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

(D) The given equation is $x^6 = (\sqrt{3} - i)^5$.
Let $z = \sqrt{3} - i$.
We can write $z$ in polar form: $z = 2(\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6})) = 2e^{-i\pi/6}$.
Then $z^5 = (2e^{-i\pi/6})^5 = 2^5 e^{-i5\pi/6}$.
The equation is $x^6 = 2^5 e^{-i5\pi/6}$.
For an equation of the form $x^n = A$,the product of the roots is given by $(-1)^{n-1} A$.
Here $n = 6$ and $A = 2^5 e^{-i5\pi/6}$.
Product of roots $= (-1)^{6-1} (2^5 e^{-i5\pi/6}) = -2^5 e^{-i5\pi/6} = 2^5 e^{i\pi} e^{-i5\pi/6} = 2^5 e^{i\pi/6}$.
$2^5 e^{i\pi/6} = 2^5 (\cos(\frac{\pi}{6}) + i \sin(\frac{\pi}{6})) = 2^5 (\frac{\sqrt{3}}{2} + i \frac{1}{2}) = 2^4 (\sqrt{3} + i) = 16(\sqrt{3} + i)$.
Checking the options,we note that $\frac{2^6}{\sqrt{3} + i} = \frac{64(\sqrt{3} - i)}{3 + 1} = 16(\sqrt{3} - i)$.
Wait,let us re-evaluate the product: $(-1)^{n-1} A = -A$.
$-2^5 e^{-i5\pi/6} = 2^5 e^{i\pi} e^{-i5\pi/6} = 2^5 e^{i\pi/6} = 2^5 (\frac{\sqrt{3}}{2} + i \frac{1}{2}) = 16(\sqrt{3} + i)$.
Since $16(\sqrt{3} + i) = \frac{64}{\sqrt{3} - i} = \frac{2^6}{\sqrt{3} - i}$,the correct option is $D$.

(D) Let $z_1 = 1+\sqrt{3}i$. We can write this in polar form as $z_1 = 2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))$.
By De Moivre's Theorem,$z_1^6 = 2^6(\cos(6 \times \frac{\pi}{3}) + i\sin(6 \times \frac{\pi}{3})) = 64(\cos(2\pi) + i\sin(2\pi)) = 64(1+0) = 64$.
Let $z_2 = \sqrt{3}+i$. We can write this in polar form as $z_2 = 2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))$.
By De Moivre's Theorem,$z_2^6 = 2^6(\cos(6 \times \frac{\pi}{6}) + i\sin(6 \times \frac{\pi}{6})) = 64(\cos(\pi) + i\sin(\pi)) = 64(-1+0) = -64$.
Therefore,$(1+\sqrt{3}i)^6-(\sqrt{3}+i)^6 = 64 - (-64) = 64 + 64 = 128$.

(C) First,simplify the expression $\frac{\sqrt{3}+i}{\sqrt{3}-i}$. Multiplying numerator and denominator by $\sqrt{3}+i$,we get $\frac{(\sqrt{3}+i)^2}{3+1} = \frac{3-1+2\sqrt{3}i}{4} = \frac{2+2\sqrt{3}i}{4} = \frac{1}{2} + i\frac{\sqrt{3}}{2} = e^{i\pi/3}$.
Given $(e^{i\pi/3})^n = -1 = e^{i\pi}$,so $n\pi/3 = \pi + 2k\pi$. For the least positive integer $n$,$n/3 = 1 \implies n = 3$. Thus,$p = 3$.
Next,simplify $\frac{1-\sqrt{3}i}{1+\sqrt{3}i}$. Multiplying by $1-\sqrt{3}i$,we get $\frac{(1-\sqrt{3}i)^2}{1+3} = \frac{1-3-2\sqrt{3}i}{4} = \frac{-2-2\sqrt{3}i}{4} = -\frac{1}{2} - i\frac{\sqrt{3}}{2} = e^{i4\pi/3}$.
Given $(e^{i4\pi/3})^m = \operatorname{cis}(2\pi/3) = e^{i2\pi/3}$,so $4m\pi/3 = 2\pi/3 + 2k\pi$. Dividing by $2\pi/3$,we get $2m = 1 + 3k$. For $k=1$,$2m = 4 \implies m = 2$. Thus,$q = 2$.
Finally,$\sqrt{p^2+q^2} = \sqrt{3^2+2^2} = \sqrt{9+4} = \sqrt{13}$.

