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

Jan 1, 2017

$\overline{\sqrt{8}} = \sqrt{8} = 2 \sqrt{2}$

#### Explanation:

In general, if $a$ and $b$ are real, then the complex conjugate of:

$a + b i$

is:

$a - b i$

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

$\overline{a + b i} = a - b i$

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

$\overline{a} = \overline{a + 0 i} = a - 0 i = a$

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

Now $\sqrt{8}$ is a real number, so:

$\overline{\sqrt{8}} = \sqrt{8}$

If you prefer, you can simplify $\sqrt{8}$ to $2 \sqrt{2}$, since:

$\sqrt{8} = \sqrt{{2}^{2} \cdot 2} = \sqrt{{2}^{2}} \cdot \sqrt{2} = 2 \sqrt{2}$

$\textcolor{w h i t e}{}$
Footnote

$\sqrt{8}$ has another conjugate, called the radical conjugate.

If $\sqrt{n}$ is irrational, and $a , b$ are rational numbers, then the radical conjugate of:

$a + b \sqrt{n}$

is:

$a - b \sqrt{n}$

This has the property that:

$\left(a + b \sqrt{n}\right) \left(a - b \sqrt{n}\right) = {a}^{2} - n {b}^{2}$

hence is often used to rationalise denominators.

The radical conjugate of $\sqrt{8}$ is $- \sqrt{8}$.

The complex conjugate is similar to the radical conjugate, but with $n = - 1$.