Why was $i$ introduced to satisfy this $\sqrt{-1}$? Can someone explain to me why $$\sqrt{-1} = i$$ I love math and I'm looking at doing it to higher levels. I know that we can NEVER have a square root of a negative number as per my reading hence if I was given $$\sqrt{-4}$$ the answer would be $$2i$$
and if I was given
$$\sqrt{4}$$ 
the answer would be $\pm2$, can I get sources explaining why $i$ was introduced?
Thanks. 

Technically, $i \neq \sqrt{-1}$. $i$ is simply a symbol that we use to denote a solution to the equation $z^2 + 1 = 0 \iff z^2 = -1$. This symbol is called the imaginary unit and there isn't really a reason why it represents a solution to $z^2 = -1$, it's more that it was defined to be that way. This imaginary unit is quite important in the field of mathematics. It extends the real numbers and generates the complex number system that has quite a significant impact on mathematics. Examples include a whole branch of analysis termed complex analysis or provides algebraic closure for polynomials in $\mathbb{C}$.
Secondly, $\sqrt{4} \neq \pm 2$. The square root function denoted by the radical sign refers to only the positive value, which is why technically $i \neq \sqrt{-1}$ and why $\sqrt{4} = 2$ and not $-2$. 
As for why the imaginary number was introduced, there are a multitude of reasons. For example, the introduction of such a number means that the field of real numbers is algebraically closed. Specifically, the algebraic closure of the real numbers is the field of complex numbers that are generated by including $i$. A lot about the usage and technicalities of the imaginary unit $i$ can be gleaned from its Wikipedia page here, that I would encourage you to read.
