# What is the square root of -16?

Sep 10, 2015

There is no Real number whose square is $- 16$.

The principal Complex square root $\sqrt{- 16} = 4 i$

$- 4 i$ is also a square root of $- 16$

#### Explanation:

If $a \in \mathbb{R}$ then ${a}^{2} \ge 0$. So there is no Real square root of $- 16$.

If $i$ is the imaginary unit, then ${i}^{2} = - 1$ and we find that:

${\left(4 i\right)}^{2} = {4}^{2} \cdot {i}^{2} = 16 \cdot - 1 = - 16$

So $4 i$ is a square root of $- 16$.

Also:

${\left(- 4 i\right)}^{2} = {\left(- 4\right)}^{2} \cdot {i}^{2} = 16 \cdot - 1 = - 16$

So $- 4 i$ is a square root of $- 16$.

If $x \in \mathbb{R}$ and $x < 0$ then $\sqrt{x}$ stands for the principal square root of $x$ defined as:

$\sqrt{x} = i \sqrt{- x}$

In our case:

$\sqrt{- 16} = i \sqrt{16} = 4 i$

Note that you do need to be slightly cautious when dealing with square roots of negative numbers. In particular, the property $\sqrt{a b} = \sqrt{a} \sqrt{b}$ fails if $a , b < 0$:

$1 = \sqrt{1} = \sqrt{- 1 \cdot - 1} \ne \sqrt{- 1} \sqrt{- 1} = {\left(\sqrt{- 1}\right)}^{2} = - 1$