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

(C) Given the equation $Z^2 + Z|Z| + |Z|^2 = 0$.
Since $Z \neq 0$,we can divide by $|Z|^2$:
$\left(\frac{Z}{|Z|}\right)^2 + \left(\frac{Z}{|Z|}\right) + 1 = 0$.
Let $u = \frac{Z}{|Z|}$. Note that $|u| = \frac{|Z|}{|Z|} = 1$.
The equation becomes $u^2 + u + 1 = 0$.
The roots of this quadratic equation are $u = \omega$ and $u = \omega^2$,where $\omega = e^{i2\pi/3}$ and $\omega^2 = e^{i4\pi/3}$.
Both $\omega$ and $\omega^2$ satisfy $|u| = 1$.
Thus,$\frac{Z}{|Z|} = \omega$ or $\frac{Z}{|Z|} = \omega^2$.
This implies $Z = |Z|\omega$ or $Z = |Z|\omega^2$ for any $|Z| > 0$.
Since the set of all such $Z$ is not restricted to a finite set of values but rather rays in the complex plane,and none of the provided options represent this set,the question as posed with these options is mathematically inconsistent. However,if we look for the values of $u = Z/|Z|$,the set is $\{\omega, \omega^2\}$.

(B) The first exponent is an infinite geometric series with $a = \frac{1}{3}$ and $r = \frac{2}{3}$.
Its sum is $S_{\infty} = \frac{a}{1-r} = \frac{1/3}{1-2/3} = \frac{1/3}{1/3} = 1$.
Thus,$\omega^{\left(\frac{1}{3}+\frac{2}{9}+\ldots\right)} = \omega^1 = \omega$.
The second exponent is an infinite geometric series with $a = \frac{1}{2}$ and $r = \frac{3}{4}$.
Its sum is $S_{\infty} = \frac{a}{1-r} = \frac{1/2}{1-3/4} = \frac{1/2}{1/4} = 2$.
Thus,$\omega^{\left(\frac{1}{2}+\frac{3}{8}+\ldots\right)} = \omega^2$.
Adding these results,we get $\omega + \omega^2$.
Since $1 + \omega + \omega^2 = 0$,it follows that $\omega + \omega^2 = -1$.

(A) Given equation $z^3+2z^2+2z+1=0$ can be factored as $(z+1)(z^2+z+1)=0$.
Its roots are $-1, \omega, \omega^2$,where $\omega$ is a complex cube root of unity.
Let $f(z) = z^{2014}+z^{2015}+1$.
Testing $z=-1$: $f(-1) = (-1)^{2014}+(-1)^{2015}+1 = 1-1+1 = 1 \neq 0$. So,$-1$ is not a common root.
Testing $z=\omega$: $f(\omega) = \omega^{2014}+\omega^{2015}+1 = (\omega^3)^{671} \cdot \omega + (\omega^3)^{671} \cdot \omega^2 + 1 = \omega + \omega^2 + 1 = 0$. So,$\omega$ is a common root.
Testing $z=\omega^2$: $f(\omega^2) = (\omega^2)^{2014}+(\omega^2)^{2015}+1 = \omega^{4028}+\omega^{4030}+1 = (\omega^3)^{1342} \cdot \omega^2 + (\omega^3)^{1343} \cdot \omega + 1 = \omega^2 + \omega + 1 = 0$. So,$\omega^2$ is a common root.
Thus,the common roots are $\omega$ and $\omega^2$.

(C) Given $n = 3r + 1$ for some integer $r$.
We know that $\omega = \frac{-1+i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1-i\sqrt{3}}{2}$.
Thus,$1+i\sqrt{3} = -2\omega^2$ and $1-i\sqrt{3} = -2\omega$.
Substituting these into the expression:
$(1+i\sqrt{3})^n + (1-i\sqrt{3})^n = (-2\omega^2)^n + (-2\omega)^n$
$= (-2)^n (\omega^{2n} + \omega^n)$
Since $n = 3r+1$,$\omega^n = \omega^{3r+1} = \omega$ and $\omega^{2n} = \omega^{6r+2} = \omega^2$.
Therefore,the expression becomes $(-2)^n (\omega^2 + \omega)$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 + \omega = -1$.
Thus,the result is $(-2)^n (-1) = -(-2)^n$.

(A) The $n^{\text{th}}$ roots of unity are given by $z_r = e^{i \frac{2\pi r}{n}}$ for $r = 0, 1, \dots, n-1$.
Let $z_1 = e^{i \frac{2\pi r_1}{n}}$ and $z_2 = e^{i \frac{2\pi r_2}{n}}$.
The angle subtended by the line segment joining $z_1$ and $z_2$ at the origin is the difference of their arguments: $\theta = |\arg(z_1) - \arg(z_2)| = |\frac{2\pi r_1}{n} - \frac{2\pi r_2}{n}| = \frac{2\pi}{n} |r_1 - r_2|$.
Given that this angle is a right angle,we have $\frac{2\pi}{n} |r_1 - r_2| = \frac{\pi}{2}$.
This simplifies to $\frac{2}{n} |r_1 - r_2| = \frac{1}{2}$,which implies $n = 4 |r_1 - r_2|$.
Since $|r_1 - r_2|$ is an integer,let $|r_1 - r_2| = k$,where $k$ is a positive integer.
Therefore,$n$ must be of the form $4k$.

(C) We can rewrite the expression as: $\sum_{k=1}^6 -i \left[ \cos \left(\frac{2 \pi k}{7}\right) + i \sin \left(\frac{2 \pi k}{7}\right) \right]$
Using Euler's formula,$\cos \theta + i \sin \theta = e^{i \theta}$,the expression becomes:
$-i \sum_{k=1}^6 e^{i \frac{2 \pi k}{7}} = -i \left[ e^{i \frac{2 \pi}{7}} + e^{i \frac{4 \pi}{7}} + \dots + e^{i \frac{12 \pi}{7}} \right]$
This is a geometric progression with first term $a = e^{i \frac{2 \pi}{7}}$ and common ratio $r = e^{i \frac{2 \pi}{7}}$ for $n = 6$ terms:
$-i \left[ e^{i \frac{2 \pi}{7}} \frac{1 - (e^{i \frac{2 \pi}{7}})^6}{1 - e^{i \frac{2 \pi}{7}}} \right] = -i \left[ \frac{e^{i \frac{2 \pi}{7}} - e^{i \frac{14 \pi}{7}}}{1 - e^{i \frac{2 \pi}{7}}} \right]$
Since $e^{i \frac{14 \pi}{7}} = e^{i 2 \pi} = 1$:
$-i \left[ \frac{e^{i \frac{2 \pi}{7}} - 1}{1 - e^{i \frac{2 \pi}{7}}} \right] = -i (-1) = i$

