write $z = 9 - 3\\sqrt{3}i$ in polar form.\n(1 point)\n$\\bigcirc \\ z = 3\\sqrt{6}\\text{cis}\\frac{11\\pi}{…

Explanation:

Step1: Calculate modulus $r$

For complex number $z=a+bi$, $r=\sqrt{a^2+b^2}$. Here $a=9$, $b=-3\sqrt{3}$. $$ r=\sqrt{9^2 + (-3\sqrt{3})^2}=\sqrt{81 + 27}=\sqrt{108}=6\sqrt{3} $$

Step2: Calculate argument $\theta$

First find reference angle: $\tan\theta_{ref}=\left|\frac{b}{a}\right|=\frac{3\sqrt{3}}{9}=\frac{\sqrt{3}}{3}$, so $\theta_{ref}=\frac{\pi}{6}$. Since $z$ is in 4th quadrant, $\theta=2\pi-\frac{\pi}{6}=\frac{11\pi}{6}$.

Step3: Write polar form

Polar form is $z=r\ cis\theta$.

Answer:

$\boldsymbol{z = 6\sqrt{3} \cis \frac{11\pi}{6}}$ (second option)