De Moivre's theorem and Roots of unity Questions in English

(D) Let $z_1 = 1-\sqrt{3}i$ and $z_2 = \sqrt{3}+i$.
Converting to polar form:
$z_1 = 2(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3})) = 2e^{-i\pi/3}$.
$z_2 = 2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6})) = 2e^{i\pi/6}$.
Now,$z_1^9 = 2^9 e^{-i3\pi} = 2^9(\cos(-3\pi) + i\sin(-3\pi)) = 2^9(-1) = -2^9$.
$z_2^9 = 2^9 e^{i3\pi/2} = 2^9(\cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2})) = 2^9(0 - i) = -i2^9$.
Then,$|z_1^9 + z_2^9| = |-2^9 - i2^9| = |2^9(-1-i)| = 2^9|-1-i|$.
$|z_1^9 + z_2^9| = 2^9 \sqrt{(-1)^2 + (-1)^2} = 2^9 \sqrt{2} = 2^9 \cdot 2^{1/2} = 2^{19/2}$.
Thus,the correct option is $D$.

(C) Given $z = \frac{-1-i \sqrt{3}}{2} = \omega$,where $\omega$ is the complex cube root of unity.
We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
Also,$\frac{1}{\omega} = \omega^2$.
The expression is $\sum_{k=1}^{2022} (\omega^k + \omega^{2k})^2 = \sum_{k=1}^{2022} (\omega^{2k} + \omega^{4k} + 2\omega^{3k})$.
Since $\omega^{3k} = 1$,this becomes $\sum_{k=1}^{2022} (\omega^{2k} + \omega^k + 2)$.
We can split the sum: $\sum_{k=1}^{2022} \omega^{2k} + \sum_{k=1}^{2022} \omega^k + \sum_{k=1}^{2022} 2$.
Since $2022$ is a multiple of $3$,the sum of powers of $\omega$ over $3$ terms is $0$.
Thus,$\sum_{k=1}^{2022} \omega^k = 0$ and $\sum_{k=1}^{2022} \omega^{2k} = 0$.
The total sum is $0 + 0 + 2 \times 2022 = 4044$.

(B) Given,$k(\cos \theta+i \sin \theta)=2+2 \sqrt{3} i$.
Comparing the modulus and argument,we have $k = \sqrt{2^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4$.
Also,$\cos \theta = \frac{2}{4} = \frac{1}{2}$ and $\sin \theta = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$,which implies $\theta = \frac{\pi}{3}$.
Now,we need to evaluate $\frac{1}{\sqrt{3}}[\cos 6 \theta+i \sin 6 \theta]$.
Substituting $\theta = \frac{\pi}{3}$,we get $6\theta = 6(\frac{\pi}{3}) = 2\pi$.
Thus,$\frac{1}{\sqrt{3}}[\cos 2\pi + i \sin 2\pi] = \frac{1}{\sqrt{3}}[1 + i(0)] = \frac{1}{\sqrt{3}}$.

(D) Let $z = \sqrt{3} + i$.
Then,the modulus $|z| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$.
The argument $\theta = \arg(z) = \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.
Thus,the polar form is $z = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$.
Using De Moivre's Theorem,$z^{10} = 2^{10}(\cos \frac{10\pi}{6} + i \sin \frac{10\pi}{6})$.
Simplifying the angle,$\frac{10\pi}{6} = \frac{5\pi}{3}$.
$z^{10} = 1024(\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3})$.
Since $\cos \frac{5\pi}{3} = \frac{1}{2}$ and $\sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2}$,
$z^{10} = 1024(\frac{1}{2} - i \frac{\sqrt{3}}{2}) = 512 - 512i\sqrt{3}$.
Comparing with $a+bi$,we get $a = 512$ and $b = -512\sqrt{3}$.

(D) We have,
$\sum_{r=1}^{16}\left(\sin \frac{2 r \pi}{17}+i \cos \frac{2 r \pi}{17}\right) = i \sum_{r=1}^{16}\left(\cos \frac{2 r \pi}{17} - i \sin \frac{2 r \pi}{17}\right)$
$= i \sum_{r=1}^{16} e^{-\frac{i 2 r \pi}{17}}$
Let $K = e^{-\frac{2 i \pi}{17}}$. Then the sum is $i \sum_{r=1}^{16} K^r$.
This is a geometric series with $16$ terms:
$i \left[ K \frac{1-K^{16}}{1-K} \right] = i \frac{K-K^{17}}{1-K}$
Since $K^{17} = e^{-2 i \pi} = \cos(2 \pi) - i \sin(2 \pi) = 1$,
The expression becomes $i \frac{K-1}{1-K} = i(-1) = -i$.

(D) Let $z = \sqrt{3}-i$. In polar form,$z = 2(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})) = 2e^{-i\frac{\pi}{6}}$.
We want to find the product of all values of $w = z^{3/7} = (2e^{-i\frac{\pi}{6} + 2k\pi i})^{3/7}$ for $k = 0, 1, 2, 3, 4, 5, 6$.
This simplifies to $w_k = 2^{3/7} e^{i(-\frac{\pi}{14} + \frac{6k\pi}{7})}$.
The product of these $7$ values is $P = \prod_{k=0}^{6} 2^{3/7} e^{i(-\frac{\pi}{14} + \frac{6k\pi}{7})}$.
$P = (2^{3/7})^7 \cdot e^{i \sum_{k=0}^{6} (-\frac{\pi}{14} + \frac{6k\pi}{7})} = 8 \cdot e^{i (-\frac{7\pi}{14} + \frac{6\pi}{7} \cdot \frac{6 \cdot 7}{2})} = 8 \cdot e^{i (-\frac{\pi}{2} + 18\pi)} = 8 \cdot e^{-i\frac{\pi}{2}} = 8(-i) = -8i$.

(C) Given $|Z|=2$ and $\operatorname{amp}(Z)=\theta$, we can write $Z=2e^{i\theta}$.
Then $Z_1 = \frac{2e^{i\theta}}{2} e^{i\alpha} = e^{i(\theta+\alpha)}$.
Now, $Z_1^n = (e^{i(\theta+\alpha)})^n = e^{in(\theta+\alpha)} = \cos(n(\theta+\alpha)) + i\sin(n(\theta+\alpha))$.
Similarly, $Z_1^{-n} = e^{-in(\theta+\alpha)} = \cos(n(\theta+\alpha)) - i\sin(n(\theta+\alpha))$.
Substituting these into the expression:
$\frac{Z_1^n-Z_1^{-n}}{Z_1^n+Z_1^{-n}} = \frac{(\cos(n(\theta+\alpha)) + i\sin(n(\theta+\alpha))) - (\cos(n(\theta+\alpha)) - i\sin(n(\theta+\alpha)))}{(\cos(n(\theta+\alpha)) + i\sin(n(\theta+\alpha))) + (\cos(n(\theta+\alpha)) - i\sin(n(\theta+\alpha)))}$
$= \frac{2i\sin(n(\theta+\alpha))}{2\cos(n(\theta+\alpha))} = i\tan(n(\theta+\alpha))$.
Thus, the correct option is $C$.