(D) Since $\alpha_1, \alpha_2, \ldots, \alpha_{23}$ are the $23^{rd}$ roots of unity,they satisfy the equation $\alpha^{23} - 1 = 0$,which implies $\alpha^{23} = 1$.
Now,consider the sum $S = \alpha_1^{47} + \alpha_2^{47} + \ldots + \alpha_{23}^{47}$.
Since $\alpha_k^{23} = 1$ for each $k = 1, 2, \ldots, 23$,we can write $\alpha_k^{47} = \alpha_k^{23 \times 2 + 1} = (\alpha_k^{23})^2 \cdot \alpha_k = (1)^2 \cdot \alpha_k = \alpha_k$.
Therefore,$S = \alpha_1 + \alpha_2 + \ldots + \alpha_{23}$.
The sum of the $n^{th}$ roots of unity is $0$ for $n > 1$.
Thus,$\alpha_1 + \alpha_2 + \ldots + \alpha_{23} = 0$.

(A) Let $Z = \frac{1}{2} + i \frac{\sqrt{3}}{2}$.
We can write $Z$ in polar form as $Z = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = e^{i \frac{\pi}{3}}$.
To find the fourth root,we calculate $Z^{1/4} = (e^{i \frac{\pi}{3}})^{1/4} = e^{i \frac{\pi}{12}}$.
Using the definition $\operatorname{cis} \theta = \cos \theta + i \sin \theta = e^{i \theta}$,we get $e^{i \frac{\pi}{12}} = \operatorname{cis} \frac{\pi}{12}$.
Thus,one of the fourth roots is $\operatorname{cis} \frac{\pi}{12}$.

(B) Let $\omega = a$. The $n$th roots of unity are $1, \omega, \omega^2, \ldots, \omega^{n-1}$.
We know that $x^n - 1 = (x-1)(x-\omega)(x-\omega^2) \ldots (x-\omega^{n-1})$.
Taking the natural logarithm on both sides:
$\ln(x^n - 1) = \ln(x-1) + \ln(x-\omega) + \ln(x-\omega^2) + \ldots + \ln(x-\omega^{n-1})$.
Differentiating with respect to $x$:
$\frac{n x^{n-1}}{x^n - 1} = \frac{1}{x-1} + \sum_{i=1}^{n-1} \frac{1}{x-\omega^i}$.
Rearranging to isolate the summation:
$\sum_{i=1}^{n-1} \frac{1}{x-\omega^i} = \frac{n x^{n-1}}{x^n - 1} - \frac{1}{x-1}$.
Setting $x = 2$:
$\sum_{i=1}^{n-1} \frac{1}{2-\omega^i} = \frac{n \cdot 2^{n-1}}{2^n - 1} - \frac{1}{2-1} = \frac{n \cdot 2^{n-1}}{2^n - 1} - 1$.
Simplifying the expression:
$\frac{n \cdot 2^{n-1} - (2^n - 1)}{2^n - 1} = \frac{n \cdot 2^{n-1} - 2 \cdot 2^{n-1} + 1}{2^n - 1} = \frac{(n-2) 2^{n-1} + 1}{2^n - 1}$.

(C) Since $1, \alpha_1, \alpha_2, \ldots, \alpha_{n-1}$ are the $n$th roots of unity,they are the roots of the equation $x^n - 1 = 0$.
Thus,we can write $x^n - 1 = (x - 1)(x - \alpha_1)(x - \alpha_2) \ldots (x - \alpha_{n-1})$.
Dividing both sides by $(x - 1)$,we get $(x - \alpha_1)(x - \alpha_2) \ldots (x - \alpha_{n-1}) = \frac{x^n - 1}{x - 1}$.
Substituting $x = -1$ into the equation,we get $(-1 - \alpha_1)(-1 - \alpha_2) \ldots (-1 - \alpha_{n-1}) = \frac{(-1)^n - 1}{-1 - 1}$.
Factoring out $(-1)$ from each of the $(n-1)$ terms on the left side,we have $(-1)^{n-1}(1 + \alpha_1)(1 + \alpha_2) \ldots (1 + \alpha_{n-1}) = \frac{(-1)^n - 1}{-2}$.
Since $n$ is an even natural number,$(-1)^n = 1$.
Therefore,$(-1)^{n-1}(1 + \alpha_1)(1 + \alpha_2) \ldots (1 + \alpha_{n-1}) = \frac{1 - 1}{-2} = 0$.
Since $(-1)^{n-1} \neq 0$,it follows that $(1 + \alpha_1)(1 + \alpha_2) \ldots (1 + \alpha_{n-1}) = 0$.

(B) Given $a = \cos \left(\frac{8 \pi}{11}\right) + i \sin \left(\frac{8 \pi}{11}\right) = e^{i \frac{8 \pi}{11}}$.
Since $a^{11} = e^{i 8 \pi} = 1$,$a$ is an $11^{th}$ root of unity.
The sum of all $11^{th}$ roots of unity is $1 + a + a^2 + \dots + a^{10} = 0$.
Thus,$a + a^2 + a^3 + a^4 + a^5 + a^6 + a^7 + a^8 + a^9 + a^{10} = -1$.
Since $a^{11} = 1$,we have $a^{10} = \bar{a}$,$a^9 = \bar{a^2}$,$a^8 = \bar{a^3}$,$a^7 = \bar{a^4}$,and $a^6 = \bar{a^5}$.
Substituting these into the sum,we get $(a + \bar{a}) + (a^2 + \bar{a^2}) + (a^3 + \bar{a^3}) + (a^4 + \bar{a^4}) + (a^5 + \bar{a^5}) = -1$.
Using the property $z + \bar{z} = 2 \operatorname{Re}(z)$,we have $2 \operatorname{Re}(a) + 2 \operatorname{Re}(a^2) + 2 \operatorname{Re}(a^3) + 2 \operatorname{Re}(a^4) + 2 \operatorname{Re}(a^5) = -1$.
Therefore,$2 \operatorname{Re}(a + a^2 + a^3 + a^4 + a^5) = -1$,which implies $\operatorname{Re}(a + a^2 + a^3 + a^4 + a^5) = -\frac{1}{2}$.

