<s>[INST] Which one is correct for $\sqrt{-16} \times \sqrt{-1}$? $4$ or $-4$? As we can find in order to evaluate $\sqrt{-16} \times \sqrt{-1}$, we can do it in two ways. 
FIRST
\begin{align*}
     \sqrt{-16} \times \sqrt{-1} &= \sqrt{(-16) \times (-1)}\\
     &= \sqrt{16}\\
     &=4
\end{align*}
SECOND
\begin{align*}
     \sqrt{-16} \times \sqrt{-1} &= \sqrt{16i^2} \times \sqrt{i^2}\\
     &= 4i \times i\\
     &=4i^2\\
     &=-4
\end{align*}
Incidentally if the above is input in  complex mode of Casio scientific calculator, the result comes out as $-4$.
Which of the above solutions is correct?
 [/INST] The 'square root function' (the function we represent by $\sqrt{\cdot}$) is actually a branch cut of the multiple valued 'square root relation'. The typical square root function returns a complex value with an argument in $[0, \pi)$.
So $\sqrt{-16}\cdot\sqrt{-1}=4i\cdot i=-4$ because $i$ has an argument of $\pi/2$. The other value that we would reasonably see as being "$\sqrt{-16}$" is $-4i$, but this complex number has an argument of $3\pi/2$.
This pattern also persists in the real numbers. We have that $\sqrt{4}=2$. But we could also reasonably say that $\sqrt{4}=-2$. But $2$ has an argument of $0$ while $-2$ has an argument of $\pi$.
</s>