$\displaystyle \sqrt{4} \bullet \sqrt{9} = 2 \bullet 3 = 6$

When I do it this way --- $\displaystyle \sqrt{4 \bullet 9} = \sqrt{36} = 6$

I get the same answer either way I write it. However if i do this:

$\displaystyle \sqrt{-1} \bullet \sqrt{-1} = 1$

or if I write:

$\displaystyle \sqrt{-1 \bullet -1}$ Either way I get answer of 1 because

$\displaystyle -1 \bullet -1 = 1$ and $\displaystyle \sqrt{1}$ is just one.

However complex numbers say that $\displaystyle i^2$ is equal to $\displaystyle -1$

How can $\displaystyle i^2$ be equal to $\displaystyle -1$ if

$\displaystyle \sqrt{-1} \bullet \sqrt{-1}$ equals $\displaystyle \sqrt{1}$ or just $\displaystyle 1$?

By my logic $\displaystyle i^2$ should be just 1. So there is obviously a rule I am missing here that would lead me to believe this. If someone could explain to me why $\displaystyle \sqrt{4 \bullet 9} = \sqrt{36} = 6$ but $\displaystyle \sqrt{-1 \bullet -1}$ doesnt not equal $\displaystyle \sqrt{1}$ or just $\displaystyle 1$ . It would be much appreciated!

Sorry in advance if my code got messed up, I havent used it before.