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?
 A: $$\sqrt a\cdot\sqrt b=\sqrt{ab}$$ only work if $a,b\ge0$
A: In the real numbers it's possible to define a square root function that is injective: for $x\ge0$, $\sqrt{x}$ is the unique nonnegative real number $y$ such that $y^2=x$.
In the complex numbers this is not really possible: an injective square root function can be defined, for instance declaring that $\sqrt{0}=0$ and for a nonzero number
$$
x=r(\cos\alpha+i\sin\alpha)\ne0,
$$
with $0\le\alpha<2\pi$ and $r>0$,
defining
$$
\sqrt{x}=\sqrt{r}(\cos(\alpha/2)+i\sin(\alpha/2)),
$$
where $\sqrt{r}$ is the unique square root of a positive number defined above. However, this function has no good algebraic property, apart from
$$
(\sqrt{x})^2=x.
$$
For instance, with this definition, $\sqrt{-1}=i$, because $-1=1(\cos\pi+i\sin\pi)$, but
$$
-1=\sqrt{-1\mathstrut}\sqrt{-1\mathstrut}\ne\sqrt{(-1)(-1)}=1
$$
which instead would be a decent property to have.
Indeed, if instead of the interval $[0,2\pi)$ we chose $[-\pi,\pi)$ for the argument, we would have to write $-1=1(\cos(-\pi)+i\sin(-\pi))$ and so the square root would suddenly become $-i$. Why choosing one interval for the argument and not another one? In any case the product rule wouldn't hold.
Thus $\sqrt{-16}\cdot\sqrt{-1}$ doesn't really make sense except perhaps to denote two values. But this opens other problems: how many values would $\sqrt{-4}+\sqrt{-4}$ have? Of course three: $-4i$, $0$ and $4i$. On the other hand, $2\sqrt{-4}$ would have only two, namely $-4i$ and $4i$.
You see that there's no way out. Well, there is, and is called “going to a Riemann surface”. Not at all elementary.
A: $$\sqrt[n]{a b} = \sqrt[n]{a} \, \sqrt[n]{b} \quad (*)$$
if $a$ and $b$ are negative, then $(*)$ works only for $n$-th roots with odd $n$,
alas $n = 2$ is even. 
A: The answer to each sqrt after removing i is +-, so for a simple explanation you have either (+)(+) or (-)(+) or (-)*(-), depending on the domain or context of the particular problem.
A: 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 $\pi$.