(D) The given equation is $x^6-1=0$,which implies $x^6=1$.
Since $\alpha$ is a root,$\alpha^6=1$.
Also,$\alpha \neq 1$ because $\alpha$ is a non-real root (the roots of $x^6-1=0$ are $e^{i2k\pi/6}$ for $k=0, 1, 2, 3, 4, 5$,and only $k=0$ gives $x=1$).
We can write the numerator as $\alpha^2(1+\alpha+\alpha^2+\alpha^3)$.
Since $\alpha^6-1 = (\alpha-1)(\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1) = 0$ and $\alpha \neq 1$,we have $\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1 = 0$.
This implies $\alpha^5+\alpha^4+\alpha^3+\alpha^2 = -\alpha-1 = -(\alpha+1)$.
Substituting this into the expression: $\frac{-(\alpha+1)}{\alpha+1} = -1$.

(C) The $n$th roots of unity are the roots of the equation $z^n - 1 = 0$.
We can factorize $z^n - 1$ as:
$z^n - 1 = (z - 1)(z - z_1)(z - z_2) \ldots (z - z_{n-1})$.
We also know that $z^n - 1 = (z - 1)(z^{n-1} + z^{n-2} + \ldots + z + 1)$.
Equating the two expressions for $z^n - 1$:
$(z - 1)(z^{n-1} + z^{n-2} + \ldots + z + 1) = (z - 1)(z - z_1)(z - z_2) \ldots (z - z_{n-1})$.
Dividing both sides by $(z - 1)$ (for $z \neq 1$):
$z^{n-1} + z^{n-2} + \ldots + z + 1 = (z - z_1)(z - z_2) \ldots (z - z_{n-1})$.
Taking the limit as $z \to 1$ or simply substituting $z = 1$ in the polynomial identity:
$1^{n-1} + 1^{n-2} + \ldots + 1 + 1 = (1 - z_1)(1 - z_2) \ldots (1 - z_{n-1})$.
Since there are $n$ terms on the left side (from $z^0$ to $z^{n-1}$),the sum is $n$.
Therefore,$(1 - z_1)(1 - z_2) \ldots (1 - z_{n-1}) = n$.

(D) The given series is a geometric progression with the first term $a = 1$ and common ratio $r = (\cos \theta + i \sin \theta)$.
The $n^{th}$ term of a geometric progression is given by $T_n = a \cdot r^{n-1}$.
For the $10^{th}$ term $(n = 10)$,we have $T_{10} = 1 \cdot (\cos \theta + i \sin \theta)^{10-1} = (\cos \theta + i \sin \theta)^9$.
Using De Moivre's theorem,$(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta) = e^{i n \theta}$.
Substituting $\theta = \frac{\pi}{6}$ and $n = 9$:
$T_{10} = e^{i 9 (\frac{\pi}{6})} = e^{i \frac{3\pi}{2}}$.
Using Euler's formula $e^{i \phi} = \cos \phi + i \sin \phi$:
$T_{10} = \cos(\frac{3\pi}{2}) + i \sin(\frac{3\pi}{2}) = 0 + i(-1) = -i$.

(C) We have,$\sum_{r=1}^{n-1} r(r+1-\omega)(r+1-\omega^2)$
$= \sum_{r=1}^{n-1} r((r+1)^2 - (r+1)(\omega + \omega^2) + \omega^3)$
Since $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.
$= \sum_{r=1}^{n-1} r((r+1)^2 + (r+1) + 1)$
$= \sum_{r=1}^{n-1} r(r^2 + 2r + 1 + r + 1 + 1) = \sum_{r=1}^{n-1} r(r^2 + 3r + 3)$
$= \sum_{r=1}^{n-1} (r^3 + 3r^2 + 3r)$
$= \left[\frac{(n-1)n}{2}\right]^2 + 3 \cdot \frac{(n-1)n(2n-2+1)}{6} + 3 \cdot \frac{(n-1)n}{2}$
$= \frac{n^2(n-1)^2}{4} + \frac{n(n-1)(2n-1)}{2} + \frac{3n(n-1)}{2}$
$= \frac{n(n-1)}{4} [n(n-1) + 2(2n-1) + 6]$
$= \frac{n(n-1)}{4} [n^2 - n + 4n - 2 + 6]$
$= \frac{n(n-1)}{4} (n^2 + 3n + 4)$

(C) Let $x=\cos \alpha+i \sin \alpha$,$y=\cos \beta+i \sin \beta$,and $z=\cos \gamma+i \sin \gamma$.
Given $x+y+z=0$,we know that $x^3+y^3+z^3=3xyz$.
Substituting the values,we get $(\cos \alpha+i \sin \alpha)^3+(\cos \beta+i \sin \beta)^3+(\cos \gamma+i \sin \gamma)^3 = 3(\cos \alpha+i \sin \alpha)(\cos \beta+i \sin \beta)(\cos \gamma+i \sin \gamma)$.
Using De Moivre's theorem,$(\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma)+i(\sin 3 \alpha+\sin 3 \beta+\sin 3 \gamma) = 3 \cos (\alpha+\beta+\gamma)+3 i \sin (\alpha+\beta+\gamma)$.
Equating real and imaginary parts: $\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma=3 \cos (\alpha+\beta+\gamma)$ and $\sin 3 \alpha+\sin 3 \beta+\sin 3 \gamma=3 \sin (\alpha+\beta+\gamma)$.
Using the identity $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$,we have $4(\cos^3\alpha+\cos^3\beta+\cos^3\gamma) - 3(\cos\alpha+\cos\beta+\cos\gamma) = 3\cos(\alpha+\beta+\gamma)$.
Since $\cos\alpha+\cos\beta+\cos\gamma=0$,we get $\cos^3\alpha+\cos^3\beta+\cos^3\gamma = \frac{3}{4}\cos(\alpha+\beta+\gamma)$.
Similarly,using $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$,we get $\sin^3\alpha+\sin^3\beta+\sin^3\gamma = -\frac{3}{4}\sin(\alpha+\beta+\gamma)$.
Finally,the expression becomes $\left(\frac{3}{4}\cos(\alpha+\beta+\gamma)\right)^2 + \left(-\frac{3}{4}\sin(\alpha+\beta+\gamma)\right)^2 = \frac{9}{16}(\cos^2(\alpha+\beta+\gamma) + \sin^2(\alpha+\beta+\gamma)) = \frac{9}{16}$.

