The problem is that every complex number, except 0, has two square roots. With positive real numbers, since every real number is either 0 or positive or negative, so the two square roots are one positive and one negative, we define the square root of a to be the positive root. The complex numbers do not have the "trichotomy" property so we cannot do that.
If you want a really valid definition of "i" start by defining the "complex numbers.
The complex numbers are defined as the set of all pairs of real numbers, (a, b) with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication defined by (a, b)*(c, d)= (ac- bd, ad+ bc). It just takes a little algebra to show that addition and multiplication are "associative" and "commutative" and that multiplication "distributes" over addition.
(0, 0) is the additive identity, (-a, -b) is the additive inverse of (a, b), (1, 0) is the multiplicative identity, and (a/(a^2+ b^2), -b/(a^2+ b^2)) is the multiplicative inverse of (a, b) as long as (a, b) is not (0, 0).
Then (a, 0)*(c, 0)= (ac- 0, 0)= (ac, 0) so we identify (a, 0) with the real number a so. In this sense, the real numbers are a subset of the complex numbers. But (0, b)*(0, d)= (0- bd, 0)= (-bd, 0). In particular, while (1, 0) is the multiplicative identity, (1, 0)*(a, b)= (a, b), identified with the number, "1", (0, b) has the property that its square is (0, b)* (0, b)= (0- b^2, 0)= (-b^2, 0) which we identify with the real number -b^2. In particular, (0, 1)*(0, 1)= (-1, 0) which we identify with the real number -1. We give the complex number (0, 1) the label "i" so i^2= -1.
Of course (a, b)= (a, 0)+ (0, b)= a+ b(0, 1)= a+ bi. With those definitions and identifications, we avoid the problem you have. i is NOT just the "square root of -1" (Which one? There are two!) but is specifically (0, 1). not (0, -1).