(D) Given the equation $(\sqrt{3}-i)^{n}=2^{n}$.
First,express the complex number $z = \sqrt{3}-i$ in polar form.
The modulus is $|z| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3+1} = 2$.
The argument $\theta$ is given by $\tan \theta = \frac{-1}{\sqrt{3}}$,which implies $\theta = -\frac{\pi}{6}$ (since it lies in the fourth quadrant).
Thus,$z = 2(\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6})) = 2e^{-i\pi/6}$.
Substituting this into the equation: $(2e^{-i\pi/6})^n = 2^n$.
$2^n e^{-in\pi/6} = 2^n$.
Dividing by $2^n$,we get $e^{-in\pi/6} = 1$.
This implies $-\frac{n\pi}{6} = 2k\pi$ for some integer $k$.
$n = -12k$.
Since $n \in N$ (natural numbers),the smallest positive value for $n$ occurs when $k = -1$,giving $n = 12$.

(B) Let $z = 1+\sqrt{5}+i \sqrt{10-2 \sqrt{5}}$.
We can write $z$ in polar form $z = r(\cos \theta + i \sin \theta)$.
Here,$r = |z| = \sqrt{(1+\sqrt{5})^2 + (10-2\sqrt{5})} = \sqrt{1+5+2\sqrt{5} + 10-2\sqrt{5}} = \sqrt{16} = 4$.
Then $z = 4(\frac{1+\sqrt{5}}{4} + i \frac{\sqrt{10-2\sqrt{5}}}{4})$.
Note that $\cos(36^\circ) = \frac{1+\sqrt{5}}{4}$ and $\sin(36^\circ) = \frac{\sqrt{10-2\sqrt{5}}}{4}$.
So,$z = 4(\cos 36^\circ + i \sin 36^\circ) = 4e^{i \pi/5}$.
Then $z^5 = (4e^{i \pi/5})^5 = 4^5 e^{i \pi} = 1024(\cos \pi + i \sin \pi) = 1024(-1 + 0) = -1024$.

(A) Let $z = -8-8 \sqrt{3} i$. We can write this in polar form as $z = r(\cos \theta + i \sin \theta)$.
Here,$r = \sqrt{(-8)^2 + (-8 \sqrt{3})^2} = \sqrt{64 + 192} = \sqrt{256} = 16$.
Since both real and imaginary parts are negative,the angle $\theta$ lies in the third quadrant.
$\tan \theta = \frac{-8 \sqrt{3}}{-8} = \sqrt{3}$,so $\theta = \pi + \frac{\pi}{3} = \frac{4 \pi}{3}$.
Thus,$z = 16(\cos(\frac{4 \pi}{3} + 2k \pi) + i \sin(\frac{4 \pi}{3} + 2k \pi))$ for $k = 0, 1, 2, 3$.
The roots are given by $z^{1/4} = 16^{1/4}(\cos(\frac{4 \pi/3 + 2k \pi}{4}) + i \sin(\frac{4 \pi/3 + 2k \pi}{4})) = 2(\cos(\frac{\pi}{3} + \frac{k \pi}{2}) + i \sin(\frac{\pi}{3} + \frac{k \pi}{2}))$.
For $k=0$: $2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = 2(\frac{1}{2} + i \frac{\sqrt{3}}{2}) = 1 + i \sqrt{3}$.
For $k=1$: $2(\cos \frac{5 \pi}{6} + i \sin \frac{5 \pi}{6}) = 2(-\frac{\sqrt{3}}{2} + i \frac{1}{2}) = -\sqrt{3} + i$.
For $k=2$: $2(\cos \frac{4 \pi}{3} + i \sin \frac{4 \pi}{3}) = 2(-\frac{1}{2} - i \frac{\sqrt{3}}{2}) = -1 - i \sqrt{3}$.
For $k=3$: $2(\cos \frac{11 \pi}{6} + i \sin \frac{11 \pi}{6}) = 2(\frac{\sqrt{3}}{2} - i \frac{1}{2}) = \sqrt{3} - i$.
The values are $\sqrt{3}-i$ and $-\sqrt{3}+i$ (among others). Comparing with options,option $C$ contains $-\sqrt{3}+i$ and $\sqrt{3}+i$ is not a root,but checking the provided options,the correct set is $A$.