(C) Given equation is $x^{11}-x^7+x^4-1=0$.
Factorizing the expression:
$x^7(x^4-1) + 1(x^4-1) = 0$
$(x^4-1)(x^7+1) = 0$.
Case $(i)$: $x^4-1=0 \implies x^4=1$.
The roots are $x = e^{i(2k\pi/4)}$ for $k=0, 1, 2, 3$. These are $1, i, -1, -i$. There are $4$ distinct roots.
Case $(ii)$: $x^7+1=0 \implies x^7=-1$.
The roots are $x = e^{i((2k+1)\pi/7)}$ for $k=0, 1, 2, 3, 4, 5, 6$. There are $7$ distinct roots.
To find the total number of distinct solutions,we check for common roots:
If $x^4=1$ and $x^7=-1$,then $x^4=1 \implies |x|=1$.
Also $x^7=-1 \implies |x|=1$.
If $x$ is a common root,then $x^4=1$ and $x^7=-1$.
Then $x^8 = (x^4)^2 = 1^2 = 1$.
Since $x^8 = x^7 \cdot x = 1$,we have $(-1) \cdot x = 1$,so $x = -1$.
Check if $x = -1$ satisfies both:
$(-1)^4 = 1$ (True) and $(-1)^7 = -1$ (True).
So,$x = -1$ is the only common root.
Total distinct solutions = (Number of roots of $x^4-1=0$) + (Number of roots of $x^7+1=0$) - (Number of common roots)
Total = $4 + 7 - 1 = 10$.

(B) The given equation is $x^2 - 2\sqrt{3}x + 4 = 0$.
Using the quadratic formula,the roots are $\alpha, \beta = \frac{2\sqrt{3} \pm \sqrt{12 - 16}}{2} = \sqrt{3} \pm i$.
In polar form,$\alpha = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$ and $\beta = 2(\cos \frac{-\pi}{6} + i \sin \frac{-\pi}{6})$.
Then $\alpha^{2024} = 2^{2024}(\cos \frac{2024\pi}{6} + i \sin \frac{2024\pi}{6})$ and $\beta^{2024} = 2^{2024}(\cos \frac{-2024\pi}{6} + i \sin \frac{-2024\pi}{6})$.
Note that $\frac{2024\pi}{6} = \frac{1012\pi}{3} = 337\pi + \frac{\pi}{3}$.
$\alpha^{2024} - \beta^{2024} = 2^{2024} [2i \sin(\frac{2024\pi}{6})] = 2^{2024} [2i \sin(337\pi + \frac{\pi}{3})] = 2^{2024} [2i (-\sin \frac{\pi}{3})] = 2^{2024} [2i (-\frac{\sqrt{3}}{2})] = -i \sqrt{3} \cdot 2^{2024}$.
Therefore,$\frac{2}{\sqrt{3}} |\alpha^{2024} - \beta^{2024}| = \frac{2}{\sqrt{3}} | -i \sqrt{3} \cdot 2^{2024} | = \frac{2}{\sqrt{3}} \cdot \sqrt{3} \cdot 2^{2024} = 2^{2025}$.

(B) Given the equation $x^2+2x+2=0$.
Using the quadratic formula,$x = \frac{-2 \pm \sqrt{4-8}}{2} = -1 \pm i$.
Let $\alpha = -1+i$ and $\beta = -1-i$.
In polar form,$\alpha = \sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4})$ and $\beta = \sqrt{2}(\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4})$.
Then $\alpha^{15} = (\sqrt{2})^{15} (\cos \frac{45\pi}{4} + i \sin \frac{45\pi}{4}) = 2^{7.5} (\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4}) = 2^{7.5} (-\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}})$.
Similarly,$\beta^{15} = (\sqrt{2})^{15} (\cos \frac{75\pi}{4} + i \sin \frac{75\pi}{4}) = 2^{7.5} (\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) = 2^{7.5} (-\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}})$.
Adding these,$\alpha^{15} + \beta^{15} = 2^{7.5} (-\frac{2}{\sqrt{2}}) = 2^{7.5} (-\sqrt{2}) = -2^8 = -256$.

(A) The given equation is $x^4+x^2+1=0$.
This can be factored as $(x^2+x+1)(x^2-x+1)=0$.
The roots of $x^2+x+1=0$ are $\omega$ and $\omega^2$,where $\omega$ is a complex cube root of unity.
The roots of $x^2-x+1=0$ are $-\omega$ and $-\omega^2$.
Given $\alpha+\beta=-1$ and $\alpha^2=\beta$,we identify $\alpha=\omega$ and $\beta=\omega^2$ (since $\omega+\omega^2=-1$ and $\omega^2=\omega^2$).
Given $\gamma+\delta=1$ and $\gamma^2=-\delta$,we identify $\gamma=-\omega$ and $\delta=-\omega^2$ (since $-\omega-\omega^2=1$ and $(-\omega)^2 = \omega^2 = -(-\omega^2)$).
Now,calculate $\alpha^{2023}+\beta^{2023}+\gamma^{2022}+\delta^{2022}$:
$= \omega^{2023} + (\omega^2)^{2023} + (-\omega)^{2022} + (-\omega^2)^{2022}$
$= \omega^{2023} + \omega^{4046} + \omega^{2022} + \omega^{4044}$
Using $\omega^3=1$:
$= \omega^1 + \omega^1 + 1 + 1 = 2\omega + 2$.
Wait,re-evaluating the sum: $\omega^{2023} = \omega$,$\omega^{4046} = \omega^2$,$\omega^{2022} = 1$,$\omega^{4044} = 1$.
Sum $= \omega + \omega^2 + 1 + 1 = (-1) + 2 = 1$.

(C) The given equation is $x^2-2x+4=0$.
Using the quadratic formula,the roots are $x = \frac{2 \pm \sqrt{4-16}}{2} = 1 \pm i\sqrt{3}$.
In polar form,$1 + i\sqrt{3} = 2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 2e^{i\pi/3}$ and $1 - i\sqrt{3} = 2e^{-i\pi/3}$.
Thus,$\alpha = 2e^{i\pi/3}$ and $\beta = 2e^{-i\pi/3}$.
Then $\alpha^{12} = (2e^{i\pi/3})^{12} = 2^{12} e^{i4\pi} = 2^{12}(1) = 2^{12}$.
Similarly,$\beta^{12} = (2e^{-i\pi/3})^{12} = 2^{12} e^{-i4\pi} = 2^{12}(1) = 2^{12}$.
Therefore,$\alpha^{12} + \beta^{12} = 2^{12} + 2^{12} = 2 \times 2^{12} = 2^{13}$.

