Answer:
[tex] D)\displaystyle \rm \frac{ \sqrt{3} }{12} cis \left( \frac{19\pi}{15} \right)[/tex]
Step-by-step explanation:
remember that
[tex] \displaystyle a \rm cis( \theta) \times b cis( \theta _{2}) = ab \cdot cis( \theta + \theta _{2})[/tex]
so we acquire:
[tex] \displaystyle \frac{ \sqrt{2} }{6} \rm cis \left( \frac{3\pi}{5} \right) \times \frac{ \sqrt{6} }{4} cis \left( \frac{2\pi}{3} \right) = \frac{ \sqrt{2} }{6} \times \frac{ \sqrt{6} }{4} \cdot cis \left( \frac{3\pi}{5} + \frac{2\pi}{3} \right)[/tex]
simplify the left hand side
[tex] \displaystyle \rm \frac{ \sqrt{12} }{24} \cdot cis \left( \frac{3\pi}{5} + \frac{2\pi}{3} \right)[/tex]
[tex] \displaystyle \rm \frac{2 \sqrt{3} }{24} \cdot cis \left( \frac{3\pi}{5} + \frac{2\pi}{3} \right)[/tex]
reduce fraction:
[tex] \displaystyle \rm \frac{ \sqrt{3} }{12} \cdot cis \left( \frac{3\pi}{5} + \frac{2\pi}{3} \right)[/tex]
simplify addition:
[tex] \displaystyle \rm \frac{ \sqrt{3} }{12} cis \left( \frac{19\pi}{15} \right)[/tex]
hence our answer is D