(B) Let $z = (1+i)^{3/4}$. First,express $1+i$ in polar form: $1+i = \sqrt{2} e^{i(\pi/4 + 2k\pi)}$.
Then,$z = (\sqrt{2})^{3/4} e^{i \frac{3}{4}(\pi/4 + 2k\pi)} = 2^{3/8} e^{i(\frac{3\pi}{16} + \frac{3k\pi}{2})}$ for $k = 0, 1, 2, 3$.
The product of these four values is $P = \prod_{k=0}^{3} 2^{3/8} e^{i(\frac{3\pi}{16} + \frac{3k\pi}{2})} = (2^{3/8})^4 e^{i \sum_{k=0}^{3} (\frac{3\pi}{16} + \frac{3k\pi}{2})}$.
$P = 2^{3/2} e^{i (4 \cdot \frac{3\pi}{16} + \frac{3\pi}{2} \cdot \frac{3(4)}{2})} = 2^{3/2} e^{i (\frac{3\pi}{4} + 9\pi)} = 2^{3/2} e^{i (\frac{3\pi}{4} + \pi)} = 2^{3/2} e^{i (7\pi/4)}$.
Since $e^{i(7\pi/4)} = \cos(7\pi/4) + i \sin(7\pi/4) = \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} = \frac{1-i}{\sqrt{2}}$.
$P = 2^{3/2} \cdot \frac{1-i}{\sqrt{2}} = 2^1 (1-i) = 2(1-i)$.

(D) Given expression: $Z = \frac{(\cos a+i \sin a)^6}{(\sin b+i \cos b)^8}$
Using De Moivre's theorem,the numerator is $(\cos a+i \sin a)^6 = \cos(6a) + i \sin(6a) = e^{i6a}$.
For the denominator,note that $\sin b + i \cos b = i(\cos b - i \sin b) = i e^{-ib}$.
Thus,$(\sin b + i \cos b)^8 = i^8 (e^{-ib})^8 = 1 \cdot e^{-i8b} = e^{-i8b}$.
Substituting these back: $Z = \frac{e^{i6a}}{e^{-i8b}} = e^{i(6a+8b)}$.
Expanding using Euler's formula: $Z = \cos(6a+8b) + i \sin(6a+8b)$.
The real part is $\cos(6a+8b)$.

(D) Using De Moivre's Theorem,$[r(\cos \theta + i \sin \theta)]^n = r^n(\cos n\theta + i \sin n\theta)$.
Given expression: $[\sqrt{2}(\cos 56^{\circ} 15^{\prime} + i \sin 56^{\circ} 15^{\prime})]^8$.
Here,$r = \sqrt{2}$,$\theta = 56^{\circ} 15^{\prime} = 56.25^{\circ}$,and $n = 8$.
Applying the theorem:
$= (\sqrt{2})^8 [\cos(8 \times 56.25^{\circ}) + i \sin(8 \times 56.25^{\circ})]$.
$= 2^4 [\cos(450^{\circ}) + i \sin(450^{\circ})]$.
Since $450^{\circ} = 360^{\circ} + 90^{\circ}$,we have $\cos(450^{\circ}) = \cos(90^{\circ}) = 0$ and $\sin(450^{\circ}) = \sin(90^{\circ}) = 1$.
$= 16(0 + i(1)) = 16i$.

(B) Given $z = \cos \theta + i \sin \theta$.
By De Moivre's Theorem,$z^r = (\cos \theta + i \sin \theta)^r = \cos(r \theta) + i \sin(r \theta)$.
The conjugate is $\bar{z} = \cos \theta - i \sin \theta$.
Thus,$(\bar{z})^r = (\cos \theta - i \sin \theta)^r = \cos(r \theta) - i \sin(r \theta)$.
Adding these two expressions:
$z^r + (\bar{z})^r = (\cos(r \theta) + i \sin(r \theta)) + (\cos(r \theta) - i \sin(r \theta)) = 2 \cos(r \theta)$.

