Question

Can someone answer the question?​

Answer 1

Answer:

$\frac{\sqrt{3} }{3} }(cos(\frac{19\pi}{12})+isin(\frac{19\pi}{12}))$

Step-by-step explanation:

Division of complex numbers in polar form is $z_1/z_2=\frac{r_1}{r_2}cis(\theta_1-\theta_2)$ where $z_1$ and $z_2$ are the complex numbers being divided, $r_1$ and $r_2$ are the moduli, $\theta_1$ and $\theta_2$ are the arguments, and $cis$ is shorthand for $cos\theta+isin\theta$. Therefore:

$\frac{9(cos(\frac{11\pi}{6})+isin(\frac{11\pi}{6})) }{3\sqrt{3}((cos\frac{\pi}{4})+isin(\frac{\pi}{4})) }$

$\frac{9}{3\sqrt{3} }cis(\frac{11\pi}{6}-\frac{\pi}{4})$

$\frac{\sqrt{3} }{3} }cis(\frac{19\pi}{12})$

$\frac{\sqrt{3} }{3} }(cos(\frac{19\pi}{12})+isin(\frac{19\pi}{12}))$