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