Question

4(cos150°+i sin150°) *1/6(cos120°+ i sin120°)

Answer 1

$4(\cos150а+i\sin150а)\cdot \dfrac{1}{6} \cdot (\cos120а+i\sin120а)=\\ \\ \\ = \dfrac{4}{6} \cdot(\cos(150а+120а)+i\sin(150а+120а))= \dfrac{2}{3} (\cos270а+i\sin270а)\\ \\ \\ = \dfrac{2}{3} \cdot(0+i\cdot(-1))=- \dfrac{2i}{3}$