Given two complex numbers $Z_a = 1 + \sqrt{3}i$ and $Z_b = 2 - 2i$, we are asked to: I. Convert $Z_a$ and $Z_b$ into polar form using principal angles. II. Evaluate $Z_a Z_b$ using the polar forms from part I. III. Evaluate $\frac{Z_a}{Z_b}$ using the polar forms from part I. IV. Determine $(Z_b)^5$ using De Moivre's theorem. V. Find all the cube roots of $Z_a$, i.e., find all solutions to $z^3 = Z_a$.

1. Problem Description

Given two complex numbers $Z_a = 1 + \sqrt{3}i$ and $Z_b = 2 - 2i$, we are asked to:

I. Convert $Z_a$ and $Z_b$ into polar form using principal angles.

II. Evaluate $Z_a Z_b$ using the polar forms from part I.

III. Evaluate $\frac{Z_a}{Z_b}$ using the polar forms from part I.

IV. Determine $(Z_b)^5$ using De Moivre's theorem.

V. Find all the cube roots of $Z_a$, i.e., find all solutions to $z^3 = Z_a$.

2. Solution Steps

I. Converting to polar form:

For a complex number $z = a + bi$, the polar form is $z = r(\cos\theta + i\sin\theta)$, where $r = \sqrt{a^2 + b^2}$ and $\theta = \arctan(\frac{b}{a})$.

For $Z_a = 1 + \sqrt{3}i$:

$r_a = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$.

$\theta_a = \arctan(\frac{\sqrt{3}}{1}) = \frac{\pi}{3}$.

So, $Z_a = 2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3})$.

For $Z_b = 2 - 2i$:

$r_b = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.

$\theta_b = \arctan(\frac{-2}{2}) = \arctan(-1) = -\frac{\pi}{4}$.

So, $Z_b = 2\sqrt{2}(\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}))$.

II. Evaluating $Z_a Z_b$:

If $Z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $Z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$, then

$Z_1 Z_2 = r_1 r_2 (\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2))$.

$Z_a Z_b = (2)(2\sqrt{2})(\cos(\frac{\pi}{3} - \frac{\pi}{4}) + i\sin(\frac{\pi}{3} - \frac{\pi}{4}))$.

$Z_a Z_b = 4\sqrt{2}(\cos(\frac{4\pi - 3\pi}{12}) + i\sin(\frac{4\pi - 3\pi}{12}))$.

$Z_a Z_b = 4\sqrt{2}(\cos(\frac{\pi}{12}) + i\sin(\frac{\pi}{12}))$.

III. Evaluating $\frac{Z_a}{Z_b}$:

If $Z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $Z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$, then

$\frac{Z_1}{Z_2} = \frac{r_1}{r_2} (\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2))$.

$\frac{Z_a}{Z_b} = \frac{2}{2\sqrt{2}}(\cos(\frac{\pi}{3} - (-\frac{\pi}{4})) + i\sin(\frac{\pi}{3} - (-\frac{\pi}{4}))) = \frac{1}{\sqrt{2}}(\cos(\frac{\pi}{3} + \frac{\pi}{4}) + i\sin(\frac{\pi}{3} + \frac{\pi}{4}))$.

$\frac{Z_a}{Z_b} = \frac{1}{\sqrt{2}}(\cos(\frac{4\pi + 3\pi}{12}) + i\sin(\frac{4\pi + 3\pi}{12})) = \frac{1}{\sqrt{2}}(\cos(\frac{7\pi}{12}) + i\sin(\frac{7\pi}{12}))$.

IV. Determining $(Z_b)^5$:

If $Z = r(\cos\theta + i\sin\theta)$, then $Z^n = r^n(\cos(n\theta) + i\sin(n\theta))$.

$(Z_b)^5 = (2\sqrt{2})^5(\cos(5(-\frac{\pi}{4})) + i\sin(5(-\frac{\pi}{4}))) = (2\sqrt{2})^5(\cos(-\frac{5\pi}{4}) + i\sin(-\frac{5\pi}{4}))$.

$(2\sqrt{2})^5 = (2^{3/2})^5 = 2^{15/2} = 2^7\sqrt{2} = 128\sqrt{2}$.

$(Z_b)^5 = 128\sqrt{2}(\cos(-\frac{5\pi}{4}) + i\sin(-\frac{5\pi}{4})) = 128\sqrt{2}(-\frac{1}{\sqrt{2}} + i(-\frac{1}{\sqrt{2}})) = 128\sqrt{2}(-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}) = 128(-1 - i) = -128 - 128i$.

V. Finding the cube roots of $Z_a$:

$Z_a = 2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))$. We want to find $z$ such that $z^3 = Z_a$.

Let $z = r(\cos\theta + i\sin\theta)$. Then $z^3 = r^3(\cos(3\theta) + i\sin(3\theta))$.

So $r^3 = 2$, which means $r = \sqrt[3]{2}$.

$3\theta = \frac{\pi}{3} + 2\pi k$ for $k = 0, 1, 2$.

$\theta = \frac{\pi}{9} + \frac{2\pi k}{3}$ for $k = 0, 1, 2$.

For $k = 0$: $\theta_0 = \frac{\pi}{9}$. $z_0 = \sqrt[3]{2}(\cos(\frac{\pi}{9}) + i\sin(\frac{\pi}{9}))$.

For $k = 1$: $\theta_1 = \frac{\pi}{9} + \frac{2\pi}{3} = \frac{\pi + 6\pi}{9} = \frac{7\pi}{9}$. $z_1 = \sqrt[3]{2}(\cos(\frac{7\pi}{9}) + i\sin(\frac{7\pi}{9}))$.

For $k = 2$: $\theta_2 = \frac{\pi}{9} + \frac{4\pi}{3} = \frac{\pi + 12\pi}{9} = \frac{13\pi}{9}$. $z_2 = \sqrt[3]{2}(\cos(\frac{13\pi}{9}) + i\sin(\frac{13\pi}{9}))$.

3. Final Answer

I. $Z_a = 2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))$, $Z_b = 2\sqrt{2}(\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}))$

II. $Z_a Z_b = 4\sqrt{2}(\cos(\frac{\pi}{12}) + i\sin(\frac{\pi}{12}))$

III. $\frac{Z_a}{Z_b} = \frac{1}{\sqrt{2}}(\cos(\frac{7\pi}{12}) + i\sin(\frac{7\pi}{12}))$

IV. $(Z_b)^5 = -128 - 128i$

V. $z_0 = \sqrt[3]{2}(\cos(\frac{\pi}{9}) + i\sin(\frac{\pi}{9}))$, $z_1 = \sqrt[3]{2}(\cos(\frac{7\pi}{9}) + i\sin(\frac{7\pi}{9}))$, $z_2 = \sqrt[3]{2}(\cos(\frac{13\pi}{9}) + i\sin(\frac{13\pi}{9}))$