## College Algebra (10th Edition)

Therefore, $\overline{zw}=\overline{z}\cdot\overline{w}$
$\overline{zw}= \overline{(a+bi)(c+di)}=$ $=\overline{(ac-bd)+(ad+bc)i}$ $=(ac-bd)-(ad+bc)i$ $\overline{z}\cdot\overline{w}=(a-bi)(c-di)$ $=ac-adi-bci-bd$ $=(ac-bd)-(ad+bc)i$