Question

How do you simplify `\sqrt (-81)`?

Answer 1

`sqrt(-81) = 9i`

Explanation:

The imaginary unit `i` satisfies `i^2=-1`

So we find:

`(9i)^2 = 9^2i^2 = 81 * (-1) = -81`

So `9i` is a square root of `-81`

Note that:

`(-9i)^2 = (-9)^2 i^2 = 81*(-1) = -81`

So `-9i` is also a square root of `-81`

What does `sqrt(-81)` mean?

`sqrt(-81)` denotes the principal square root of `-81`, but which is the principal one?

By convention and definition, if `n < 0` then:

`sqrt(n) = sqrt(-n)i`

So we could simply say:

`sqrt(-81) = sqrt(81)i = 9i`

Notes

Why not use `sqrt(ab) = sqrt(a)sqrt(b)` ?

Because it does not always work, especially when you are dealing with negative and/or complex values.

For example:

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

It is possible to use `sqrt(ab) = sqrt(a)sqrt(b)` for real numbers `a, b` if at least one of `a` and `b` is non-negative, essentially because of the convention:

`sqrt(n) = sqrt(-n)i" "` if `n < 0`