What is the complex conjugate of #8 - sqrt3#?

1 Answer
Jul 23, 2017

Answer:

Complex conjugate of #8-sqrt3# is #8-sqrt3#, itself.

Explanation:

Complex conjugate of a number #a+ib#, where #a#and #b# are two real numbers and #i# is imaginary number such that #i^2=-1#, is #a-ib#

Here we just have a real number #8-sqrt3# i.e. our complex number does not have imaginary part, whose sign changes in its conjugate.

Hence complex conjugate of #8-sqrt3# is #8-sqrt3#, itself.