TITLE: If "$i^2 = -1$" is an imaginary number, $i$ why is there no imaginary number for "$|x| = -1$"?
QUESTION [6 upvotes]: The equation "$i^2 = -1$" has no real solution, so there's an entire number system of imaginary numbers that satisfy this.
So why is there not imaginary numbers that satisfy "$|x| = -1$"? There aren't any numbers, real or imaginary, that satisfy that equation, so shouldn't that make a new system of imaginary numbers?

REPLY [2 votes]: There are many problems if we define a number to be solution of this equation (some of which are already given in comments above):

Let $x$ be a number such that $|x|=-1$. Then $x^2=1\implies(x-1)(x+1)=0$. Since both the terms on LHS are non-zero, the number system you're defining isn't even an integral domain.

$i^2=-1$ makes sense: since $-1$ shows one step backward on number line ($180^\circ$ rotation), and $i$ shows $90^\circ$ rotation in argand plane (which is essentially an extension of number line in 2D). There is no significance to the one you're suggesting. A point can't have negative distance from the origin.

Again as suggested in comments, it won't be  a metric then.