Roots Across the Complex Numbers

Question

Why is it the case that an even root (square root, quartic, etc) can be positive or negative across the complex numbers, but is limited to postive in the reals? Is there a good mathematical reason for this, or is it simply notation?

Answer 1

Essentially, the reason is that complex multiplication, and by extension exponentiation, are far more complicated than in the reals. There is a rotational aspect, and for an $n$th root, there are $n$ answers. It would be silly to declare that only one of these is the "actual" answer, whereas in the reals it makes much more sense. Glad to see that you have joined the site. Please continue asking questions.

Answer 2

Adding to William Grannis' answer about it not making sense to choose "the actual root": here's another reason why.

For positive $x$, we can easily distinguish between $\sqrt x$ and $-\sqrt x$, and define $\sqrt x$ to be the positive one.

Now think about

$$i^2=-1$$

This has two solutions. We "choose" one of them to be $i$, and call the other one $-i$, but there's absolutely no way to say which one we "chose". $i$ and $-i$ have completely identical properties, other than one being the negative of the other.

That is: if we said "Let's define $j=-i$, and rewrite all our mathematics in terms of $j$", nothing would change except for the name of the imaginary unit.

I've always found this a slightly alarming feature of the complex number system.

But alarming or not: it makes no sense to prefer a "positive imaginary" value to a "negative imaginary" one, since ultimately it's arbitrary which is which.