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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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    $\begingroup$ This is an excellent question to ask. This section of the Wikipedia page should help. $\endgroup$ – Zev Chonoles Sep 11 '11 at 4:12
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    $\begingroup$ Quite related... $\endgroup$ – J. M. is a poor mathematician Sep 11 '11 at 4:15
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    $\begingroup$ 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 $x+(x^2+1)$. $\endgroup$ – Arturo Magidin Sep 11 '11 at 4:19
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    $\begingroup$ At least one of these comments should be posted as an answer so it can get accepted. $\endgroup$ – joriki Sep 11 '11 at 5:14
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    $\begingroup$ 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. $\endgroup$ – GEdgar Sep 11 '11 at 13:04
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I've made my comment above into an answer.


This is an excellent question to ask. This section of the Wikipedia page should help.

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

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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 $-\mathbf{i}$.

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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.

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