Question

Can you find the square root of `-1/2` ?

Answer 1

You can't technically, but it's `1/2i`

Explanation:

You can't square root a negative. Because we come up with a problem here it is. Remeber a when you square root a number you get a number when multiplied by itself gives us the number that we just square rooted

E. g. `sqrt(16) = 4 * 4 or 4^2`

Quick Example

When we try to square root a negative we realise this isn't possible. Because when you square root a number the root has to be the same number. So to complete the problem below we'll have to split the `-16` into `-1` and `16` (`-1 * 16= -16`).

E. g. `sqrt(-16) = sqrt(-1*16)`

Now we know that 16 is a square number, so when its square root is... 4 of course! However, we cannot square the negative one so we rewrite it as `i` which is called an imaginary number as there is no square root of any negative.

Now onto the question
`sqrt(-1/2)`

Let's split it into `-1 and 1/2`

`sqrt(-1*1/2)`

This gives us `sqrt(1/2i)`

Answer 2

`sqrt(-1/2) = sqrt(2)/2i`

Explanation:

Assuming you are asking about the square root of `(-1/2)` ...

In common with all non-zero numbers, `-1/2` has two square roots, but since `-1/2` is negative, those square roots are non-real Complex numbers.

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

Hence we find:

`(sqrt(2)/2i)^2 = (sqrt(2)/2)^2 * i^2 = 1/2 * -1 = -1/2`

So one square root of `-1/2` is `sqrt(2)/2i`

We also have:

`(-sqrt(2)/2i)^2 = (-sqrt(2)/2)^2 * i^2 = 1/2 * -1 = -1/2`

So the other square root is `-sqrt(2)/2i`

By convention, when we write `sqrt(-1/2)`, we mean `sqrt(2)/2i`, which is known as the principal square root of `-1/2`.

Footnote

Why not simply write:

`sqrt(-1/2) = sqrt(-1 * 1/2) = sqrt(-1) * sqrt(1/2) = isqrt(1/2)`

I tend to avoid being too hasty with the "rule" `sqrt(ab) = sqrt(a)sqrt(b)`, because it does not always work, once you are starting to deal with square roots of negative and/or complex numbers.

For example:

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

On the other hand, the convention that if `x < 0` then:

`sqrt(x) = isqrt(-x)`

is safe.