Answer
a) $4\sqrt 2 [\cos (31\pi/12)+i \sin (31\pi/12)]$
b) $2 \sqrt 2 [\cos (13\pi/12)+i \sin (13\pi/12)]$
c) $\dfrac{1}{4}[\cos (-11\pi/6)+i \sin (-11\pi/6)]$
Work Step by Step
a) Here, we have $z=4 [\cos (\dfrac{11\pi}{6})+i \sin (\dfrac{11\pi}{6})]$ and $w=\sqrt 2 [\cos (3\pi/4)+i \sin (3\pi/4)]$
$zw=4 [\cos(\dfrac{11\pi}{6})+i \sin (\dfrac{11\pi}{6})] \times \sqrt 2 [\cos (3\pi/4)+i \sin (3\pi/4)]=4\sqrt 2 [\cos (31\pi/12)+i \sin (31\pi/12)]$
b) Here, we have $z=4 [\cos (\dfrac{11\pi}{6})+i \sin (\dfrac{11\pi}{6})]$ and $w=\sqrt 2 [\cos (3\pi/4)+i \sin (3\pi/4)]$
$\dfrac{z}{w}=\dfrac{4 [\cos (11\pi/6)+i \sin (11\pi/6)]}{\sqrt 2 [\cos (3\pi/4)+i \sin (3\pi/4)]}=2 \sqrt 2 [\cos (13\pi/12)+i \sin (13\pi/12)]$
c) Here, we have $z=4 [\cos (\dfrac{11\pi}{6})+i \sin (\dfrac{11\pi}{6})]$ and $w=\sqrt 2 [\cos (3\pi/4)+i \sin (3\pi/4)]$
Thus, we have $\dfrac{1}{z}=\dfrac{1}{4}[\cos (-11\pi/6)+i \sin (-11\pi/6)]$