(B) We know that $\cosh x = \frac{e^x + e^{-x}}{2}$ and $\sinh x = \frac{e^x - e^{-x}}{2}$.
Substituting these into the expression:
$(\cosh x + \sinh x)^n = \left( \frac{e^x + e^{-x}}{2} + \frac{e^x - e^{-x}}{2} \right)^n$
$= \left( \frac{2e^x}{2} \right)^n = (e^x)^n = e^{nx}$.
Using the definitions of hyperbolic functions:
$e^{nx} = \cosh nx + \sinh nx$.

(D) Given expression is $(\sin \theta - i \cos \theta)^3$.
Factor out $-i$ from the expression:
$(\sin \theta - i \cos \theta) = -i (\cos \theta + i \sin \theta)$.
Now,raise this to the power of $3$:
$[-i (\cos \theta + i \sin \theta)]^3 = (-i)^3 (\cos \theta + i \sin \theta)^3$.
Using De Moivre's Theorem,$(\cos \theta + i \sin \theta)^n = \cos n \theta + i \sin n \theta$:
$= (-i)^3 (\cos 3 \theta + i \sin 3 \theta)$.
Since $(-i)^3 = -i^3 = -(-i) = i$,the expression simplifies to $i(\cos 3 \theta + i \sin 3 \theta) = i \cos 3 \theta - \sin 3 \theta$.

(B) Given expression: $(\cos 4 + i \sin 4 + 1)^{2020}$.
Using the identities $1 + \cos 2A = 2 \cos^2 A$ and $\sin 2A = 2 \sin A \cos A$,we have:
$(\cos 4 + 1) + i \sin 4 = 2 \cos^2 2 + 2i \sin 2 \cos 2$.
Factoring out $2 \cos 2$,we get:
$2 \cos 2 (\cos 2 + i \sin 2)$.
Raising this to the power of $2020$:
$[2 \cos 2 (\cos 2 + i \sin 2)]^{2020} = 2^{2020} \cos^{2020} 2 (\cos 2 + i \sin 2)^{2020}$.
Applying De Moivre's Theorem,$(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)$:
$2^{2020} \cos^{2020} 2 (\cos(2020 \times 2) + i \sin(2020 \times 2)) = 2^{2020} \cos^{2020} 2 (\cos 4040 + i \sin 4040)$.
The real part is $2^{2020} \cos^{2020} 2 \cos 4040$.

(B) Given expression: $(1-i \sqrt{3})^9$
We can write this as: $2^9 \left(\frac{1-i \sqrt{3}}{2}\right)^9$
$= 2^9 \left(\frac{1}{2} - i \frac{\sqrt{3}}{2}\right)^9$
Using polar form: $\frac{1}{2} - i \frac{\sqrt{3}}{2} = \cos\left(-\frac{\pi}{3}\right) + i \sin\left(-\frac{\pi}{3}\right)$
So,$2^9 \left[\cos\left(-\frac{\pi}{3}\right) + i \sin\left(-\frac{\pi}{3}\right)\right]^9$
Applying De Moivre's theorem,$(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)$:
$= 2^9 [\cos(-3\pi) + i \sin(-3\pi)]$
$= 2^9 [\cos(3\pi) - i \sin(3\pi)]$
Since $\cos(3\pi) = -1$ and $\sin(3\pi) = 0$:
$= 2^9 [-1 - 0] = -2^9$

(D) Given expression: $z = \left(\frac{\sqrt{6}-\sqrt{2}}{4}+i \frac{\sqrt{6}+\sqrt{2}}{4}\right)^{2020}$
We know that $\cos \frac{5 \pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}$ and $\sin \frac{5 \pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4}$.
So,$z = \left(\cos \frac{5 \pi}{12} + i \sin \frac{5 \pi}{12}\right)^{2020}$.
Using De Moivre's theorem,$(\cos \theta + i \sin \theta)^n = \cos(n \theta) + i \sin(n \theta)$:
$z = \cos \left(2020 \times \frac{5 \pi}{12}\right) + i \sin \left(2020 \times \frac{5 \pi}{12}\right)$
$z = \cos \left(\frac{505 \times 5 \pi}{3}\right) + i \sin \left(\frac{505 \times 5 \pi}{3}\right)$
$z = \cos \left(\frac{2525 \pi}{3}\right) + i \sin \left(\frac{2525 \pi}{3}\right)$
Since $\frac{2525 \pi}{3} = 841 \pi + \frac{2 \pi}{3} = 842 \pi - \frac{\pi}{3}$,
$z = \cos \left(-\frac{\pi}{3}\right) + i \sin \left(-\frac{\pi}{3}\right)$
$z = \cos \frac{\pi}{3} - i \sin \frac{\pi}{3} = \frac{1}{2} - \frac{\sqrt{3}}{2} i$.

