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

2 Answers
Oct 10, 2017

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)#

Oct 13, 2017

#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.