Answer
$zw=2\sqrt 2(cos255^\circ+i\ sin255^\circ), \frac{z}{w}=\frac{\sqrt 2}{2}(cos15^\circ+i\ sin15^\circ)$
Work Step by Step
Given $z=1-i=\sqrt 2(cos315^\circ+i\ sin315^\circ)$ and $w=1-\sqrt 3i= 2(cos300^\circ+i\ sin300^\circ)$
1. $zw=2\sqrt 2(cos(315+300)^\circ+i\ sin(315+300)^\circ)=2\sqrt 2(cos255^\circ+i\ sin255^\circ)$
2. $\frac{z}{w}=\frac{\sqrt 2}{2}(cos(315-300)^\circ+i\ sin(315-300)^\circ)=\frac{\sqrt 2}{2}(cos15^\circ+i\ sin15^\circ)$