(C) Let $z = \sqrt{3} + i$. We can write $z$ in polar form as $z = 2(\cos(\frac{\pi}{6}) + i \sin(\frac{\pi}{6}))$.
Then $z^{2n} = 2^{2n}(\cos(\frac{n\pi}{3}) + i \sin(\frac{n\pi}{3}))$.
Similarly,$(\sqrt{3}-i)^{2n} = 2^{2n}(\cos(\frac{n\pi}{3}) - i \sin(\frac{n\pi}{3}))$.
Adding these,we get $2^{2n} \times 2 \cos(\frac{n\pi}{3}) = 2^{2n+1} \cos(\frac{n\pi}{3})$.
Since $n \neq 3K$,$n$ is not a multiple of $3$.
If $n = 3K \pm 1$,then $\cos(\frac{n\pi}{3}) = \cos(\frac{(3K \pm 1)\pi}{3}) = \cos(K\pi \pm \frac{\pi}{3}) = (-1)^K \cos(\frac{\pi}{3}) = (-1)^K \times \frac{1}{2}$.
This does not lead to a simple power of $2$ unless we re-evaluate the expression.
Actually,for $n$ not a multiple of $3$,the expression simplifies based on the specific values.
Given the options,the correct form is $(-1)^{n+1} 2^{n+1}$ is not listed,but evaluating for $n=1$: $(\sqrt{3}+i)^2 + (\sqrt{3}-i)^2 = (3-1+2\sqrt{3}i) + (3-1-2\sqrt{3}i) = 4$.
For $n=1$,option $B$ gives $(-1)^2 2^{2(1)+1} = 8$ (Incorrect).
Wait,let's re-check: $(\sqrt{3}+i)^2 = 2+2\sqrt{3}i$. $(\sqrt{3}-i)^2 = 2-2\sqrt{3}i$. Sum is $4$.
Option $C$ for $n=1$: $(-1)^2 2^{2(1)} = 4$. This matches.
Thus,the correct option is $C$.

(A) Let $z = 1 - i \sqrt{3}$.
In polar form,$z = r(\cos \theta + i \sin \theta)$,where $r = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2$.
The angle $\theta$ is given by $\tan \theta = \frac{-\sqrt{3}}{1} = -\sqrt{3}$. Since $z$ is in the fourth quadrant,$\theta = -\frac{\pi}{3}$.
Thus,$z = 2(\cos(-\frac{\pi}{3}) + i \sin(-\frac{\pi}{3}))$.
The cube roots are given by $w_k = \sqrt[3]{2} \left( \cos \left( \frac{-\pi/3 + 2k\pi}{3} \right) + i \sin \left( \frac{-\pi/3 + 2k\pi}{3} \right) \right)$ for $k = 0, 1, 2$.
For $k=0$: $\theta_0 = -\frac{\pi}{9} = -20^\circ$ (Fourth quadrant).
For $k=1$: $\theta_1 = \frac{-\pi/3 + 2\pi}{3} = \frac{5\pi}{9} = 100^\circ$ (Second quadrant).
For $k=2$: $\theta_2 = \frac{-\pi/3 + 4\pi}{3} = \frac{11\pi}{9} = 220^\circ$ (Third quadrant).
None of the roots lie in the first quadrant.

(B) Let $z = \sqrt{3} + i$. Converting to polar form,$r = \sqrt{(\sqrt{3})^2 + 1^2} = 2$. The argument $\theta = \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.
So,$z = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$.
By De Moivre's Theorem,$z^{10} = 2^{10}(\cos \frac{10\pi}{6} + i \sin \frac{10\pi}{6}) = 1024(\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3})$.
Since $\cos \frac{5\pi}{3} = \frac{1}{2}$ and $\sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2}$,we have $z^{10} = 1024(\frac{1}{2} - i \frac{\sqrt{3}}{2}) = 512 - 512i\sqrt{3}$.
Similarly,for $\bar{z} = \sqrt{3} - i$,$\bar{z}^{10} = 1024(\cos(-\frac{5\pi}{3}) + i \sin(-\frac{5\pi}{3})) = 1024(\frac{1}{2} + i \frac{\sqrt{3}}{2}) = 512 + 512i\sqrt{3}$.
Adding these,$(\sqrt{3}+i)^{10} + (\sqrt{3}-i)^{10} = (512 - 512i\sqrt{3}) + (512 + 512i\sqrt{3}) = 1024$.

(C) Let $z = -1 - \sqrt{3}i$.
First,express $z$ in polar form:
$|z| = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2$.
$\theta = \text{arg}(z) = \pi + \tan^{-1}(\frac{\sqrt{3}}{1}) = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$.
So,$z = 2e^{i(4\pi/3 + 2k\pi)}$ for $k = 0, 1, 2, 3$.
Then $z^{3/4} = (2e^{i(4\pi/3 + 2k\pi)})^{3/4} = 2^{3/4} e^{i(3/4)(4\pi/3 + 2k\pi)} = 2^{3/4} e^{i(\pi + 3k\pi/2)}$.
For the value to be real,the imaginary part must be zero,i.e.,$\sin(\pi + \frac{3k\pi}{2}) = 0$.
This implies $\pi + \frac{3k\pi}{2} = n\pi$ for some integer $n$.
Dividing by $\pi$,we get $1 + \frac{3k}{2} = n$,or $2 + 3k = 2n$.
Since $2n$ is even and $2$ is even,$3k$ must be even,which means $k$ must be even.
For $k \in \{0, 1, 2, 3\}$,the even values are $k = 0$ and $k = 2$.
For $k=0$,$\text{arg} = \pi$,$e^{i\pi} = -1$ (real).
For $k=2$,$\text{arg} = \pi + 3\pi = 4\pi$,$e^{i4\pi} = 1$ (real).
Thus,there are $2$ real values.

(C) Let $z = 1 - i \sqrt{3}$.
First,find the polar form of $z$.
The modulus $r = |z| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2$.
The argument $\theta = \tan^{-1}(\frac{-\sqrt{3}}{1}) = -\frac{\pi}{3}$.
So,$z = 2(\cos(-\frac{\pi}{3}) + i \sin(-\frac{\pi}{3}))$.
Using De Moivre's Theorem,$z^{2025} = 2^{2025}(\cos(2025 \times -\frac{\pi}{3}) + i \sin(2025 \times -\frac{\pi}{3}))$.
$2025 \times -\frac{\pi}{3} = -675\pi$.
Since $-675\pi = -674\pi - \pi$,the angle is coterminal with $-\pi$.
$z^{2025} = 2^{2025}(\cos(-\pi) + i \sin(-\pi)) = 2^{2025}(-1 + 0i) = -2^{2025}$.

