Definition of multiplication of complex numbers I know that given two complex numbers $z_1 = a + bi$ and $z_2 = c + di$, the multiplication of these two numbers is defined as
$$
z_1*z_2 = (ac - bd) + i(ad + cb)
$$
I also know that I can easily derive this formula by applying the distributive property of multiplication and considering $i^2 = -1$.
But $i^2 = -1$ is a consequence of the definition of multiplication.
So, the question is: How mathematicians defined the multiplication of complex numbers whether they didn't know the value of $i^2$?

Indeed, $i^2=-1$ is coherent with the definition of multiplication, but things appeared in the reverse order.
While working on the resolution of algebraic equations (roots of polynomials), the need arose to consider square roots of negative numbers. Hence the "imaginary" symbol $i=\sqrt{-1}$, which is enough to represent the square root of any negative number $\sqrt{-a}$ as $\sqrt ai$.
From there, the algebra of complex numbers easily follows, treating $i$ as a variable and using the rule $i^2=-1$:
$$(a+bi)+(c+di)=a+bi+c+di=(a+b)+(c+d)i,$$
$$(a+bi)(c+di)=ac+adi+bci+bdi^2=ac+adi+bci-bd=(ac-bd)+(ad+bc)i.$$

Division is not much more difficult to invent, using a trick to turn the denominator to a real number,
$$\frac{a+bi}{c+di}=\frac{(a+bi)(c-di)}{(c+di)(c-di)}=\frac{(a+bi)(c-di)}{c^2+d^2}.$$

Now, what about the square root $\sqrt{a+bi}$ ?
Well, let us try and solve $a+bi=(c+di)^2=c^2-d^2+2cdi$, or $$a=c^2-d^2\\b=2cd.$$
Then,
$$a^2+b^2=c^4-2c^2d^2+d^4+4c^2d^2=c^4+2c^2d^2+d^4=(c^2+d^2)^2,$$
so that
$$c^2=\frac{\sqrt{a^2+b^2}+a}2,\\d^2=\frac{\sqrt{a^2+b^2}-a}2.$$
As you can see, $c$ and $d$ are always real numbers, so that no new symbol needs to be introduced to take square roots. From the last formulas, we have $$\sqrt i=\frac{1+i}{\sqrt2}.$$
