Answer
See below.
Work Step by Step
$\overline{(a_1+b_1i)+(a_2+b_2i)}=\overline{a_1+a_2+i(b_2+b_1)}=a_1+a_2-i(b_2+b_1)$
$\overline{(a_1+b_1i)}+\overline{(a_2+b_2i)}=(a_1-b_1i)+(a_2-b_2i)=a_1+a_2-i(b_2+b_1)$
These are equal, thus we proved what we had to.