(C) Given that $Z = r(\cos \theta + i \sin \theta)$ is a solution of $x^3 = i$.
By De Moivre's Theorem,$Z^3 = r^3(\cos 3\theta + i \sin 3\theta)$.
Since $Z^3 = i$,we have $r^3(\cos 3\theta + i \sin 3\theta) = i$.
We know that $i = \cos(\pi / 2) + i \sin(\pi / 2)$.
Thus,$r^9(\cos \theta + i \sin \theta)^9 = (r^3(\cos \theta + i \sin \theta)^3)^3$.
Since $Z^3 = i$,this expression becomes $(i)^3$.
Calculating the value: $i^3 = i^2 \times i = -1 \times i = -i$.

(B) We have $(-64 i)^{5 / 6} = (64)^{5 / 6} \times (-i)^{5 / 6}$.
Since $-i = \cos(\frac{3\pi}{2}) + i \sin(\frac{3\pi}{2}) = e^{i(\frac{3\pi}{2} + 2k\pi)}$,we have:
$(-64 i)^{5 / 6} = 32 \times (e^{i(\frac{3\pi}{2} + 2k\pi)})^{5 / 6} = 32 \times e^{i(\frac{15\pi}{12} + \frac{10k\pi}{6})}$.
For $k = 3$,the exponent becomes $i(\frac{15\pi}{12} + 5\pi) = i(\frac{5\pi}{4} + 5\pi) = i(\frac{25\pi}{4})$.
Since $\frac{25\pi}{4} = 6\pi + \frac{\pi}{4}$,we have $e^{i(6\pi + \pi/4)} = e^{i\pi/4} = \cos(\frac{\pi}{4}) + i \sin(\frac{\pi}{4})$.
Thus,the value is $32(\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}) = 16\sqrt{2}(1+i)$.

(C) Let $Z = (\sqrt{3} - i)^{\frac{1}{6}}$,so $Z^6 = \sqrt{3} - i$.
Express $\sqrt{3} - i$ in polar form:
$Z^6 = 2 \left( \frac{\sqrt{3}}{2} - i \frac{1}{2} \right) = 2 \left( \cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6}) \right) = 2 \operatorname{cis}(-\frac{\pi}{6})$.
Using the general form,$Z^6 = 2 \operatorname{cis}(2K\pi - \frac{\pi}{6})$ for $K = 0, 1, 2, 3, 4, 5$.
Taking the $6^{th}$ root,$Z = 2^{\frac{1}{6}} \operatorname{cis} \left( \frac{2K\pi - \frac{\pi}{6}}{6} \right) = 2^{\frac{1}{6}} \operatorname{cis} \left( \frac{K\pi}{3} - \frac{\pi}{36} \right)$.
For $K = 5$:
$Z = 2^{\frac{1}{6}} \operatorname{cis} \left( \frac{5\pi}{3} - \frac{\pi}{36} \right) = 2^{\frac{1}{6}} \operatorname{cis} \left( \frac{60\pi - \pi}{36} \right) = 2^{\frac{1}{6}} \operatorname{cis} \left( \frac{59\pi}{36} \right)$.

(D) Let $Z = (\sqrt{3}-i)^{\frac{2}{5}}$.
First,express $\sqrt{3}-i$ in polar form: $\sqrt{3}-i = 2(\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6}))$.
Then,$Z = [2(\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6}))]^{\frac{2}{5}} = 2^{\frac{2}{5}}(\cos(-\frac{\pi}{15}) + i \sin(-\frac{\pi}{15}))$.
Alternatively,using the property $(\sqrt{3}-i)^2 = 3 - 1 - 2\sqrt{3}i = 2 - 2\sqrt{3}i = 4(\frac{1}{2} - i\frac{\sqrt{3}}{2}) = 4(\cos(-\frac{\pi}{3}) + i \sin(-\frac{\pi}{3}))$.
Taking the fifth root,one value is $4^{\frac{1}{5}}(\cos(-\frac{\pi}{15}) + i \sin(-\frac{\pi}{15}))$.
However,checking the options,we evaluate $2^{-\frac{3}{5}}(1+i\sqrt{3}) = 2^{-\frac{3}{5}} \cdot 2(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 2^{\frac{2}{5}}(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3})$.
This corresponds to the principal value of the expression.

(C) Given the equation $\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^m=1$.
First,simplify the base: $\frac{\sqrt{3}+i}{\sqrt{3}-i} = \frac{(\sqrt{3}+i)(\sqrt{3}+i)}{(\sqrt{3}-i)(\sqrt{3}+i)} = \frac{3-1+2\sqrt{3}i}{3+1} = \frac{2+2\sqrt{3}i}{4} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$.
This is equal to $\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right) = e^{i\pi/3}$.
So,the equation becomes $(e^{i\pi/3})^m = 1$,which implies $e^{im\pi/3} = 1$.
This holds if $\frac{m\pi}{3} = 2n\pi$ for some integer $n$,meaning $m = 6n$.
We are given $2022 < m < 2029$.
Checking multiples of $6$: $2022/6 = 337$ and $2028/6 = 338$.
The only multiple of $6$ in the range $(2022, 2029)$ is $2028$.
Thus,$m = 2028$.

(B) Given expression: $(-32 i)^{\frac{2}{5}}$
We can write $-i$ as $\cos \frac{3 \pi}{2} + i \sin \frac{3 \pi}{2}$.
So,$(-32 i)^{\frac{2}{5}} = (32)^{\frac{2}{5}} (-i)^{\frac{2}{5}} = 4 (\cos \frac{3 \pi}{2} + i \sin \frac{3 \pi}{2})^{\frac{2}{5}}$.
Using De Moivre's theorem,$(\cos \theta + i \sin \theta)^n = \cos(n \theta) + i \sin(n \theta)$,we consider the principal value:
$= 4 (\cos(\frac{3 \pi}{2} \times \frac{2}{5}) + i \sin(\frac{3 \pi}{2} \times \frac{2}{5}))$
$= 4 (\cos \frac{3 \pi}{5} + i \sin \frac{3 \pi}{5})$
$= 4 \operatorname{cis} \frac{3 \pi}{5}$.

(A) We know that $\sum_{k=0}^{6} \cos \frac{2 \pi k}{7} = 0$ and $\sum_{k=0}^{6} \sin \frac{2 \pi k}{7} = 0$.
Since $\cos 0 = 1$ and $\sin 0 = 0$,we have $\sum_{k=1}^{6} \cos \frac{2 \pi k}{7} = -1$ and $\sum_{k=1}^{6} \sin \frac{2 \pi k}{7} = 0$.
The given expression is $\sum_{k=1}^{6} (\sin \frac{2 \pi k}{7} - i \cos \frac{2 \pi k}{7}) = \sum_{k=1}^{6} \sin \frac{2 \pi k}{7} - i \sum_{k=1}^{6} \cos \frac{2 \pi k}{7}$.
Substituting the values,we get $0 - i(-1) = i$.

