Complex number 2

$\large\sqrt{-16} \times\ \sqrt{-1}$

$= \large \sqrt{16} \sqrt{-1} \times\ \sqrt{-1}$

$\large=4 \times\ i \times\ i$

$\large =4i^2$

$\large =-4$

$\sqrt{-16} \times \sqrt{-1}$

$= \sqrt{-1} \sqrt{16} \times i$

$= 4 i^2$

$= 4(-1)$

$= \boxed{-4}$

$\sqrt{-16}(\sqrt{-1})=i\sqrt{16}(i\sqrt{1})=4i(i)=i^2(4) = -4$

Here, $\sqrt{a} \times \sqrt{b}$ , where $a$ and $b$ are negative, is not equal to $\sqrt{ab}$ ! $\begin{aligned} \sqrt{-16} \times \sqrt{-1} &= 4i \times i \\ &= 4i^2 \\ &= \boxed{-4} \end{aligned}$

Therefore, the answer is -4 .