# 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.

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This is an excellent question to ask. This section of the Wikipedia page should help. – Zev Chonoles Sep 11 '11 at 4:12
Quite related... – J. M. Sep 11 '11 at 4:15
If you define the complex numbers to be the set of all pairs of real numbers $(a,b)$ with componentwise addition and multiplication given by $(a,b)*(c,d) = (ac-bd,ad+bc)$, then $i$ is by definition the pair $(0,1)$. If you define the complex numbers to be the quotient $\mathbb{R}[x]/(x^2+1)$, then $i$ is by definition the element $x+(x^2+1)$. – Arturo Magidin Sep 11 '11 at 4:19
At least one of these comments should be posted as an answer so it can get accepted. – joriki Sep 11 '11 at 5:14
Equation $x^2+1=0$ has two solutions in the complex numbers. There is no algebraic property that can tell the two solutions apart. If I choose one of them for $i$ and you choose the other one for $i$, we won't ever be able to tell the difference. so: Just let $i$ be one of the solutions and go on from there. – GEdgar Sep 11 '11 at 13:04

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let $e^{z\pi}=-1$, then $z=(2k-1)i, k\in \mathbb N$

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