(C) Given $n \in N$,let $z = \left(\frac{1+\cos \theta+i \sin \theta}{1+\cos \theta-i \sin \theta}\right)^n$.
Using the identities $1+\cos \theta = 2 \cos^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$,we get:
$z = \left(\frac{2 \cos^2 \frac{\theta}{2} + 2i \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2} - 2i \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)^n$
$z = \left(\frac{2 \cos \frac{\theta}{2} (\cos \frac{\theta}{2} + i \sin \frac{\theta}{2})}{2 \cos \frac{\theta}{2} (\cos \frac{\theta}{2} - i \sin \frac{\theta}{2})}\right)^n$
$z = \left(\frac{\cos \frac{\theta}{2} + i \sin \frac{\theta}{2}}{\cos \frac{\theta}{2} - i \sin \frac{\theta}{2}}\right)^n$
Using the exponential form $e^{i\phi} = \cos \phi + i \sin \phi$,we have:
$z = \left(\frac{e^{i\theta/2}}{e^{-i\theta/2}}\right)^n = (e^{i\theta})^n = e^{in\theta}$
$z = \cos (n \theta) + i \sin (n \theta)$.

(B) Given that $z^{\frac{1}{5}} = 2 \cos \left(\frac{7 \pi}{5}\right) + 2i \sin \left(\frac{7 \pi}{5}\right)$.
By De Moivre's theorem,if $w = r(\cos \theta + i \sin \theta)$ is a root of $z^{\frac{1}{5}}$,then $z = w^5 = r^5(\cos(5\theta) + i \sin(5\theta))$.
Here,$r = 2$ and $\theta = \frac{7 \pi}{5}$.
So,$z = 2^5 \left(\cos \left(5 \times \frac{7 \pi}{5}\right) + i \sin \left(5 \times \frac{7 \pi}{5}\right)\right)$.
$z = 32(\cos(7 \pi) + i \sin(7 \pi))$.
Since $\cos(7 \pi) = -1$ and $\sin(7 \pi) = 0$,we have $z = 32(-1 + 0i) = -32$.

(D) Given $|a|=1$ and $\arg (a)=\theta$,we have $a = \cos \theta + i \sin \theta = e^{i\theta}$.
The equation is $\left(\frac{1+i z}{1-i z}\right)^4 = e^{i\theta}$.
Taking the fourth root,$\frac{1+i z}{1-i z} = e^{i\left(\frac{2k\pi+\theta}{4}\right)} = \cos \phi + i \sin \phi$,where $\phi = \frac{2k\pi+\theta}{4}$ and $k=0,1,2,3$.
Let $\omega = \frac{2k\pi+\theta}{8}$. Then $\phi = 2\omega$.
Using componendo and dividendo on $\frac{1+iz}{1-iz} = \cos 2\omega + i \sin 2\omega$:
$\frac{1}{iz} = \frac{\cos 2\omega + i \sin 2\omega + 1}{\cos 2\omega + i \sin 2\omega - 1} = \frac{2\cos^2 \omega + 2i \sin \omega \cos \omega}{-2\sin^2 \omega + 2i \sin \omega \cos \omega} = \frac{2\cos \omega (\cos \omega + i \sin \omega)}{2i \sin \omega (\cos \omega + i \sin \omega)} = \frac{\cos \omega}{i \sin \omega} = \frac{1}{i \tan \omega}$.
Thus,$z = \tan \omega = \tan \left(\frac{2k\pi+\theta}{8}\right)$ for $k=0,1,2,3$.

(B) Let $z = \frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}$.
Using the half-angle formulas $1+\cos \theta = 2 \cos^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$,we get:
$z = \frac{2 \cos^2 \frac{\pi}{16} - i (2 \sin \frac{\pi}{16} \cos \frac{\pi}{16})}{2 \cos^2 \frac{\pi}{16} + i (2 \sin \frac{\pi}{16} \cos \frac{\pi}{16})}$
$z = \frac{2 \cos \frac{\pi}{16} (\cos \frac{\pi}{16} - i \sin \frac{\pi}{16})}{2 \cos \frac{\pi}{16} (\cos \frac{\pi}{16} + i \sin \frac{\pi}{16})}$
$z = \frac{\cos \frac{\pi}{16} - i \sin \frac{\pi}{16}}{\cos \frac{\pi}{16} + i \sin \frac{\pi}{16}} = \frac{e^{-i \pi / 16}}{e^{i \pi / 16}} = e^{-i \pi / 8}$.
Now,$z^{12} = (e^{-i \pi / 8})^{12} = e^{-i 12 \pi / 8} = e^{-i 3 \pi / 2}$.
$e^{-i 3 \pi / 2} = \cos(-\frac{3 \pi}{2}) + i \sin(-\frac{3 \pi}{2}) = \cos(\frac{3 \pi}{2}) - i \sin(\frac{3 \pi}{2}) = 0 - i(1) = -i$ is incorrect,let's re-evaluate: $e^{-i 3 \pi / 2} = \cos(\frac{3 \pi}{2}) + i \sin(\frac{3 \pi}{2}) = 0 + i(1) = i$.

(D) Let $z = 1 - i \sqrt{3}$.
First,express $z$ in polar form: $|z| = \sqrt{1^2 + (-\sqrt{3})^2} = 2$.
The argument $\theta$ satisfies $\tan \theta = \frac{-\sqrt{3}}{1} = -\sqrt{3}$. Since $z$ is in the fourth quadrant,$\theta = -\frac{\pi}{3}$ or $2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$.
Thus,$z = 2(\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3})$.
The cube roots of $z$ are given by $z^{1/3} = 2^{1/3} \left[ \cos \left( \frac{5\pi/3 + 2k\pi}{3} \right) + i \sin \left( \frac{5\pi/3 + 2k\pi}{3} \right) \right]$ for $k = 0, 1, 2$.
The arguments are $\theta_k = \frac{5\pi + 6k\pi}{9}$.
For $k=0$,$\theta_0 = \frac{5\pi}{9}$.
For $k=1$,$\theta_1 = \frac{11\pi}{9}$.
For $k=2$,$\theta_2 = \frac{17\pi}{9}$.
All these are positive and less than $2\pi$.
The sum is $\frac{5\pi}{9} + \frac{11\pi}{9} + \frac{17\pi}{9} = \frac{33\pi}{9} = \frac{11\pi}{3}$.

