why is $\sqrt{-1} = i$ and not $\pm i$? [duplicate]

this is something that came up when working with one of my students today and it has been bothering me since. It is more of a maths question than a pedagogical question so i figured i would ask here instead of MESE.

Why is $\sqrt{-1} = i$ and not $\sqrt{-1}=\pm i$?

With positive numbers the square root function always returns both a positive and negative number, is it different for negative numbers?

The square root function doesn't return two values for positive numbers, or it wouldn't be a function.

It's a fact that, if $x$ is a positive real number, there are two real numbers whose square is $x$. The positive one is denoted by $\sqrt{x}$, so the negative one is $-\sqrt{x}$.

In this way the function has the pleasant property that, for $x,y>0$, $\sqrt{xy\mathstrut}=\sqrt{x\mathstrut}\sqrt{y\mathstrut}$.

In the complex numbers, for every nonzero $x\in\mathbb{C}$ there are two complex numbers whose square is $x$. However, it's not possible to define a square root function with the property above, that is, $\sqrt{xy\mathstrut}=\sqrt{x\mathstrut}\sqrt{y\mathstrut}$.

Maybe you'd want to stretch the notion of function, to allow multiple values; but then, how many values should you assign to the expression $$ \sqrt{2}+\sqrt{3}+\sqrt{5} $$ and similar ones? You couldn't make the more obvious simplifications: from $$ x+\sqrt{2}=\sqrt{2} $$ you'd get three values for $x$.

When I introduce the complex numbers, I never use $\sqrt{-1}$, but rather I say that $i^2=-1$, which is a quite different statement, exactly because it's impossible to define a square root function that has sensible algebraic properties.

If $x$ is a non-negative real number, there's an unambiguous interpretation for the expression $\sqrt{x}$, namely, the non-negative square root of $x$. (The $\pm$ signs aren't "part of the square root function", which is why they have to be included explicitly when "solving an equation by taking square roots", e.g., passing from $x^{2} = 1$ to $x = \pm 1$.)

When one tries to extend the square root function to the complex numbers, there are tricky domain issues. Rather than write "$\sqrt{-1} = \pm i$" (which can't be true if the radical symbol denotes a function), it's safer to stick with $(\pm i)^{2} = -1$ (which is unambiguously true) until one is invested in understanding the fine points of functions of a complex variable.

In case you or your student are curious: Each non-zero complex number has two square roots, but there is no continuous choice of square root on the complex plane.

To get a continuous "branch of square root" it's necessary to remove enough of the plane that "the domain of the square root doesn't encircle the origin". The customary choice is to remove the non-positive reals. (Ironically, this explicitly excludes $-1$ from the domain of the square root.)

A common alternative choice is to remove the non-positive imaginary axis. In this event, $\sqrt{-1} = i$ for the continuous branch of square root that satisfies $\sqrt{1} = 1$.

The take-away points are:

• If $\sqrt{\ }$ denotes a function, then it must be single-valued (no $\pm$).

• When allowing complex numbers (other than non-negative reals) under a radical sign, You Really Need To Be Careful.