To solve for the modulus $r$ and argument $\theta$ of the complex number $z = \frac{v^3}{w^2}$, we will follow these steps:

Step 1: Express $v$ and $w$ in standard form

Given: $v = -1 - i$ $w = -3 - 3\sqrt{3}i$

Step 2: Compute the modulus and argument of $v$ and $w$

1. Modulus of $v$: $|v| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}$

2. Argument of $v$: $\theta_v = \text{arg}(-1 - i) = \tan^{-1}\left(\frac{-1}{-1}\right) = \tan^{-1}(1) = -\frac{3\pi}{4}$ (since $v$ lies in the third quadrant).

3. Modulus of $w$: $|w| = \sqrt{(-3)^2 + (-3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6$

4. Argument of $w$: $\theta_w = \text{arg}(-3 - 3\sqrt{3}i) = \tan^{-1}\left(\frac{-3\sqrt{3}}{-3}\right) = \tan^{-1}(\sqrt{3}) = -\frac{2\pi}{3}$ (since $w$ lies in the third quadrant).

Step 3: Compute the modulus and argument of $z$

Using the properties of modulus and arguments of complex numbers:

1. Modulus of $z$: $|z| = \left|\frac{v^3}{w^2}\right| = \frac{|v|^3}{|w|^2} = \frac{(\sqrt{2})^3}{6^2} = \frac{2\sqrt{2}}{36} = \frac{\sqrt{2}}{18}$
2. Argument of $z$: $\theta_z = 3\theta_v - 2\theta_w = 3\left(-\frac{3\pi}{4}\right) - 2\left(-\frac{2\pi}{3}\right) = -\frac{9\pi}{4} + \frac{4\pi}{3} = -\frac{27\pi + 16\pi}{12} = -\frac{11\pi}{12}$

Thus, the modulus and argument of $z$ are: $r = \frac{\sqrt{2}}{18}, \quad \theta = -\frac{11\pi}{12}$

Would you like further details or have any questions?

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Here are 5 questions to deepen your understanding:

1. How do you calculate the modulus of a complex number?
2. Why does the argument depend on which quadrant the complex number lies in?
3. Can the argument of a complex number exceed $2\pi$ or go below $0$?
4. How does raising a complex number to a power affect its modulus and argument?
5. What happens to the argument of a quotient of two complex numbers?

Tip: Always express angles for arguments in radians unless specifically asked for degrees, as radians are standard in most complex number problems.