Question

What is the square root of -16?

Answer 1

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

The principal Complex square root `sqrt(-16) = 4i`

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

Explanation:

If `a in RR` then `a^2 >= 0`. So there is no Real square root of `-16`.

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

`(4i)^2 = 4^2*i^2 = 16 * -1 = -16`

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

Also:

`(-4i)^2 = (-4)^2*i^2 = 16 * -1 = -16`

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

If `x in RR` 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) = 4i`

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

`1 = sqrt(1) = sqrt(-1 * -1) != sqrt(-1)sqrt(-1) = (sqrt(-1))^2 = -1`