## Algebra and Trigonometry 10th Edition

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