Is $i$ equal to $\sqrt{-1}$?
Some of my math teachers said that $i=\sqrt{-1}$ is the wrong definition for $i$ and that the correct definition is $i^2=-1$. If the second definition if true, then does it mean $\sqrt{-1}= ±i$? Which of the two definitions is true?
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$\begingroup$From a formal algebraic perspective, $\Bbb{C}$ is the field $\Bbb{R}[x]/(x^2+1)$ which is the field you get by adjoining to $\Bbb{R}$ the roots of $x^2 + 1$ and all the linear combinations of the form $a+ib$, where $a, b \in \Bbb{R}$. You really need to add just one of them (we call it $i$), because the other is just $0 + i(-1)$. Notice that it doesn't matter which we pick as $i$: if I pick one and you picked the other, we would never notice we made different choices.
Infact, the Galois group (automorphisms preserving the original field, thus fixing $\Bbb{R}$ in this case) of $\Bbb{C}$ has exactly two automorphisms: the identity, sending $i$ to itself, and the conjugation, sending $i$ to its opposite. By definition, automorphisms preserve structure, thus confirming the choice of $i$ doesn't really matter.
$\endgroup$ 7 $\begingroup$The square root symbol does not make sense when applied to a negative number. Granted, people often write it that way for convenience. $i$ is defined by the equation $i^2=-1$. Yes, if there is a field containing a root of $x^2+1$ then if $i$ is such a root so is $-i$. When we define a field such as $\mathbb C$ we are implicitly choosing a root of $x^2+1$.
$\endgroup$ 5 $\begingroup$$\sqrt{\cdot}$ is seen in most cases as a function in $[0,\infty)$, so it have a unique value associated to each non-negative real. In this tradition we generally define $i:=\sqrt{-1}$, that can be thought informally as a kind of extension to the classical square root function.
Indeed we can extend the definition of the square root to any complex number, setting it as the principal value of $\sqrt z:=e^{1/2\ln z}$ with the same result.
However is true that $(-i)^2=-1$.
$\endgroup$ $\begingroup$$\sqrt{-1}$ is denoted as $i$ and $i$ is a solution of the algebraic equation $x^2+1=0$. That is why $i^2=-1$.
One could define $i=-\sqrt{-1}$ and using this notation he could built the Complex analysis. That is up to him.
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