Answer
$-\frac{1}{6}-\frac{\sqrt{3}}{6}i$
Work Step by Step
First, we use the division theorem to divide the absolute values and subtract the arguments:
$\frac{3(cis 305^{\circ})}{9(cis65^{\circ})}
\\=\frac{1}{3}cis(305^{\circ}-65^{\circ})
\\=\frac{1}{3}cis240^{\circ}$
Next, we change the expression into its equivalent form:
$=\frac{1}{3} cis 240^{\circ}
\\=\frac{1}{3}(\cos240^{\circ}+i\sin240^{\circ})$
Since we know that $\cos240^{\circ}=-\frac{1}{2}$ and $\sin240^{\circ}=-\frac{\sqrt{3}}{2}$, we can substitute these values in the expression and simplify:
$\frac{1}{3} (\cos240^{\circ}+i\sin240^{\circ})
\\=\frac{1}{3}(-\frac{1}{2}-\frac{\sqrt{3}}{2}i)
\\=-\frac{1}{6}-\frac{\sqrt{3}}{6}i$
