What is the precise definition of $i$? This may seem like an extraordinarily trivial question and yet it has completely confounded me.  The technical definition of $i$ is
$$i^2=-1$$
But there are two numbers which fulfill this requirement:
$$\sqrt{-1},-\sqrt{-1}$$
Wouldn't a more precise definition of $i$ simply be $\sqrt{-1}$?
Thank you and forgive the elementary nature of the question.
 A: I've made my comment above into an answer.

This is an excellent question to ask. This section of the Wikipedia page should help.
A: let $e^{z\pi}=-1$, then $z=(2k-1)i, k\in \mathbb N$ 
A: There is no property of the arithmetic and calculus of the complex numbers possessed by $\mathbf{i}$ that is not also possessed by $-\mathbf{i}$; one is just as good as the other.
For example, if you started with $-\mathbf{i}$ as the complex unit and went through the usual development of arithmetic, you would find that the principal square root is $\sqrt{-1} = -\mathbf{i}$.
In fact, a key feature of the complex numbers is complex conjugation, an operation that swaps $\mathbf{i}$ and $-\mathbf{i}$.
A: From this answer:

The imaginary number $i$ was discovered/invented/revealed as a solution the equation
  $$
x^2+1=0\tag{$\ast$}
$$
  That equation, and that equation alone, is the link between $i$ and the reals. However, there is another solution to $(\ast)$: $-i=-1\times i$ also satisfies $(\ast)$. Basically, this is because $(-1)^2=1$.
Since the link between $i$ and the reals is the same equation that is satisfied by $-i$, one can easily wonder, "which is which?" When we choose a solution to $(\ast)$, do we get $i$ or $-i$? In a sense, it doesn't matter; both satisfy $(\ast)$ and $i^2=-1$ is the property we use to do math in $\mathbb{C}$.
On the other hand, complex conjugation, swapping $i\leftrightarrow-i$, is an important isomorphism of $\mathbb{C}$. For instance, any polynomial with real coefficients that has $x+iy$ as a root, must also have $x-iy$ as a root. There are many ways to show this, but one of the most basic is by swapping $i\leftrightarrow-i$. This isomorphism does not affect the real coefficients of the polynomial, but it swaps $x+iy\leftrightarrow x-iy$.
In the end, we call one root of $(\ast)$, $i$, and the other, $-i$; it really doesn't matter which. They both satisfy $(\ast)$ and that is what is important.