(C) Consider the expression: $\left[\frac{\left(1+\cos \frac{\pi}{12}\right)+i \sin \frac{\pi}{12}}{\left(1+\cos \frac{\pi}{12}\right)-i \sin \frac{\pi}{12}}\right]^{72}$
Using the half-angle identities $1+\cos \theta = 2 \cos^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$,we get:
$\left[\frac{2 \cos^2 \frac{\pi}{24} + i 2 \sin \frac{\pi}{24} \cos \frac{\pi}{24}}{2 \cos^2 \frac{\pi}{24} - i 2 \sin \frac{\pi}{24} \cos \frac{\pi}{24}}\right]^{72}$
Factoring out $2 \cos \frac{\pi}{24}$ from the numerator and denominator:
$\left[\frac{2 \cos \frac{\pi}{24} (\cos \frac{\pi}{24} + i \sin \frac{\pi}{24})}{2 \cos \frac{\pi}{24} (\cos \frac{\pi}{24} - i \sin \frac{\pi}{24})}\right]^{72} = \left(\frac{\cos \frac{\pi}{24} + i \sin \frac{\pi}{24}}{\cos \frac{\pi}{24} - i \sin \frac{\pi}{24}}\right)^{72}$
Using the property $\frac{e^{i\theta}}{e^{-i\theta}} = e^{i2\theta}$,we have:
$\left(e^{i \frac{\pi}{24}} / e^{-i \frac{\pi}{24}}\right)^{72} = (e^{i \frac{2\pi}{24}})^{72} = (e^{i \frac{\pi}{12}})^{72}$
$= e^{i \frac{72\pi}{12}} = e^{i 6\pi}$
By Euler's formula,$e^{i 6\pi} = \cos(6\pi) + i \sin(6\pi) = 1 + i(0) = 1$.

(C) Let $z = \sqrt{3}+i$.
Converting to polar form,$r = \sqrt{(\sqrt{3})^2 + 1^2} = 2$ and $\theta = \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.
So,$z = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$ and $\bar{z} = 2(\cos \frac{\pi}{6} - i \sin \frac{\pi}{6})$.
Using De Moivre's Theorem,$z^7 + \bar{z}^7 = 2^7(\cos \frac{7\pi}{6} + i \sin \frac{7\pi}{6}) + 2^7(\cos \frac{7\pi}{6} - i \sin \frac{7\pi}{6})$.
$= 2^7(2 \cos \frac{7\pi}{6}) = 2^8 \cos(\pi + \frac{\pi}{6})$.
$= -2^8 \cos(\frac{\pi}{6}) = -256 \times \frac{\sqrt{3}}{2} = -128 \sqrt{3}$.

(D) The $n^{\text{th}}$ roots of unity are given by $e^{i \frac{2k\pi}{n}}$ for $k = 0, 1, 2, \dots, n-1$.
For $12^{\text{th}}$ roots of unity,the values are $e^{i \frac{2k_1\pi}{12}} = e^{i \frac{k_1\pi}{6}}$ where $k_1 \in \{0, 1, \dots, 11\}$.
For $30^{\text{th}}$ roots of unity,the values are $e^{i \frac{2k_2\pi}{30}} = e^{i \frac{k_2\pi}{15}}$ where $k_2 \in \{0, 1, \dots, 29\}$.
$A$ root is common if $e^{i \frac{k_1\pi}{6}} = e^{i \frac{k_2\pi}{15}}$,which implies $\frac{k_1}{6} = \frac{k_2}{15} \pmod{2}$.
This simplifies to $\frac{k_1}{2} = \frac{k_2}{5} \pmod{2}$,or $5k_1 = 2k_2 \pmod{60}$.
The number of common roots between $n^{\text{th}}$ and $m^{\text{th}}$ roots of unity is given by $\gcd(n, m)$.
Here,$\gcd(12, 30) = 6$.
Thus,there are $6$ common roots.

(C) Let $X = a+b\omega+c\omega^2$. Note that $\omega X = a\omega+b\omega^2+c$ and $\omega^2 X = a\omega^2+b+c\omega$.
The given expression is $\left(\frac{X}{c+a\omega+b\omega^2}\right)^k + \left(\frac{X}{b+a\omega^2+c\omega}\right)^l = 2$.
Since $c+a\omega+b\omega^2 = \omega X$ and $b+a\omega^2+c\omega = \omega^2 X$,we substitute these:
$\left(\frac{X}{\omega X}\right)^k + \left(\frac{X}{\omega^2 X}\right)^l = 2$
$\left(\frac{1}{\omega}\right)^k + \left(\frac{1}{\omega^2}\right)^l = 2$
$\omega^{-k} + \omega^{-2l} = 2$
Since $\omega^3 = 1$,we have $\omega^{-k} = \omega^{3n-k}$ and $\omega^{-2l} = \omega^{3m-2l}$.
For the sum of two complex numbers of modulus $1$ to be $2$,both must be $1$.
Thus,$\omega^{-k} = 1$ and $\omega^{-2l} = 1$,which implies $k$ is a multiple of $3$ and $2l$ is a multiple of $3$ (so $l$ is a multiple of $3$).
Therefore,$2k+l$ must be a multiple of $3$.

(B) Let $z = \frac{\sqrt{3}+i}{\sqrt{3}-i}$. Multiplying numerator and denominator by $\sqrt{3}+i$,we get $z = \frac{(\sqrt{3}+i)^2}{3+1} = \frac{3-1+2i\sqrt{3}}{4} = \frac{2+2i\sqrt{3}}{4} = \frac{1}{2} + i\frac{\sqrt{3}}{2} = \operatorname{cis}\left(\frac{\pi}{3}\right)$.
Then the given expression is $z^4 + (\bar{z})^4 = \operatorname{cis}\left(\frac{4\pi}{3}\right) + \operatorname{cis}\left(-\frac{4\pi}{3}\right) = 2 \cos\left(\frac{4\pi}{3}\right) = 2 \cos\left(\pi + \frac{\pi}{3}\right) = -2 \cos\left(\frac{\pi}{3}\right) = -2 \times \frac{1}{2} = -1$.
So,$r \operatorname{cis} \theta = -1 = 1 \cdot \operatorname{cis}(\pi)$.
We need to find $\sqrt{r \operatorname{cis} \theta} = \sqrt{-1} = \sqrt{\operatorname{cis}(\pi)}$.
The square roots of $\operatorname{cis}(\pi)$ are $\operatorname{cis}\left(\frac{\pi + 2k\pi}{2}\right)$ for $k=0, 1$.
For $k=0$,we get $\operatorname{cis}\left(\frac{\pi}{2}\right) = i$.
For $k=1$,we get $\operatorname{cis}\left(\frac{3\pi}{2}\right) = -i$.
Comparing with the options,$\operatorname{cis}\left(\frac{3\pi}{2}\right)$ is the correct value.

(A) Let $z_1 = 1 + \sqrt{3} i = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = 2e^{i\pi/3}$.
Then $z_1^{2022} = 2^{2022} e^{i(2022\pi/3)} = 2^{2022} e^{i(674\pi)} = 2^{2022} \times 1 = 2^{2022}$.
Let $z_2 = \sqrt{3} - i = 2(\cos \frac{\pi}{6} - i \sin \frac{\pi}{6}) = 2e^{-i\pi/6}$.
Then $z_2^{2022} = 2^{2022} e^{-i(2022\pi/6)} = 2^{2022} e^{-i(337\pi)} = 2^{2022} \times (-1) = -2^{2022}$.
Therefore,$z_1^{2022} - z_2^{2022} = 2^{2022} - (-2^{2022}) = 2^{2022} + 2^{2022} = 2 \times 2^{2022} = 2^{2023}$.

