# How do you express #sqrt(-4/5)# as a product of a real number and i?

##### 2 Answers

#### Explanation:

We will be using two facts:

#i = sqrt(-1)# #sqrt(ab) = sqrt(a) sqrt(b)# if#a>=0# or#b>=0#

With these, we have

Note that just as with non-complex square roots, if we are not speaking of the principal square root, we have

Thus

#### Explanation:

Normally speaking, every non-zero number has two square roots. In fact, in a technical sense,

When we use the square root symbol

If we can do this then the square root we denote by

If

One of the nice *bonus* properties of square roots of positive numbers is that

If

These are unambiguous and helpful definitions, leading to the answer given above.

Note however that with this definition, the property

It does get more complicated when you are dealing with square roots of Complex numbers in general:

If

#sqrt((sqrt(a^2+b^2)+a)/2) + (sqrt((sqrt(a^2+b^2)-a)/2)) i#

#-sqrt((sqrt(a^2+b^2)+a)/2) - (sqrt((sqrt(a^2+b^2)-a)/2)) i#

We would probably pick the first of these as the *principal* square root as both its Real and imaginary parts are positive.

If

#sqrt((sqrt(a^2+b^2)+a)/2) - (sqrt((sqrt(a^2+b^2)-a)/2)) i#

#-sqrt((sqrt(a^2+b^2)+a)/2) + (sqrt((sqrt(a^2+b^2)-a)/2)) i#

In this case it is not obvious which one should be called the *principal* square root.

The answer depends on whether you like to consider your Complex numbers to have an angle in the range