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

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Answer 1 · 2017-01-01T09:02:55

#bar(sqrt(8)) = sqrt(8) = 2sqrt(2)#

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

#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 #sqrt(8)# is a real number, so:

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

If you prefer, you can simplify #sqrt(8)# to #2sqrt(2)#, since:

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

#color(white)()#
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+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 #sqrt(8)# is #-sqrt(8)#.

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