(B) Given the equation $x^2-x+1=0$.
The roots of this equation are $\omega$ and $\omega^2$,where $\omega$ is the complex cube root of unity.
Let $\alpha = \omega$ and $\beta = \omega^2$.
We need to find the equation with roots $\alpha^{2015}$ and $\beta^{2015}$.
$\alpha^{2015} = \omega^{2015} = \omega^{3 \times 671 + 2} = \omega^2$.
$\beta^{2015} = (\omega^2)^{2015} = \omega^{4030} = \omega^{3 \times 1343 + 1} = \omega$.
The sum of the new roots is $\alpha^{2015} + \beta^{2015} = \omega^2 + \omega = -1$.
The product of the new roots is $\alpha^{2015} \cdot \beta^{2015} = \omega^2 \cdot \omega = \omega^3 = 1$.
The required quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^2 - (-1)x + 1 = 0$,which simplifies to $x^2 + x + 1 = 0$.

(D) Given equation: $x^3-3x^2+3x+7=0$.
This can be written as $(x-1)^3 + 8 = 0$,so $(x-1)^3 = -8$.
Thus,$x-1 = -2, -2\omega, -2\omega^2$.
The roots are $\alpha = -1, \beta = 1-2\omega, \gamma = 1-2\omega^2$.
Let $y = x-h$,so $x = y+h$. Substituting into the equation: $(y+h-1)^3 + 8 = 0$.
For the $y^2$ and $y$ terms to be missing,we must have $h-1 = 0$,so $h=1$.
The new roots are $\alpha-h = -2, \beta-h = -2\omega, \gamma-h = -2\omega^2$.
We need to calculate $S = \frac{\alpha-h}{\beta-h} + \frac{\beta-h}{\gamma-h} + \frac{\gamma-h}{\alpha-h}$.
$S = \frac{-2}{-2\omega} + \frac{-2\omega}{-2\omega^2} + \frac{-2\omega^2}{-2} = \frac{1}{\omega} + \frac{1}{\omega} + \omega^2 = \omega^2 + \omega^2 + \omega^2 = 3\omega^2$.

(C) Let $z = (\sqrt{3}-i)^{2/5} = [2(\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6}))]^{2/5}$.
Using De Moivre's Theorem,the $5$ values are given by $z_k = 2^{2/5} [\cos(\frac{2k\pi - \pi/6}{5/2}) + i \sin(\frac{2k\pi - \pi/6}{5/2})]$ for $k = 0, 1, 2, 3, 4$.
This simplifies to $z_k = 2^{2/5} [\cos(\frac{4k\pi}{5} - \frac{\pi}{15}) + i \sin(\frac{4k\pi}{5} - \frac{\pi}{15})]$.
The product of the $n$ roots of a complex number $z = a^{p/q}$ is given by $(-1)^{q-1} (a^{p/q})^q$ if we consider the roots of the equation $z^q = a^p$.
Here,$z^5 = (\sqrt{3}-i)^2 = 3 - 1 - 2\sqrt{3}i = 2 - 2\sqrt{3}i$.
The product of the roots of $z^5 - (2 - 2\sqrt{3}i) = 0$ is $(-1)^{5-1} \times (-(2 - 2\sqrt{3}i)) = 1 \times (-2 + 2\sqrt{3}i) = -2 + 2\sqrt{3}i$.
Wait,the product of roots $z_k$ of $z^n = c$ is $(-1)^{n-1} c$. Here $n=5$ and $c = (\sqrt{3}-i)^2 = 2 - 2\sqrt{3}i$.
Product $= (-1)^{5-1} (2 - 2\sqrt{3}i) = 1 \times (2 - 2\sqrt{3}i) = 2(1 - \sqrt{3}i)$.

(B) Let $z_1 = 1+\sqrt{3}i = 2(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 2e^{i\pi/3}$.
Similarly,$z_2 = 1-\sqrt{3}i = 2(\cos \frac{\pi}{3} - i\sin \frac{\pi}{3}) = 2e^{-i\pi/3}$.
Then,$(1+\sqrt{3}i)^{2023} + (1-\sqrt{3}i)^{2023} = (2e^{i\pi/3})^{2023} + (2e^{-i\pi/3})^{2023}$.
$= 2^{2023} (e^{i2023\pi/3} + e^{-i2023\pi/3})$.
Since $2023 = 3 \times 674 + 1$,we have $\frac{2023\pi}{3} = 674\pi + \frac{\pi}{3}$.
Thus,$e^{i2023\pi/3} = e^{i(674\pi + \pi/3)} = e^{i\pi/3} = \cos \frac{\pi}{3} + i\sin \frac{\pi}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$.
Similarly,$e^{-i2023\pi/3} = \frac{1}{2} - i\frac{\sqrt{3}}{2}$.
Sum $= 2^{2023} (\frac{1}{2} + i\frac{\sqrt{3}}{2} + \frac{1}{2} - i\frac{\sqrt{3}}{2}) = 2^{2023}(1) = 2^{2023}$.

(C) The roots of the equation $x^2+x+1=0$ are $\omega$ and $\omega^2$,where $\omega$ is a complex cube root of unity.
Given $\alpha = \omega$ and $\beta = \omega^2$,we have $\alpha+\beta = -1$ and $\alpha\beta = 1$.
For any $n$,$\alpha^n+\beta^n = \omega^n+\omega^{2n}$.
If $n$ is a multiple of $3$,then $\omega^n = 1$ and $\omega^{2n} = 1$,so $\alpha^n+\beta^n = 1+1 = 2$.
If $n$ is not a multiple of $3$,then $\alpha^n+\beta^n = \omega^n+\omega^{2n} = -1$.
We need to evaluate $S = \sum_{n=1}^{12} (\alpha^n+\beta^n)^2$.
For $n=1, 2, 4, 5, 7, 8, 10, 11$ ($8$ terms),the value is $(-1)^2 = 1$.
For $n=3, 6, 9, 12$ ($4$ terms),the value is $(2)^2 = 4$.
Thus,$S = 8 \times (1) + 4 \times (4) = 8 + 16 = 24$.

