# When taking a square root of a negative number, why is the result i instead of positive and negative i? [closed]

Why does the result only equal the principal root? For example, why is the square root of -16 just 4i, instead of positive and negative 4i?

When I think about this, I see sqrt -16 as sqrt 16 * sqrt -1. The sqrt 16 is positive and negative 4 and square root -1 is i, I get the result pos and neg 4i. All my textbooks just list 4i. Why is that?

Forgive my syntax, but when the question is stated: simplify sqrt -16, I still believe the answer is + and - 4i. Yet, my texts show only 4i and do not include -4i. Is this simply for convention sake? Or a higher reason than I am missing? The solutions that I have seen thus far indicate, that it is convention, rather than pure mathematics that this is so.

Speaking with colleagues, I have been told that unless a positive/negative symbol is placed before the square root of -16 the expression can only be equivalent to 4i. Is this accurate? And if so, why?

## closed as unclear what you're asking by Namaste, JonMark Perry, ahulpke, Brian Borchers, NarasimhamJan 20 '18 at 18:48

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• There is no "the square root" of a complex number, as a nonzero complex number has two square roots. I'd say both "$4i$ is a square root of $-16$" and "$-4i$ is a square root of $-16$". – Lord Shark the Unknown Jan 13 '18 at 4:56
• "The sqrt 16 is positive and negative 4" Huh? Any reference for this meaning of $\sqrt{16}$? – Did Jan 13 '18 at 7:08
• I thought that that what was I was trying to convey, that both 4i and -4i are square roots of -16. Am I still stating that idea incorrectly? – Nikki Jan 13 '18 at 7:51
• You have put @Did in your comment if you want him pinged. – spaceisdarkgreen Jan 13 '18 at 8:02
• @Nikki I mean, expressions evaluate. Like wolframalpha.com/input/?i=sqrt(-16)*sqrt(-25) . It's a nice feature, no? It's true that you often have to think about both roots, but it's nice to be able to say something like: "clearly the root that gives us our attracting fixed point is $-\frac{b}{2} + \sqrt{(b/2)^2-4c}$ because (reasoning)" and actually know which one you're talking about. – spaceisdarkgreen Jan 13 '18 at 8:40

You are correct that both $4i$ and $-4i$ are square roots of $-16,$ just as $4$ and $-4$ are both square roots of $16.$
Nonetheless we like sometimes to pretend that "the square root" is a function. One convention for defining a function on the complex numbers that results in a square root is to write a complex number $z$ in polar coordinates as $$z = x+iy = r\cos\theta + i r\sin\theta$$ for $r\ge 0$ and $0 \le \theta < 2\pi$ and then define $$\sqrt{z} = \sqrt{r}\cos(\theta/2) + i\sqrt{r} \sin(\theta/2),$$ which you can check has the property that $(\sqrt{z})^2 = z.$ This results in every square root being in the upper half-plane.
By this definition, $$\sqrt{-16} = 4i.$$ However, this function has some odd properties. For one, it is discontinuous at the positive real axis since a point just below the positive real axis will get mapped to a point just above the negative real axis whereas points on the positive real axis are mapped to points on the positive real axis. Also, it doesn't have many of the arithmetical properties we are used to, and this makes calculation tricky. For instance, we have $$\sqrt{-4} \sqrt{-9}=(2i)(3i) = -6 \ne \sqrt{(-4)(-9)} = 6.$$