Why are the coefficients of the complementary function of a second order differential equation with complex roots conjugate pairs?

Question

Once I have the complementary function of a second order differential equation with complex roots; $y=Ae^{(a+bi)x} + Be^{(a-bi)x}$

I expanded this out to $e^{ax} ((A+B)\cos(bx) + i(A-B)\sin(bx))$

Apparently $A+B$ can be written as $C$ and $i(A-B)$ can be written as $D$ where $C$ and $D$ are real numbers as $A$ and $B$ are conjugate pairs. The textbook explains why they are conjugate by using an example question but I wanted to know more generally why the coefficients $A$ and $B$ are conjugate pairs. Thanks for any help.

Answer 1

$y$ is a real function, $y=\bar y$. Comparing coefficients in $$ Ae^{(a+bi)x}+Be^{(a-bi)x}=y(x)=\overline{y(x)}=\bar Ae^{(a-bi)x}+\bar Be^{(a+bi)x} $$ gives directly $B=\bar A$.