TITLE: why is $\sqrt{-1} = i$ and not $\pm i$?
QUESTION [10 upvotes]: 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?

REPLY [2 votes]: There are many abstract ways to construct the complex numbers, and however you do it, there is a complex number which squares to $-1$, and to which you happen to give a name, often $i$. Once you have named $i$, it becomes obvious that $-i$ also has square $-1$, and that all complex numbers have the form $a +bi$ for unique real $a$ and $b.$ It really does not matter which $i$ you choose in the first place, you could equally well have chosen $-i$ instead, and this is reflected by the fact that the mapping which sends $a + bi$ to $a-bi$ for $a,b \in \mathbb{R}$ preserves all the additive and multiplicative structure of the complex field (ie is an automorphism of the complex field fixing every element of $\mathbb{R}$). The point is that there is no mystical significance to the symbol $i$. If you make the other choice and give $-i$ a new symbol $j$, you can work equally well with expressions of the form $a + bj$ for $a,b \in \mathbb{R}$, and never know the difference.
When you start to represent complex numbers geometrically on the Argand diagram, you do tend to make a choice, but that is a choice of using clockwise or counterclockwise direction for rotations. It is usual to represent the complex number $\cos \theta + i\sin \theta$ by the point $(\cos \theta, \sin \theta)$ in $\mathbb{R}^{2}.$  But if we consistently used $(\cos \theta, -\sin \theta)$ instead we would have no difficulty.