What's the thing with $\sqrt{-1} = i$

What's the thing with $\sqrt{-1} = i$? Do they really teach this in the US? It makes very little sense, because $-i$ is also a square root of $-1$, and the choice of which root to label as $i$ is arbitrary. So saying $\sqrt{-1} = i$ is plainly false!

So why do people say $\sqrt{-1} = i$? Is this how it's taught in the US?

Solution 1:

I believe this is a common misconception. Here in Sweden we (or at least I) was taught that $i^2 = -1$, not that $\sqrt{-1} = i$. They are two fundamentally different statements. Of course most of the time one chooses $\sqrt{-1} = i$ as the principal branch of the square root.

Solution 2:

$\sqrt{-1}$ is an imprecise notation. There are several ways of making it precise, some of which involve what comes down to an arbitrary choice of square root of $-1$ and some of which don't. This is because there are several related ways to construct $\mathbb{C}$ from $\mathbb{R}$:

• As the ring $F = \mathbb{R}[x]/(x^2 + 1)$ (edit: following in the discussion in the comments, perhaps it would be better to say "an algebraic closure of $\mathbb{R}$."). The Galois group $\text{Gal}(F/\mathbb{R})$ has order $2$, and complex conjugation $x \mapsto -x$ is its only nontrivial element. The Galois symmetry here forbids us from distinguishing between $x$ and $-x$.

• As the ring $\mathbb{C} = \mathbb{R}[\imath]/(\imath^2 + 1)$, where we fix the choice $\imath$ of generator. This is what is typically meant by $\mathbb{C}$. This field has a distinguished element $\imath$ satisfying $\imath^2 = -1$, and this is what we usually mean by $i$. $\mathbb{C}$ is isomorphic to $F$, but not canonically so, since $\imath$ can be sent to either $x$ or $-x$ and there is no way to choose between these.

• As the ring $R$ of linear endomorphisms $\mathbb{R}^2 \to \mathbb{R}^2$ of non-negative determinant preserving an inner product $\langle \cdot, \cdot \rangle$. There are two elements of $R$ squaring to $-1$, corresponding to a rotation and its inverse, and there is no way to choose between them unless $\mathbb{R}^2$ is also equipped with an orientation, which is an identification of its exterior square with $\mathbb{R}$. This choice of orientation corresponds to the difference between the first and the second constructions above.

But most people will not bother to stop and talk about such subtleties as $F, \mathbb{C}, R$ being non-canonically isomorphic, so they say $\sqrt{-1} = i$ because the truth is (for practical purposes) unnecessarily complicated.

Solution 3:

As even the capitalistic Wikipedia acknowledges, there are three laws of dialectics:

1. The law of the unity and conflict of opposites;
2. The law of the passage of quantitative changes into qualitative changes;
3. The law of the negation of the negation.

Thus, from law 1, we see that i and -i are both equal and opposite. From law 2, we learn that $\mathbb C$ is qualitatively different from $\mathbb R$. From law 3, we attain the dynamic equilibrium between i and -i that was so powerfully expounded by the revolutionary hero Évariste Galois.

This at least is how we taught such matters in the USSR.