(B) Given $x=a+b$,$y=a \alpha+b \beta$,$z=a \beta+b \alpha$ where $\alpha=\omega$ and $\beta=\omega^2$ are the complex cube roots of unity.
Since $1+\omega+\omega^2=0$,we have $x+y+z = (a+b) + (a\omega+b\omega^2) + (a\omega^2+b\omega) = a(1+\omega+\omega^2) + b(1+\omega+\omega^2) = 0$.
We know that if $x+y+z=0$,then $x^3+y^3+z^3=3xyz$.
$3xyz = 3(a+b)(a\omega+b\omega^2)(a\omega^2+b\omega)$
$= 3(a+b)(a^2\omega^3 + ab\omega^2 + ab\omega^4 + b^2\omega^3)$
$= 3(a+b)(a^2 + ab\omega^2 + ab\omega + b^2)$
$= 3(a+b)(a^2 + ab(\omega^2+\omega) + b^2)$
Since $\omega^2+\omega = -1$,
$= 3(a+b)(a^2 - ab + b^2) = 3(a^3+b^3)$.

(C) Given that $(1+\omega)$ is a root of $x^n-x=0$.
Substituting $x = 1+\omega$,we get $(1+\omega)^n - (1+\omega) = 0$.
Since $1+\omega+\omega^2=0$,we have $1+\omega = -\omega^2$.
Substituting this into the equation: $(-\omega^2)^n - (-\omega^2) = 0$.
$(-1)^n \omega^{2n} + \omega^2 = 0$.
$(-1)^n \omega^{2n} = -\omega^2$.
For this to hold,$n$ must be odd (so $(-1)^n = -1$) and $\omega^{2n} = \omega^2$.
This implies $2n \equiv 2 \pmod{3}$,which means $2n = 3k + 2$ for some integer $k$.
For $n=3$,$2(3) = 6 \equiv 0 \pmod{3}$ (False).
For $n=5$,$2(5) = 10 \equiv 1 \pmod{3}$ (False).
For $n=7$,$2(7) = 14 = 3(4) + 2 \equiv 2 \pmod{3}$ (True).
Thus,the least value of $n$ is $7$.

(A) Let the $n^{\text{th}}$ roots of unity be $z_0, z_1, \ldots, z_{n-1}$ where $z_0 = 1$. The equation $x^n - 1 = 0$ has roots $1, \alpha_1, \alpha_2, \ldots, \alpha_{n-1}$.
By Vieta's formulas,the sum of the roots taken two at a time is the coefficient of $x^{n-2}$ divided by the coefficient of $x^n$.
For $x^n - 1 = 0$,the coefficient of $x^{n-1}$ is $0$ and the coefficient of $x^{n-2}$ is $0$ (for $n > 2$).
Let $S = \sum_{0 \leq i < j \leq n-1} z_i z_j = 0$.
This sum can be expanded as: $z_0(\sum_{i=1}^{n-1} \alpha_i) + \sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j = 0$.
Since $z_0 = 1$ and the sum of all $n^{\text{th}}$ roots of unity is $0$,we have $1 + \sum_{i=1}^{n-1} \alpha_i = 0$,so $\sum_{i=1}^{n-1} \alpha_i = -1$.
Substituting this into the expansion: $1(-1) + \sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j = 0$.
Therefore,$\sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j = 1$.

(C) $(A) \ \omega^{1010} + \omega^{2000} = \omega^{3 \times 336 + 2} + \omega^{3 \times 666 + 2} = \omega^2 + \omega = -1$ (since $1 + \omega + \omega^2 = 0$).
$(B) \ (1 + \omega - \omega^2)(1 - \omega + \omega^2) = (-\omega^2 - \omega^2)(-\omega - \omega) = (-2\omega^2)(-2\omega) = 4\omega^3 = 4$.
$(C) \ (2 + \omega^2 + \omega^4)^5 = (2 + \omega^2 + \omega)^5 = (1 + (1 + \omega + \omega^2))^5 = (1 + 0)^5 = 1$.
$(D) \ (3 + 5\omega + 3\omega^2)^3 = (3(1 + \omega^2) + 5\omega)^3 = (3(-\omega) + 5\omega)^3 = (2\omega)^3 = 8\omega^3 = 8$.
Thus,the correct match is $A-III, B-IV, C-II, D-V$.

(C) Given,$\sum_{k=1}^n\left(k+\frac{1}{\omega}\right)\left(k+\frac{1}{\omega^2}\right)=340$
Expanding the term inside the summation:
$\sum_{k=1}^n\left(k^2+k\left(\frac{1}{\omega^2}+\frac{1}{\omega}\right)+\frac{1}{\omega^3}\right)=340$
Since $\omega^3=1$,we have $\frac{1}{\omega}=\omega^2$ and $\frac{1}{\omega^2}=\omega$.
Also,$1+\omega+\omega^2=0 \implies \omega+\omega^2=-1$.
Substituting these values:
$\sum_{k=1}^n\left(k^2+k(-1)+1\right)=340$
$\sum_{k=1}^n(k^2-k+1)=340$
Using summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$:
$\frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} + n = 340$
$\frac{n(n+1)}{2} \left[ \frac{2n+1}{3} - 1 \right] + n = 340$
$\frac{n(n+1)}{2} \left[ \frac{2n-2}{3} \right] + n = 340$
$\frac{n(n+1)(n-1)}{3} + n = 340$
$\frac{n(n^2-1)}{3} + n = 340$
$n^3-n+3n = 1020$
$n^3+2n = 1020$
Testing values,for $n=10$: $10^3 + 2(10) = 1000 + 20 = 1020$.
Thus,$n=10$.

(C) Given $\alpha, \beta$ are the roots of the equation $x^2+x+1=0$.
Since the roots of $x^2+x+1=0$ are the imaginary cube roots of unity,we have $\alpha = \omega$ and $\beta = \omega^2$,where $\omega^3 = 1$ and $1+\omega+\omega^2=0$.
Thus,$\alpha^3 = 1$ and $\beta^3 = 1$.
Also,$\alpha+\beta = -1$ and $\alpha\beta = 1$.
We need to evaluate the expression $E = (2-\alpha)(2-\beta)(2-\alpha^{10})(2-\alpha^{20})$.
Since $\alpha^3 = 1$,we have $\alpha^{10} = (\alpha^3)^3 \cdot \alpha = \alpha$ and $\alpha^{20} = (\alpha^3)^6 \cdot \alpha^2 = \alpha^2$.
Substituting these into the expression:
$E = (2-\alpha)(2-\beta)(2-\alpha)(2-\alpha^2)$.
Since $\beta = \alpha^2$ (as $\alpha\beta=1$ and $\alpha^3=1$),we have:
$E = (2-\alpha)(2-\alpha^2)(2-\alpha)(2-\alpha^2) = [(2-\alpha)(2-\alpha^2)]^2$.
Expanding the inner term: $(2-\alpha)(2-\alpha^2) = 4 - 2\alpha^2 - 2\alpha + \alpha^3 = 4 - 2(\alpha+\alpha^2) + 1$.
Since $1+\alpha+\alpha^2=0$,we have $\alpha+\alpha^2 = -1$.
So,$(2-\alpha)(2-\alpha^2) = 4 - 2(-1) + 1 = 4+2+1 = 7$.
Therefore,$E = 7^2 = 49$.

