# What is the complex conjugate of #sqrt(8)#?

##### 1 Answer

#### Answer:

#### Explanation:

In general, if

#a+bi#

is:

#a-bi#

Complex conjugates are often denoted by placing a bar over an expression, so we can write:

#bar(a+bi) = a-bi#

Any real number is also a complex number, but with a zero imaginary part. So we have:

#bar(a) = bar(a+0i) = a-0i = a#

That is, the complex conjugate of any real number is itself.

Now

#bar(sqrt(8)) = sqrt(8)#

If you prefer, you can simplify

#sqrt(8) = sqrt(2^2*2) = sqrt(2^2)*sqrt(2) = 2sqrt(2)#

**Footnote**

If

#a+bsqrt(n)#

is:

#a-bsqrt(n)#

This has the property that:

#(a+bsqrt(n))(a-bsqrt(n)) = a^2-n b^2#

hence is often used to rationalise denominators.

The radical conjugate of

The complex conjugate is similar to the radical conjugate, but with