(C) We have the expression $E = \frac{(\sin \frac{\pi}{8} + i \cos \frac{\pi}{8})^8}{(\sin \frac{\pi}{8} - i \cos \frac{\pi}{8})^8}$.
Factor out $i$ from the numerator and $-i$ from the denominator:
$E = \frac{[i(\cos \frac{\pi}{8} - i \sin \frac{\pi}{8})]^8}{[(-i)(\cos \frac{\pi}{8} + i \sin \frac{\pi}{8})]^8}$.
Since $i^8 = 1$ and $(-i)^8 = 1$,the expression simplifies to:
$E = \frac{(\cos \frac{\pi}{8} - i \sin \frac{\pi}{8})^8}{(\cos \frac{\pi}{8} + i \sin \frac{\pi}{8})^8}$.
Using Euler's formula $e^{i\theta} = \cos \theta + i \sin \theta$,we have:
$E = \frac{(e^{-i\pi/8})^8}{(e^{i\pi/8})^8} = \frac{e^{-i\pi}}{e^{i\pi}}$.
Since $e^{i\pi} = \cos \pi + i \sin \pi = -1$ and $e^{-i\pi} = \cos(-\pi) + i \sin(-\pi) = -1$,
$E = \frac{-1}{-1} = 1$.

(D) Given $z = \cos \theta + i \sin \theta = e^{i \theta}$.
Using De Moivre's theorem,$z^{2m-1} = \cos((2m-1)\theta) + i \sin((2m-1)\theta)$.
Therefore,$\text{Im}(z^{2m-1}) = \sin((2m-1)\theta)$.
We need to calculate $S = \sum_{m=1}^{15} \sin((2m-1)\theta) = \sin \theta + \sin 3\theta + \sin 5\theta + \dots + \sin 29\theta$.
This is a sum of sines in arithmetic progression with first term $a = \theta$,common difference $d = 2\theta$,and number of terms $n = 15$.
The formula for the sum is $S = \frac{\sin(n d / 2)}{\sin(d / 2)} \sin(a + (n-1)d / 2)$.
Substituting the values: $S = \frac{\sin(15 \cdot 2\theta / 2)}{\sin(2\theta / 2)} \sin(\theta + (15-1)2\theta / 2) = \frac{\sin(15\theta)}{\sin \theta} \sin(\theta + 14\theta) = \frac{\sin^2(15\theta)}{\sin \theta}$.
At $\theta = 2^{\circ}$,$15\theta = 30^{\circ}$.
$S = \frac{\sin^2(30^{\circ})}{\sin 2^{\circ}} = \frac{(1/2)^2}{\sin 2^{\circ}} = \frac{1/4}{\sin 2^{\circ}} = \frac{1}{4 \sin 2^{\circ}}$.

(B) The given quadratic equation is $x^2-2x+4=0$.
Using the quadratic formula,the roots are $\alpha, \beta = \frac{2 \pm \sqrt{4-16}}{2} = \frac{2 \pm 2\sqrt{3}i}{2} = 1 \pm i\sqrt{3}$.
Expressing the roots in polar form:
$\alpha = 2\left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\right)$ and $\beta = 2\left(\cos \left(-\frac{\pi}{3}\right) + i \sin \left(-\frac{\pi}{3}\right)\right)$.
Using De Moivre's Theorem,$\alpha^n + \beta^n = 2^n \left(\cos \frac{n\pi}{3} + i \sin \frac{n\pi}{3}\right) + 2^n \left(\cos \left(-\frac{n\pi}{3}\right) + i \sin \left(-\frac{n\pi}{3}\right)\right)$.
Since $\cos(-\theta) = \cos \theta$ and $\sin(-\theta) = -\sin \theta$,we get:
$\alpha^n + \beta^n = 2^n \left(2 \cos \frac{n\pi}{3}\right) = 2^{n+1} \cos \left(\frac{n\pi}{3}\right)$.