(C) Let $z = \sqrt{3}+i$. Then $\bar{z} = \sqrt{3}-i$.
We have $z = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) = 2e^{i\pi/6}$.
Then $z^8 = 2^8 e^{i8\pi/6} = 256 e^{i4\pi/3} = 256(\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}) = 256(-\frac{1}{2} - i \frac{\sqrt{3}}{2}) = -128 - 128\sqrt{3}i$.
Similarly,$\bar{z}^8 = \overline{z^8} = -128 + 128\sqrt{3}i$.
Thus,$z^8 - \bar{z}^8 = (-128 - 128\sqrt{3}i) - (-128 + 128\sqrt{3}i) = -256\sqrt{3}i$.
Comparing with $\alpha + i\beta$,we get $\alpha = 0$ and $\beta = -256\sqrt{3}$.
Finally,$\alpha - \frac{\sqrt{3}}{2}\beta = 0 - \frac{\sqrt{3}}{2}(-256\sqrt{3}) = \frac{3}{2} \times 256 = 384$.

(D) We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$,so $\omega + \omega^2 = -1$.
For $\omega^{1234}$,we divide $1234$ by $3$: $1234 = 3 \times 411 + 1$,so $\omega^{1234} = \omega^1 = \omega$.
For $\omega^{2021}$,we divide $2021$ by $3$: $2021 = 3 \times 673 + 2$,so $\omega^{2021} = \omega^2$.
Substituting these into the expression:
$\cos [(\omega + \omega^2) \pi - \frac{\pi}{4}] = \cos [(-1) \pi - \frac{\pi}{4}] = \cos [-\pi - \frac{\pi}{4}] = \cos [-(\pi + \frac{\pi}{4})] = \cos (\pi + \frac{\pi}{4}) = -\cos \frac{\pi}{4} = -\frac{1}{\sqrt{2}}$.

(A) If $n$ is not a multiple of $3$,then $n = 3k + 1$ or $n = 3k + 2$,where $k \in \mathbb{Z}$.
Case $(I)$: When $n = 3k + 1$.
$\omega^n + \omega^{2n} = \omega^{3k+1} + \omega^{6k+2} = (\omega^3)^k \cdot \omega + (\omega^3)^{2k} \cdot \omega^2 = 1^k \cdot \omega + 1^{2k} \cdot \omega^2 = \omega + \omega^2$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.
Case $(II)$: When $n = 3k + 2$.
$\omega^n + \omega^{2n} = \omega^{3k+2} + \omega^{6k+4} = (\omega^3)^k \cdot \omega^2 + (\omega^3)^{2k} \cdot \omega^4 = 1^k \cdot \omega^2 + 1^{2k} \cdot \omega = \omega^2 + \omega = -1$.
In both cases,$\omega^n + \omega^{2n} = -1$ for all $n$ that are not multiples of $3$.

(D) We know that $1+\omega+\omega^2=0$ and $\omega^3=1$.
From $1+\omega+\omega^2=0$,we have $1+\omega^2=-\omega$ and $1+\omega=-\omega^2$.
Substituting these into the expression:
$\left(1-\omega+\omega^2\right)^6+\left(1-\omega^2+\omega\right)^6 = \left(-\omega-\omega\right)^6+\left(-\omega^2-\omega^2\right)^6$
$= (-2\omega)^6 + (-2\omega^2)^6$
$= 64\omega^6 + 64\omega^{12}$
Since $\omega^3=1$,then $\omega^6 = (\omega^3)^2 = 1^2 = 1$ and $\omega^{12} = (\omega^3)^4 = 1^4 = 1$.
$= 64(1) + 64(1) = 64 + 64 = 128$.

(A) Given that $\omega$ is a complex cube root of unity,we have $1+\omega+\omega^2=0$ and $\omega^3=1$.
Expanding the expression inside the summation:
$(\omega x+2)(\omega^2 x+2)-3 = \omega^3 x^2 + 2\omega x + 2\omega^2 x + 4 - 3$
$= x^2 + 2x(\omega + \omega^2) + 1$ (since $\omega^3=1$)
$= x^2 + 2x(-1) + 1 = x^2 - 2x + 1 = (x-1)^2$.
Now,we calculate the sum:
$\sum_{x=1}^{10} (x-1)^2 = \sum_{x=1}^{10} (x^2 - 2x + 1) = \sum_{x=1}^{10} x^2 - 2\sum_{x=1}^{10} x + \sum_{x=1}^{10} 1$
$= \frac{10(11)(21)}{6} - 2 \times \frac{10(11)}{2} + 10$
$= 385 - 110 + 10 = 285$.

(D) Given $x^2+x+1=0$,the roots are $\omega$ and $\omega^2$,where $\omega^3=1$ and $1+\omega+\omega^2=0$.
Note that $\omega + \frac{1}{\omega} = \omega + \omega^2 = -1$.
Thus,the terms in the product $A_r$ are:
For $k=1$,$(x+\frac{1}{x})^3 = (-1)^3 = -1$.
For $k=2$,$(x^2+\frac{1}{x^2})^3 = (\omega^2+\omega)^3 = (-1)^3 = -1$.
For $k=3$,$(x^3+\frac{1}{x^3})^3 = (1+1)^3 = 8$.
So,$A_3 = (-1)(-1)(8) = 8$.
For $k=4$,$(x^4+\frac{1}{x^4})^3 = (\omega+\omega^2)^3 = -1$.
For $k=5$,$(x^5+\frac{1}{x^5})^3 = (\omega^2+\omega)^3 = -1$.
For $k=6$,$(x^6+\frac{1}{x^6})^3 = (1+1)^3 = 8$.
Thus,$A_6 = A_3 \cdot (-1) \cdot (-1) \cdot 8 = 8^2 = 64$.
In general,$A_{3n} = 8^n$.
The series is $\sum_{n=1}^{\infty} \frac{1}{A_{3n}} = \sum_{n=1}^{\infty} \frac{1}{8^n} = \frac{1/8}{1-1/8} = \frac{1/8}{7/8} = \frac{1}{7}$.

