QuestionUse $z=9 \sqrt{3}+9 i$ and $w=-\frac{\sqrt{3}}{6}-\frac{1}{6} i$ to compute the quantity. Express your answers in polar form $r \operatorname{cis}(\theta)$. The exercise should be worked out without the aid of a calculator. Give your answer in trigonometric form, with $0 \leq \theta<2 \pi$ : $$\frac{w}{z^{2}}=\square \operatorname{cis}(\square)$$ Submit Question