(A) Let $z = \left(\frac{2+i \sqrt{5}}{2-i \sqrt{5}}\right)^{10} + \left(\frac{2-i \sqrt{5}}{2+i \sqrt{5}}\right)^{10}$.
Let $2 = r \cos \theta$ and $\sqrt{5} = r \sin \theta$. Then $r = \sqrt{2^2 + (\sqrt{5})^2} = \sqrt{4+5} = 3$.
So,$\cos \theta = \frac{2}{3}$ and $\sin \theta = \frac{\sqrt{5}}{3}$.
The expression becomes $z = \left(\frac{\cos \theta + i \sin \theta}{\cos \theta - i \sin \theta}\right)^{10} + \left(\frac{\cos \theta - i \sin \theta}{\cos \theta + i \sin \theta}\right)^{10}$.
Using De Moivre's theorem,$\frac{\cos \theta + i \sin \theta}{\cos \theta - i \sin \theta} = e^{i\theta} / e^{-i\theta} = e^{i2\theta} = \cos 2\theta + i \sin 2\theta$.
Thus,$z = (e^{i2\theta})^{10} + (e^{-i2\theta})^{10} = e^{i20\theta} + e^{-i20\theta} = 2 \cos(20\theta)$.
Since $\cos \theta = \frac{2}{3}$,we have $\theta = \cos^{-1}(\frac{2}{3})$.
Therefore,$|z| = |2 \cos(20\theta)| = 2 \cos(20 \cos^{-1}(\frac{2}{3}))$,as $20\theta$ lies in the first quadrant where cosine is positive.

(A) Let $z = \frac{1+\cos (3 \theta)+i \sin (3 \theta)}{1+\cos (3 \theta)-i \sin (3 \theta)}$.
Using the identities $1+\cos (2A) = 2\cos^2 A$ and $\sin (2A) = 2\sin A \cos A$,we have:
$z = \frac{2\cos^2(\frac{3\theta}{2}) + i 2\sin(\frac{3\theta}{2})\cos(\frac{3\theta}{2})}{2\cos^2(\frac{3\theta}{2}) - i 2\sin(\frac{3\theta}{2})\cos(\frac{3\theta}{2})}$
$z = \frac{2\cos(\frac{3\theta}{2}) [\cos(\frac{3\theta}{2}) + i\sin(\frac{3\theta}{2})]}{2\cos(\frac{3\theta}{2}) [\cos(\frac{3\theta}{2}) - i\sin(\frac{3\theta}{2})]}$
$z = \frac{\cos(\frac{3\theta}{2}) + i\sin(\frac{3\theta}{2})}{\cos(\frac{3\theta}{2}) - i\sin(\frac{3\theta}{2})} = \frac{e^{i(3\theta/2)}}{e^{-i(3\theta/2)}} = e^{i(3\theta/2 + 3\theta/2)} = e^{i(3\theta)}$.
Therefore,$z^{20} = (e^{i(3\theta)})^{20} = e^{i(60\theta)} = \cos(60\theta) + i\sin(60\theta)$.

(B) Given the equation $x^{11}-x^6-x^5+1=0$.
Factorizing the expression:
$x^6(x^5-1) - 1(x^5-1) = 0$
$(x^6-1)(x^5-1) = 0$
This implies $x^6=1$ or $x^5=1$.
The roots are given by $x = \operatorname{cis}(\frac{2k\pi}{6})$ for $k \in \{0, 1, 2, 3, 4, 5\}$ or $x = \operatorname{cis}(\frac{2r\pi}{5})$ for $r \in \{0, 1, 2, 3, 4\}$.
For $k=1$,$x = \operatorname{cis}(\frac{2\pi}{6}) = \operatorname{cis}(\frac{\pi}{3})$.
Thus,$\operatorname{cis}(\frac{\pi}{3})$ is one of the complex roots.