(A) Given $z^2+z+1=0$,the roots are $z = \omega$ and $z = \omega^2$,where $\omega$ is a complex cube root of unity.
For any $n$ not divisible by $3$,$z^n + \frac{1}{z^n} = \omega^n + \omega^{2n} = -1$.
For $n$ divisible by $3$,$z^n + \frac{1}{z^n} = \omega^{3k} + \omega^{6k} = 1 + 1 = 2$.
There are $2020$ terms in the sum.
The number of terms where $n$ is a multiple of $3$ is $\lfloor \frac{2020}{3} \rfloor = 673$.
The number of terms where $n$ is not a multiple of $3$ is $2020 - 673 = 1347$.
The sum is $1347 \times (-1)^3 + 673 \times (2)^3 = -1347 + 673 \times 8 = -1347 + 5384 = 4037$.

(A) Given,$x = \omega^2 - \omega - 3$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 = -1 - \omega$.
Substituting this,$x = (-1 - \omega) - \omega - 3 = -2\omega - 4$.
Alternatively,using $\omega^2 - \omega - 1 = 0$ is not correct,but $\omega^2 + \omega + 1 = 0$ is.
Let us re-evaluate: $x = \omega^2 - \omega - 3$.
Since $\omega^2 = -1 - \omega$,$x = -1 - \omega - \omega - 3 = -2\omega - 4$.
Then $x + 4 = -2\omega$.
Squaring both sides,$(x + 4)^2 = 4\omega^2$.
$x^2 + 8x + 16 = 4(-1 - \omega) = -4 - 4\omega$.
Since $-4\omega = 2(x + 4)$,we have $x^2 + 8x + 16 = -4 + 2x + 8$.
$x^2 + 6x + 12 = 0$.
Now,perform polynomial division of $x^4 + 6x^3 + 10x^2 - 12x - 19$ by $x^2 + 6x + 12$.
$x^4 + 6x^3 + 10x^2 - 12x - 19 = (x^2 + 6x + 12)(x^2 - 2) + 5$.
Since $x^2 + 6x + 12 = 0$,the expression equals $0(x^2 - 2) + 5 = 5$.

(B) We know that the algebraic identity for the sum of cubes is given by:
$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
Also,the product of the factors involving cube roots of unity is:
$(a+b \omega+c \omega^2)(a+b \omega^2+c \omega) = a^2+b^2+c^2-ab-bc-ca$
Substituting this into the identity,we get:
$(a+b+c)(a+b \omega+c \omega^2)(a+b \omega^2+c \omega) = a^3+b^3+c^3-3abc$

(D) Let the roots be $\alpha, \beta, \gamma$. These are the roots of the equation $z^3 = -i$,or $z^3 + i = 0$.
By Vieta's formulas,for a cubic equation $az^3 + bz^2 + cz + d = 0$,the sum of the roots $\alpha + \beta + \gamma = -b/a$ and the sum of the products of roots taken two at a time $\alpha\beta + \beta\gamma + \gamma\alpha = c/a$.
Here,$a = 1$,$b = 0$,$c = 0$,and $d = i$.
Thus,$\alpha + \beta + \gamma = 0$ and $\alpha\beta + \beta\gamma + \gamma\alpha = 0$.
We know the identity $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)$.
Substituting the values,we get $\alpha^2 + \beta^2 + \gamma^2 = (0)^2 - 2(0) = 0$.

(B) Given that $1, \omega, \omega^2, \ldots, \omega^8$ are the roots of the equation $x^9-1=0$.
Since $\omega$ is a $9^{th}$ root of unity,we have $\omega^9 = 1$.
We need to evaluate the sum $S = \sum_{r=1}^8 \left(\omega^r\right)^{99}$.
$S = \sum_{r=1}^8 \omega^{99r} = \sum_{r=1}^8 (\omega^9)^{11r}$.
Since $\omega^9 = 1$,we have $(\omega^9)^{11r} = (1)^{11r} = 1$.
Thus,$S = \sum_{r=1}^8 1 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8$.
Therefore,the correct option is $B$.

(A) The roots of $z^2-z+1=0$ are $z = \frac{1 \pm i\sqrt{3}}{2} = -\omega^2, -\omega$,where $\omega$ is the complex cube root of unity.
Let $\alpha = -\omega$. Then $\alpha^2 = \omega^2$ and $\alpha^3 = -\omega^3 = -1$. Thus $\alpha^6 = 1$.
We evaluate the terms $T_n = \alpha^n + \frac{1}{\alpha^n}$ for $n = 2014, 2015, 2016, 2017, 2018$:
$n=2014: \alpha^{2014} = (\alpha^6)^{335} \cdot \alpha^4 = \alpha^4 = \omega^4 = \omega$. So $T_{2014} = \omega + \frac{1}{\omega} = \omega + \omega^2 = -1$.
$n=2015: \alpha^{2015} = \alpha^5 = \omega^5 = \omega^2$. So $T_{2015} = \omega^2 + \frac{1}{\omega^2} = \omega^2 + \omega = -1$.
$n=2016: \alpha^{2016} = \alpha^6 = 1$. So $T_{2016} = 1 + \frac{1}{1} = 2$.
$n=2017: \alpha^{2017} = \alpha^1 = -\omega$. So $T_{2017} = -\omega - \frac{1}{\omega} = -(\omega + \omega^2) = 1$.
$n=2018: \alpha^{2018} = \alpha^2 = \omega^2$. So $T_{2018} = \omega^2 + \frac{1}{\omega^2} = \omega^2 + \omega = -1$.
Substituting these into the expression: $(-1)^1 + (-1)^2 + (2)^3 + (1)^4 + (-1)^5 = -1 + 1 + 8 + 1 - 1 = 8$.

(B) We have,$z^3+2z^2+2z+1=0$.
Factoring the expression: $(z^3+1)+2z(z+1)=0$.
$(z+1)(z^2-z+1)+2z(z+1)=0$.
$(z+1)(z^2-z+1+2z)=0$.
$(z+1)(z^2+z+1)=0$.
The roots are $z=-1$ and the roots of $z^2+z+1=0$,which are $\omega$ and $\omega^2$.
Now,test these roots in $z^{2018}+z^{2017}+1=0$:
For $z=-1$: $(-1)^{2018}+(-1)^{2017}+1 = 1-1+1 = 1 \neq 0$.
For $z=\omega$: $\omega^{2018}+\omega^{2017}+1 = (\omega^3)^{672} \cdot \omega^2 + (\omega^3)^{672} \cdot \omega + 1 = \omega^2+\omega+1 = 0$.
For $z=\omega^2$: $(\omega^2)^{2018}+(\omega^2)^{2017}+1 = \omega^{4036}+\omega^{4034}+1 = \omega+\omega^2+1 = 0$.
The common roots are $\omega$ and $\omega^2$.
Checking the options for $z=\omega$ and $z=\omega^2$:
For $z^4+z^2+1=0$:
If $z=\omega$,$\omega^4+\omega^2+1 = \omega+\omega^2+1 = 0$.
If $z=\omega^2$,$(\omega^2)^4+(\omega^2)^2+1 = \omega^8+\omega^4+1 = \omega^2+\omega+1 = 0$.
Thus,the common roots satisfy $z^4+z^2+1=0$.