# What is the complex conjugate of 8 - sqrt3?

Complex conjugate of $8 - \sqrt{3}$ is $8 - \sqrt{3}$, itself.
Complex conjugate of a number $a + i b$, where $a$and $b$ are two real numbers and $i$ is imaginary number such that ${i}^{2} = - 1$, is $a - i b$
Here we just have a real number $8 - \sqrt{3}$ i.e. our complex number does not have imaginary part, whose sign changes in its conjugate.
Hence complex conjugate of $8 - \sqrt{3}$ is $8 - \sqrt{3}$, itself.