I'm not sure if I recall this correctly, but I thought there was a reason why you shouldn't write $i=\sqrt{-1}$. And if this is not true, then I wonder: Why would you define $i$ as $i^2=-1$, why wouldn't you define it as $i=\sqrt{-1}$.
I was thinking that the reason to not write $i=\sqrt{-1}$, because otherwise you could argue that $$i^2=\sqrt{-1}\sqrt{-1}=\sqrt{-1\cdot -1}=\sqrt{1}=1$$
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$\begingroup$None of what you wrote is the definition of the imaginary unit. The proper definition is as follows:
A complex number $z \in \mathbb C$ is an ordered pair of real numbers $(a,b) \in \mathbb R^2$ such that the following rules for addition and multiplication hold: For any $z = (a,b)$ and $w = (c,d)$ in $\mathbb C$, $$\begin{align*} w+z &= (a,b) + (c,d) = (a+c, b+d), \\ w \cdot z &= (a,b) \cdot (c,d) = (ac - bd, ad + bc). \end{align*}$$ From this definition, the imaginary unit $i$ is defined as the ordered pair $i = (0,1)$. It is then easy to show that $i^2 = -1$ using the above rules.
$\endgroup$ 8 $\begingroup$You can write $i=\sqrt{-1}$ as long as you understand that all this means is that $i^2=-1$.
$\endgroup$ 7 $\begingroup$Before you can say $i=\sqrt{−1}$ you need to make sure that there is some context in which $-1$ has an existent square root. For example, when speaking in context of real numbers, $\sqrt{-1}$ is not defined. Rather, defining $i^2=-1$ ensures that you don't let $i"="-i$, as $i^2=(-i)^2=-1$. See more at the accepted answer's "$\mathbf{Edit}$" of Is the square root of a negative number defined?.
$\endgroup$ $\begingroup$While with nonnegative real numbers $a,$ we have the convention that $\sqrt{a}$ is the nonnegative number $b$ such that $b^2=a,$ there is no such convention for what the principal square root of other numbers "should" be. In particular, for any given non-zero complex $w$, there are two complex solutions to $z^2=w$--that is, there are two square roots of $w$--but only when $w$ is positive do we have a canonical choice for the principal square root of $w,$ $\sqrt{w}.$
$\endgroup$ 2 $\begingroup$If you have done some abstract algebra, another definition of $\mathbf{C}$ is that it is the quotient ring $\mathbf{R}[X]/(X^2+1)$. The imaginary unit is then the coset of $X$ in this ring, and we then have $\overline{X}^2 + 1 = \overline{X^2+1} = \overline{0}$, so $\overline{X}^2 = -1$. (Relate this definition with the one given by heropup.)
$\endgroup$ $\begingroup$Complex numbers are not an ordered ring, so in other words giving them an order doesn't make sense. In fact neither of those definitions defines the element uniquely. Since if $i^2=-1$ then $(-i)^2=(-1(i))^2=(-1)i(-1)i=(-1)(-1)i^2=1(-1)=-1$
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