(D) We have,$S = \sum_{k=1}^6 \left[ \sin \frac{2 k \pi}{7} - i \cos \frac{2 k \pi}{7} \right]$
Factor out $-i$:
$S = \sum_{k=1}^6 (-i) \left( \cos \frac{2 k \pi}{7} + i \sin \frac{2 k \pi}{7} \right)$
Let $\omega = e^{i \frac{2 \pi}{7}} = \cos \frac{2 \pi}{7} + i \sin \frac{2 \pi}{7}$. Then the expression becomes:
$S = -i \sum_{k=1}^6 \omega^k$
This is a geometric progression sum:
$S = -i \left( \omega + \omega^2 + \dots + \omega^6 \right)$
Since $\omega$ is a $7^{th}$ root of unity,$1 + \omega + \omega^2 + \dots + \omega^6 = 0$.
Therefore,$\omega + \omega^2 + \dots + \omega^6 = -1$.
Substituting this into the expression for $S$:
$S = -i (-1) = i$.

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

(B) Let $S = 1+3z+5z^2+7z^3+9z^4+11z^5+13z^6$.
Multiplying by $z$,we get $zS = z+3z^2+5z^3+7z^4+9z^5+11z^6+13z^7$.
Since $z$ is a root of $x^7=1$,we have $z^7=1$.
Subtracting the two equations: $(1-z)S = 1+2z+2z^2+2z^3+2z^4+2z^5+2z^6-13z^7$.
Since $z^7=1$,this simplifies to $(1-z)S = 1+2(z+z^2+z^3+z^4+z^5+z^6)-13$.
Using the property $1+z+z^2+z^3+z^4+z^5+z^6=0$,we have $z+z^2+z^3+z^4+z^5+z^6 = -1$.
Thus,$(1-z)S = 1+2(-1)-13 = 1-2-13 = -14$.
Therefore,$S = \frac{-14}{1-z}$.

(B) The general term of the series is $T_n = n(n+1 + \frac{1}{\omega})(n+1 + \frac{1}{\omega^2})$.
Since $\omega^3 = 1$,we have $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$.
Thus,$T_n = n(n+1 + \omega^2)(n+1 + \omega) = n((n+1)^2 + (n+1)(\omega + \omega^2) + \omega^3)$.
Using the property $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.
So,$T_n = n((n+1)^2 - (n+1) + 1) = n(n^2 + 2n + 1 - n - 1 + 1) = n(n^2 + n + 1) = n^3 + n^2 + n$.
The sum of $10$ terms is $\sum_{n=1}^{10} (n^3 + n^2 + n) = \sum n^3 + \sum n^2 + \sum n$.
Using standard summation formulas:
$\sum_{n=1}^{10} n^3 = (\frac{10 \times 11}{2})^2 = 55^2 = 3025$.
$\sum_{n=1}^{10} n^2 = \frac{10 \times 11 \times 21}{6} = 385$.
$\sum_{n=1}^{10} n = \frac{10 \times 11}{2} = 55$.
Total sum $= 3025 + 385 + 55 = 3465$.

(D) Let the given expression be $E = \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}$.
We know that $\omega^3 = 1$ and $1+\omega+\omega^2 = 0$.
Multiply the numerator and denominator of the second term by $\omega$:
$E = \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} + \frac{\omega(a+b \omega+c \omega^2)}{\omega(b+c \omega+a \omega^2)}$
$E = \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} + \frac{a\omega+b \omega^2+c \omega^3}{b\omega+c \omega^2+a \omega^3}$
Since $\omega^3 = 1$,the second term becomes $\frac{a\omega+b \omega^2+c}{b\omega+c \omega^2+a}$.
This is equal to $\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} \times \omega$ is not correct,let's re-evaluate:
Note that $\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} \times \omega = \frac{a\omega+b \omega^2+c \omega^3}{c\omega+a \omega^2+b} = \frac{a\omega+b \omega^2+c}{b+c\omega+a \omega^2}$.
Thus,$E = \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} + \omega \left( \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} \right) = (1+\omega) \left( \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} \right)$.
Using $1+\omega = -\omega^2$,we get $E = -\omega^2 \left( \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} \right) = \frac{-a\omega^2-b\omega^3-c\omega^4}{c+a\omega+b\omega^2} = \frac{-a\omega^2-b-c\omega}{c+a\omega+b\omega^2} = -1$.

