Answer
(a) $ 36(cos130^\circ+i\ sin130^\circ)$
(b) $ 2\sqrt 3+2i$
(c) $ -\frac{27\sqrt 3}{2}+\frac{27}{2}i$
Work Step by Step
Given $w= 12(cos80^\circ+i\ sin80^\circ)$ and $z= 3(cos50^\circ+i\ sin50^\circ)$, we have:
(a) $wz=(12)(3)(cos(80+50)^\circ+i\ sin(80+50)^\circ)=36(cos130^\circ+i\ sin130^\circ)$
(b) $\frac{w}{z}=(\frac{12}{3})(cos(80-50)^\circ+i\ sin(80-50)^\circ)=4(\frac{\sqrt 3}{2}+i\ \frac{1}{2})=2\sqrt 3+2i$
(c) $z^3=3^3(cos(3\cdot50)^\circ+i\ sin(3\cdot50)^\circ)=27(-\frac{\sqrt 3}{2}+i\ \frac{1}{2})=-\frac{27\sqrt 3}{2}+\frac{27}{2}i$
