what is the square root of -16?\n-8i\n-4i\n4i\n8i

Explanation:

Step1: Recall the definition of square - root and imaginary unit

The square root of a negative number can be found using the fact that $\sqrt{-1}=i$. Let $\sqrt{-16}=x$. Then $x^{2}=-16$. We know that $16 = 4\times4$, and $\sqrt{-16}=\sqrt{16\times(- 1)}$.

Step2: Use the property of square - roots $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ (for $a\geq0,b\geq0$ with an extension for complex numbers)

$\sqrt{-16}=\sqrt{16}\times\sqrt{-1}$. Since $\sqrt{16} = 4$ and $\sqrt{-1}=i$, we have $\sqrt{-16}=4i$. Also, $(-4i)^{2}=(-4)^{2}\times i^{2}=16\times(-1)=-16$. So the square roots of $-16$ are $\pm4i$. Among the given options, the correct one is $4i$.

Answer:

C. $4i$