Answer
$zw = -64(cos(\frac{\pi}{6})-isin(\frac{\pi}{6}))$
$\frac{z}{w} = cos(\frac{\pi}{6})-isin(\frac{\pi}{6}) $
$\frac{1}{z}=\frac{1}{8}(cos(\frac{\pi}{6})+isin(\frac{\pi}{6}))$
Work Step by Step
$r = abs(z) =\sqrt (a^2 +b^2)$
$theta = arctan(\frac{b}{a})$
$z=r(cos(theta) +isin(theta))$
$z = 8(cos(\frac{\pi}{6})-isin(\frac{\pi}{6}))$
$w = 8(cos(\frac{\pi}{2})-isin(\frac{\pi}{2})$
Use normal algebra for $zw$ and $\frac{z}{w}$ .
$zw$ = $8(cos(\frac{\pi}{6})-isin(\frac{\pi}{6}))$ X $8(cos(\frac{\pi}{2})-isin(\frac{\pi}{2})$
$zw$ = $8(cos(\frac{\pi}{6} + \frac{\pi}{2} )-isin(\frac{\pi}{6} + \frac{\pi}{2}))$
For $\frac{1}{z}:$
$\frac{1}{z} = \frac{1}{8(cos(\frac{\pi}{6})-isin(\frac{\pi}{6}))}
$
but we know that $\frac{1}{cos(theta)-isin(theta)} = cos(theta)+isin(theta)$
Therefore $\frac{1}{z}=\frac{1}{8}(cos(\frac{\pi}{6})+isin(\frac{\pi}{6}))$