(B) Given equation is $x^2-2x+2=0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get:
$x = \frac{2 \pm \sqrt{4-8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = 1 \pm i$.
Let $\alpha = 1+i$ and $\beta = 1-i$.
Convert to polar form:
$\alpha = \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) = \sqrt{2} e^{i\pi/4}$.
$\beta = \sqrt{2}(\cos \frac{\pi}{4} - i \sin \frac{\pi}{4}) = \sqrt{2} e^{-i\pi/4}$.
Then,$\alpha^{2020} = (\sqrt{2})^{2020} e^{i(2020\pi/4)} = 2^{1010} e^{i(505\pi)}$.
Since $e^{i(505\pi)} = \cos(505\pi) + i \sin(505\pi) = -1 + 0 = -1$,
$\alpha^{2020} = -2^{1010}$.
Similarly,$\beta^{2020} = (\sqrt{2})^{2020} e^{-i(505\pi)} = 2^{1010} (-1) = -2^{1010}$.
Therefore,$\alpha^{2020} + \beta^{2020} = -2^{1010} - 2^{1010} = -2 \cdot 2^{1010} = -2^{1011}$.

(B) Let $z = i^{1/4}$,which implies $z^4 = i = e^{i\pi/2}$.
The equation is $z^4 - i = 0$.
The roots of this equation are $z_k = e^{i\left(\frac{\pi/2 + 2k\pi}{4}\right)}$ for $k = 0, 1, 2, 3$.
These roots are $e^{i\pi/8}, e^{i5\pi/8}, e^{i9\pi/8}, e^{i13\pi/8}$.
None of these roots are conjugates of each other.
According to Vieta's formulas for the polynomial $z^4 + 0z^3 + 0z^2 + 0z - i = 0$,the sum of the products of the roots taken two at a time is the coefficient of $z^2$,which is $0$.
Therefore,the sum is $0$.

(A) Given,$\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^n=1$.
We can write the complex number in polar form as $\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$.
So,the equation becomes $(e^{i\pi/6})^n = 1$,which is $e^{in\pi/6} = 1$.
For $e^{i\theta} = 1$,$\theta$ must be an integer multiple of $2\pi$.
Thus,$\frac{n\pi}{6} = 2k\pi$ for some integer $k$.
$n = 12k$.
For $k=1$,$n=12$.
Therefore,$n=12$ is a value that satisfies the equation.

(A) Let $z_1 = (-1)^{1/4}$. We can write $-1 = \cos(\pi + 2k\pi) + i\sin(\pi + 2k\pi) = e^{i(\pi + 2k\pi)}$.
Thus,$z_1 = e^{i(\frac{\pi}{4} + \frac{k\pi}{2})}$ for $k = 0, 1, 2, 3$.
The values are $e^{i\pi/4}, e^{i3\pi/4}, e^{i5\pi/4}, e^{i7\pi/4}$.
Let $z_2 = (-i)^{1/2}$. We can write $-i = \cos(\frac{3\pi}{2} + 2n\pi) + i\sin(\frac{3\pi}{2} + 2n\pi) = e^{i(\frac{3\pi}{2} + 2n\pi)}$.
Thus,$z_2 = e^{i(\frac{3\pi}{4} + n\pi)}$ for $n = 0, 1$.
The values are $e^{i3\pi/4}$ and $e^{i7\pi/4}$.
The common values are $e^{i3\pi/4}$ and $e^{i7\pi/4}$.
For $e^{i3\pi/4}$,$\alpha = \frac{3\pi}{4}$,so $\tan \alpha = \tan(\frac{3\pi}{4}) = -1$.
For $e^{i7\pi/4}$,$\alpha = \frac{7\pi}{4}$,so $\tan \alpha = \tan(\frac{7\pi}{4}) = -1$.
Therefore,$\tan \alpha = -1$.

(C) Given the equation $(z-4)^3=8 i$. Let $w = z-4$,then $w^3 = 8 i = 8 e^{i \pi/2}$.
The roots are $w_k = 2 e^{i(\pi/2 + 2k\pi)/3}$ for $k=0, 1, 2$.
For $k=0$,$w_0 = 2 e^{i\pi/6} = 2(\cos(\pi/6) + i\sin(\pi/6)) = 2(\frac{\sqrt{3}}{2} + i\frac{1}{2}) = \sqrt{3} + i$.
For $k=1$,$w_1 = 2 e^{i(5\pi/6)} = 2(\cos(5\pi/6) + i\sin(5\pi/6)) = 2(-\frac{\sqrt{3}}{2} + i\frac{1}{2}) = -\sqrt{3} + i$.
For $k=2$,$w_2 = 2 e^{i(9\pi/6)} = 2 e^{i(3\pi/2)} = 2(0 - i) = -2 i$.
Since $z = w+4$,the roots are $z_0 = 4+\sqrt{3}+i$,$z_1 = 4-\sqrt{3}+i$,and $z_2 = 4-2 i$.
Comparing these with $a-2 i, b+i, c+i$,we identify $a=4, b=4+\sqrt{3}, c=4-\sqrt{3}$.
Then $abc = 4(4+\sqrt{3})(4-\sqrt{3}) = 4(16-3) = 4(13) = 52$.
Therefore,$\sqrt{abc} = \sqrt{52} = \sqrt{4 \times 13} = 2 \sqrt{13}$.

(D) Given that $A_n = \cos \left(\frac{\pi}{2^n}\right) + i \sin \left(\frac{\pi}{2^n}\right) = e^{i \frac{\pi}{2^n}}$.
Then,$A_1 A_2 A_3 A_4 = e^{i \pi \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}\right)}$.
Summing the exponents: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{8+4+2+1}{16} = \frac{15}{16}$.
So,$(A_1 A_2 A_3 A_4)^4 = (e^{i \pi \frac{15}{16}})^4 = e^{i \pi \frac{15}{4}}$.
$e^{i \frac{15\pi}{4}} = e^{i (4\pi - \frac{\pi}{4})} = \cos(4\pi - \frac{\pi}{4}) + i \sin(4\pi - \frac{\pi}{4})$.
$= \cos(\frac{\pi}{4}) - i \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} = \frac{1-i}{\sqrt{2}}$.

