What is the complex conjugate of #sqrt(8)#?
1 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