit circle.

hmakholm left over Monica
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Think about.

$$-1 = e^{\pi i}$$

$$\sqrt[n]{-1} = \left(e^{\pi i}\right)^{1/n} = e^{\pi i/n}$$

What could you say about, when $n = \pi$?

That answer was meant to be the general case..

Enrico M.
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    OP asked about $\pi$-th root, not $n$-th root. – Friedrich Philipp Mar 19 '16 at 23:04
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    He was stating in general, allowing the OP to make the connection himself as opposed to just giving the answer.... – Eleven-Eleven Mar 19 '16 at 23:07
  • The rules about exponentiation of positive real numbers cannot be applied to complex numbers as such hence the double identity in your post is invalid. – Did Mar 23 '16 at 06:29
  • @Did I think instead it holds, since I applied De Moivre law. https://it.wikipedia.org/wiki/Formula_di_de_Moivre – Enrico M. Mar 23 '16 at 11:09
  • $e^{\pi i/n}$ is the value of one branch of $z \mapsto z^{1/n}$ at $-1 = e^{\pi i}$. If $1/n$ happens to be an integer, it's the only branch, otherwise there are several branches (finitely many if $1/n$ is rational, infinitely many if it's irrational). De Moivre's formula only considers integer exponents. You can generalise it to non-integer exponents, but then you must take care to be explicit about the considered branch of the power. – Daniel Fischer Mar 23 '16 at 11:25
  • The very page you quote explains that $(e^{ix})^n=e^{inx}$ holds "con $n$ intero". Here, you would need it for $n=1/\pi$, not an integer. – Did Mar 23 '16 at 12:45
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Let's start by defining square roots in terms of exponents.

$\sqrt{2} = 2^{\frac{1}{2}}$ because if you multiply it by itself then as we want exponents to add in general you would get,

$2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^1 = 2$

which is what we would expect considering the square root of a number is the number that when you multiply it by itself you get the original number. So you could define any root, $x$, as raising the number to the exponent $\frac{1}{x}$ because $\frac{1}{x}$ is the multiplicative inverse of $x$ and we wanted a root to be the reverse of raising that number to that power.

Now we get a problem that $\pi$ is irrational, but if you allow all rational roots then you can fill in the missing roots by taking limits since we want exponentiation to be continuous. For irrational powers we generally use the exponential to fill in the holes and define,

$a^q = e^{\ln a^q} = e^{q \ln a}$

So for your question we have,

$-1^{\frac{1}{\pi}} = e^{\frac{\ln -1}{\pi}}$

Now we have a problem that $\ln$ is usually initially defined only for positive numbers. It can be defined for negative numbers although it becomes multiple valued in a way similar to how you have to decide which root you want when taking square roots. The usual value (called the principal branch cut) for $-1$ is $i\pi$ because $e^{i\pi} = -1$. So we have,

$e^{\frac{\ln -1}{\pi}} = e^{i} = \cos 1 + i \sin 1$

where in the last line I used the formula, $e^{i\theta} = \cos \theta + i \sin \theta$. This is approximately (using a calculator to find $\cos 1$ and $\sin 1$),

$0.54 + 0.84i$

and gives you one pi root of $-1$.

Mehdi2277
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