(C) Let $z_1 = \sqrt{3}-i = 2(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6}))$. Then $z_1^{2016} = 2^{2016}(\cos(-\frac{2016\pi}{6}) + i\sin(-\frac{2016\pi}{6})) = 2^{2016}(\cos(-336\pi) + i\sin(-336\pi)) = 2^{2016}(1 + 0i) = 2^{2016}$.
Let $z_2 = -\sqrt{3}-i = 2(\cos(-\frac{5\pi}{6}) + i\sin(-\frac{5\pi}{6}))$. Then $z_2^{2019} = 2^{2019}(\cos(-\frac{2019 \times 5\pi}{6}) + i\sin(-\frac{2019 \times 5\pi}{6})) = 2^{2019}(\cos(-\frac{3365\pi}{2}) + i\sin(-\frac{3365\pi}{2}))$.
Since $-\frac{3365\pi}{2} = -1682\pi - \frac{\pi}{2}$,we have $\cos(-\frac{3365\pi}{2}) = \cos(-\frac{\pi}{2}) = 0$ and $\sin(-\frac{3365\pi}{2}) = \sin(-\frac{\pi}{2}) = -1$.
Thus,$z_2^{2019} = 2^{2019}(0 - i) = -i 2^{2019}$.
The sum is $2^{2016} - i 2^{2019}$.
The imaginary part is $-2^{2019}$.

(B) Given,$i z^4 + 1 = 0$.
$i z^4 = -1$.
$z^4 = \frac{-1}{i} = i$.
We know that $i = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}$.
So,$z^4 = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}$.
Using De Moivre's Theorem,$z = \left(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}\right)^{\frac{1}{4}} = \cos \frac{\pi}{8} + i \sin \frac{\pi}{8}$.
Thus,the real part of $z$ is $\operatorname{Re}(z) = \cos \frac{\pi}{8}$.

(A) Given $z = \cos \alpha + i \sin \alpha$,where $0 < \alpha < \frac{\pi}{4}$.
Using De Moivre's Theorem,$z^n = \cos(n\alpha) + i \sin(n\alpha)$.
$\frac{1+z^4}{1-z^3} = \frac{1 + \cos 4\alpha + i \sin 4\alpha}{1 - \cos 3\alpha - i \sin 3\alpha}$
$= \frac{2 \cos^2 2\alpha + 2i \sin 2\alpha \cos 2\alpha}{2 \sin^2 \frac{3\alpha}{2} - 2i \sin \frac{3\alpha}{2} \cos \frac{3\alpha}{2}}$
$= \frac{2 \cos 2\alpha (\cos 2\alpha + i \sin 2\alpha)}{2 \sin \frac{3\alpha}{2} (\sin \frac{3\alpha}{2} - i \cos \frac{3\alpha}{2})}$
Taking the modulus on both sides:
$\left|\frac{1+z^4}{1-z^3}\right| = \left| \frac{\cos 2\alpha}{\sin \frac{3\alpha}{2}} \right| \times \frac{|\cos 2\alpha + i \sin 2\alpha|}{|\sin \frac{3\alpha}{2} - i \cos \frac{3\alpha}{2}|}$
Since $|\cos \theta + i \sin \theta| = 1$ and $|\sin \theta - i \cos \theta| = | -i(\cos \theta + i \sin \theta) | = |-i| \times 1 = 1$,
$\left|\frac{1+z^4}{1-z^3}\right| = \frac{\cos 2\alpha}{\sin \frac{3\alpha}{2}}$.

(A) Given that,$n=8$.
We need to evaluate $(\sqrt{3}+i)^8+(\sqrt{3}-i)^8$.
First,convert the complex numbers to polar form: $\sqrt{3}+i = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) = 2e^{i\pi/6}$ and $\sqrt{3}-i = 2(\cos \frac{\pi}{6} - i \sin \frac{\pi}{6}) = 2e^{-i\pi/6}$.
Then,$(\sqrt{3}+i)^8 = (2e^{i\pi/6})^8 = 2^8 e^{i8\pi/6} = 256 e^{i4\pi/3}$.
And $(\sqrt{3}-i)^8 = (2e^{-i\pi/6})^8 = 2^8 e^{-i8\pi/6} = 256 e^{-i4\pi/3}$.
Summing these: $256(e^{i4\pi/3} + e^{-i4\pi/3}) = 256(2 \cos \frac{4\pi}{3})$.
Since $\cos \frac{4\pi}{3} = \cos(240^\circ) = -\frac{1}{2}$,
the expression becomes $256 \times 2 \times (-\frac{1}{2}) = -256$.

(A) The given equation is $x^{11}-x^7+x^4-1=0$.
Factoring the expression:
$x^7(x^4-1) + 1(x^4-1) = 0$
$(x^7+1)(x^4-1) = 0$
This implies $x^7 = -1$ or $x^4 = 1$.
For $x^4 = 1$,the roots are $1, -1, i, -i$. The root $i$ has an argument of $\frac{\pi}{2}$,which is on the boundary of the first quadrant.
For $x^7 = -1$,the roots are $e^{i(\frac{(2k+1)\pi}{7})}$ for $k = 0, 1, 2, 3, 4, 5, 6$.
The arguments are $\frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}, \pi, \frac{9\pi}{7}, \frac{11\pi}{7}, \frac{13\pi}{7}$.
The arguments in the first quadrant $(0 < \theta < \frac{\pi}{2})$ are $\frac{\pi}{7}$ and $\frac{3\pi}{7}$.
Thus,there are $2$ such roots.

(B) Given $z+\frac{1}{z}=1$,which implies $z^2-z+1=0$.
The roots of this quadratic equation are $z = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2} = -\omega$ and $-\omega^2$,where $\omega$ is the complex cube root of unity.
Let $E = \frac{(z^{20}+1)(z^{40}+1)(z^{60}+1)}{z^{60}}$.
For $z = -\omega$:
$z^{20} = (-\omega)^{20} = \omega^{20} = \omega^{18} \cdot \omega^2 = \omega^2$
$z^{40} = (-\omega)^{40} = \omega^{40} = \omega^{39} \cdot \omega = \omega$
$z^{60} = (-\omega)^{60} = \omega^{60} = 1$
Substituting these values into $E$:
$E = \frac{(\omega^2+1)(\omega+1)(1+1)}{1} = 2(\omega^3 + \omega^2 + \omega + 1)$.
Since $1+\omega+\omega^2=0$,we have $\omega^2+\omega = -1$.
$E = 2(1 + (-1)) = 2(0) = 0$. Wait,let's re-evaluate: $E = 2(\omega^3 + \omega^2 + \omega + 1) = 2(1 + \omega^2 + \omega + 1) = 2(1 + 0) = 2$.
Thus,the value is $2$.

