Is the $\pm$ sign used when finding the root of a negative number?

Question

If $\sqrt{64}$ is equal to $\pm{}8$, is $-64$ equal to $\pm{}8i$, or just $8i$?

Answer 1

$\sqrt{a}$, for real $a$, is almost always defined to be the positive solution of the equation $x^2 - a = 0$, so $\sqrt{64}$ is $8$ and not $\pm 8$. The reason the square root only takes on one value is because it is a function and so each element in the domain can be mapped to at most one element in the codomain.

As for your second question, $i$ is defined to as a number satisfying the equation $i^2 + 1 = 0$ and so we can say that $\sqrt{-1} = i$.

Following from this, we have $\sqrt{-64} = \sqrt{-1} \sqrt{64} = i\sqrt{64} = 8i$.

Answer 2

Well, in theory, the square root of any number should return both its negative and positive root. Meaning $\sqrt{x^2}=\pm x$. But if you think about the geometric meaning of the square root, it’s finding the side length which makes a square of area $x^2$. So some people think that the square root should only be a positive answer since length cannot be negative. In the case where we do want both the negative and positive root, here’s what you can do. $\sqrt{-16}=\sqrt{-1} \cdot \sqrt{16}$ $\sqrt{-1}=i$, $\sqrt{16}=\pm 4$, which gives us $i \cdot \pm 4=\pm 4i$