Question about this exponent problem.

Question

So I was playing around with exponents, and happened to come across a problem that seems to have 3 solutions.

The expression is $(-1)^{\frac{-1}{2}}$. Depending on how I write the fraction and take the root, I get three results.

$(-1)^{\frac{-1}{2}}=\sqrt{\frac{1}{-1}}=\sqrt{-1}=i$

$(-1)^{\frac{-1}{2}}=\sqrt{\frac{1}{-1}}=\sqrt{-\frac{1}{1}}=-1 $

$(-1)^{\frac{-1}{2}}=\sqrt{\frac{1}{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\frac{1}{i}=-i$

Answer 1

By definition, when dealing with complex numbers $a^b = e^{b \log(a)}$ where we can use any of the branches of $\log(a)$ (thus, in general, this is a multivalued function). In this case $b = -1/2$ and $\log(-1) = (2n+1) \pi i$ for arbitrary integers $n$, so $$(-1)^{-1/2} = e^{(-n-1/2)\pi i}$$
which has two values, $i$ and $-i$, depending on whether $n$ is odd or even.

Answer 2

Your usual rules for exponents of real numbers do not work with complex numbers. In the reals we define $a^b=e^{b \log a}$ and all works well. In the complex numbers we have $e^{2\pi i}=1$ so the $\log$ is only defined up to the addition of an integer multiple of $2\pi i$.

Answer 3

Your first is completely correct.

Your second is not, because $$\sqrt{-1}\neq -\sqrt 1$$