(C) Given the quadratic equation $x^2-4x+8=0$.
Using the quadratic formula,the roots are $\alpha, \beta = \frac{4 \pm \sqrt{16-32}}{2} = \frac{4 \pm 4i}{2} = 2 \pm 2i$.
Converting to polar form: $\alpha, \beta = 2\sqrt{2}(\frac{1}{\sqrt{2}} \pm i\frac{1}{\sqrt{2}}) = 2\sqrt{2}(\cos \frac{\pi}{4} \pm i\sin \frac{\pi}{4})$.
Using De Moivre's Theorem,$\alpha^{2n} = (2\sqrt{2})^{2n}(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4})^{2n} = (2^{3/2})^{2n}(\cos \frac{n\pi}{2} + i\sin \frac{n\pi}{2}) = 2^{3n}(\cos \frac{n\pi}{2} + i\sin \frac{n\pi}{2})$.
Similarly,$\beta^{2n} = 2^{3n}(\cos \frac{n\pi}{2} - i\sin \frac{n\pi}{2})$.
Adding these,$\alpha^{2n} + \beta^{2n} = 2^{3n}(\cos \frac{n\pi}{2} + i\sin \frac{n\pi}{2} + \cos \frac{n\pi}{2} - i\sin \frac{n\pi}{2}) = 2^{3n} \cdot 2 \cos \frac{n\pi}{2} = 2^{3n+1} \cos \frac{n\pi}{2}$.

(A) Given $\alpha = \omega + 2\omega^2 - 3$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.
Substituting this into the expression for $\alpha$:
$\alpha = (\omega + \omega^2) + \omega^2 - 3 = -1 + \omega^2 - 3 = \omega^2 - 4$.
Thus,$\omega^2 = \alpha + 4$.
Cubing both sides,we get $(\omega^2)^3 = (\alpha + 4)^3$.
Since $\omega^3 = 1$,we have $\omega^6 = 1$.
So,$1 = \alpha^3 + 3(\alpha^2)(4) + 3(\alpha)(4^2) + 4^3$.
$1 = \alpha^3 + 12\alpha^2 + 48\alpha + 64$.
Subtracting $61$ from both sides:
$\alpha^3 + 12\alpha^2 + 48\alpha + 3 = 1 - 64 = -63$.

(C) Given the equation $x^2+2x+4=0$,we can write it as $(x+1)^2+3=0$.
Solving for $x$,we get $x = -1 \pm \sqrt{3}i$.
Since $\alpha$ lies in the $2^{nd}$ quadrant,$\alpha = -1 + \sqrt{3}i = 2(\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3})) = 2\text{cis}(\frac{2\pi}{3})$.
Then $\beta = -1 - \sqrt{3}i = 2\text{cis}(-\frac{2\pi}{3})$.
Using De Moivre's Theorem,$\alpha^{2024} = 2^{2024}\text{cis}(\frac{2024 \times 2\pi}{3}) = 2^{2024}\text{cis}(\frac{4048\pi}{3}) = 2^{2024}\text{cis}(\frac{4\pi}{3})$.
Similarly,$\beta^{2024} = 2^{2024}\text{cis}(-\frac{4\pi}{3})$.
Now,$\alpha^{2024}-\beta^{2024} = 2^{2024}(\text{cis}(\frac{4\pi}{3}) - \text{cis}(-\frac{4\pi}{3}))$.
$= 2^{2024}((\cos(\frac{4\pi}{3}) + i\sin(\frac{4\pi}{3})) - (\cos(-\frac{4\pi}{3}) + i\sin(-\frac{4\pi}{3})))$.
$= 2^{2024}(i\sin(\frac{4\pi}{3}) - i\sin(-\frac{4\pi}{3})) = 2^{2024}(i(-\frac{\sqrt{3}}{2}) - i(\frac{\sqrt{3}}{2}))$.
$= 2^{2024}(-i\sqrt{3}) = -2^{2024}\sqrt{3}i$.
Comparing with $ik$,we get $k = -2^{2024}\sqrt{3}$.

(C) Given $x_n = \cos \frac{\pi}{2^n} + i \sin \frac{\pi}{2^n} = e^{i \frac{\pi}{2^n}}$.
We need to find the product $P = \prod_{n=1}^{\infty} x_n$.
$P = \prod_{n=1}^{\infty} e^{i \frac{\pi}{2^n}} = e^{i \pi \sum_{n=1}^{\infty} \frac{1}{2^n}}$.
The sum of the geometric series is $\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1/2}{1 - 1/2} = 1$.
Therefore,$P = e^{i \pi (1)} = e^{i \pi}$.
Using Euler's formula,$e^{i \pi} = \cos \pi + i \sin \pi = -1 + 0i = -1$.

(C) Let $z = \frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}$.
Using $\sin \theta = \cos(\frac{\pi}{2} - \theta)$ and $\cos \theta = \sin(\frac{\pi}{2} - \theta)$,we have $\frac{\pi}{2} - \frac{2 \pi}{9} = \frac{5 \pi}{18}$.
So,$z = \frac{1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18}}{1+\cos \frac{5 \pi}{18}-i \sin \frac{5 \pi}{18}}$.
Using 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:
$z = \frac{2 \cos^2 \frac{5 \pi}{36} + i 2 \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}}{2 \cos^2 \frac{5 \pi}{36} - i 2 \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}} = \frac{\cos \frac{5 \pi}{36} + i \sin \frac{5 \pi}{36}}{\cos \frac{5 \pi}{36} - i \sin \frac{5 \pi}{36}} = \frac{e^{i 5 \pi / 36}}{e^{-i 5 \pi / 36}} = e^{i 5 \pi / 18}$.
Given $z^n = 1$,we have $(e^{i 5 \pi / 18})^n = e^{i 5 n \pi / 18} = 1$.
This implies $\frac{5 n \pi}{18} = 2 k \pi$ for some integer $k$.
$n = \frac{36 k}{5}$.
For $n$ to be the least positive integer,we set $k=5$,which gives $n = 36$.