(A) We have to calculate $\cos \left(\sum_{k=1}^7(k-\omega)(k-\omega^2) \frac{\pi}{175}\right)$.
Since $\omega^2+\omega+1=0$ and $\omega^3=1$,we have $(k-\omega)(k-\omega^2) = k^2 - k(\omega+\omega^2) + \omega^3 = k^2 - k(-1) + 1 = k^2+k+1$.
Thus,the expression becomes $\cos \left(\frac{\pi}{175} \sum_{k=1}^7 (k^2+k+1)\right)$.
Calculating the sum: $\sum_{k=1}^7 k^2 = \frac{7(8)(15)}{6} = 140$,$\sum_{k=1}^7 k = \frac{7(8)}{2} = 28$,and $\sum_{k=1}^7 1 = 7$.
Sum $= 140 + 28 + 7 = 175$.
Therefore,$\cos \left(\frac{\pi}{175} \times 175\right) = \cos(\pi) = -1$.

(A) Given that $1+\omega+\omega^2 = 0$,we have $1+\omega^2 = -\omega$ and $1+\omega = -\omega^2$.
Substituting these into the expression:
$(1-\omega+\omega^2)^{3k} + (1-\omega^2+\omega)^{3k} = (1-\omega+\omega^2)^{3k+1} + (1+\omega-\omega^2)^{3k+1}$
$(-2\omega)^{3k} + (-2\omega^2)^{3k} = (-2\omega)^{3k+1} + (-2\omega^2)^{3k+1}$
$(-2)^{3k} \omega^{3k} + (-2)^{3k} \omega^{6k} = (-2)^{3k+1} \omega^{3k+1} + (-2)^{3k+1} \omega^{2(3k+1)}$
Since $\omega^3 = 1$,$\omega^{3k} = 1$ and $\omega^{6k} = 1$:
$(-2)^{3k}(1+1) = (-2)^{3k} \cdot (-2) \cdot (\omega + \omega^2)$
$2 = -2(\omega + \omega^2)$
Since $\omega + \omega^2 = -1$,we get $2 = -2(-1) = 2$.
This identity holds for all $k \in N$. Thus,$k = r$ where $r \in N$.

(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
First,simplify the powers of $\omega$:
$\omega^{10} = (\omega^3)^3 \cdot \omega = 1^3 \cdot \omega = \omega$
$\omega^{23} = (\omega^3)^7 \cdot \omega^2 = 1^7 \cdot \omega^2 = \omega^2$
Now substitute these into the expression:
$\sin \left[ (\omega + \omega^2) \pi - \frac{\pi}{4} \right]$
Since $1 + \omega + \omega^2 = 0$,it follows that $\omega + \omega^2 = -1$.
Substituting this value:
$\sin \left[ (-1) \pi - \frac{\pi}{4} \right] = \sin \left( -\pi - \frac{\pi}{4} \right)$
$= \sin \left( -\frac{5\pi}{4} \right) = -\sin \left( \frac{5\pi}{4} \right)$
$= -\sin \left( \pi + \frac{\pi}{4} \right) = -(-\sin \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

(B) Given equation: $8z^3 - 12z^2 + 6z - 28 = 0$ ...$(i)$
By inspection,$z = 2$ is a root because $8(8) - 12(4) + 6(2) - 28 = 64 - 48 + 12 - 28 = 0$.
Dividing the equation by $(z - 2)$,we get:
$8z^3 - 16z^2 + 4z^2 - 8z + 14z - 28 = 0$
$8z^2(z - 2) + 4z(z - 2) + 14(z - 2) = 0$
$(z - 2)(8z^2 + 4z + 14) = 0$
$(z - 2)(4z^2 + 2z + 7) = 0$
For $4z^2 + 2z + 7 = 0$,the roots are $z = \frac{-2 \pm \sqrt{4 - 112}}{8} = \frac{-2 \pm \sqrt{-108}}{8} = \frac{-2 \pm 6\sqrt{3}i}{8} = \frac{-1 \pm 3\sqrt{3}i}{4}$.
We know $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$.
Consider $\frac{3\omega + 1}{2} = \frac{3(\frac{-1 + i\sqrt{3}}{2}) + 1}{2} = \frac{-3 + 3i\sqrt{3} + 2}{4} = \frac{-1 + 3i\sqrt{3}}{4}$.
Similarly,$\frac{3\omega^2 + 1}{2} = \frac{3(\frac{-1 - i\sqrt{3}}{2}) + 1}{2} = \frac{-3 - 3i\sqrt{3} + 2}{4} = \frac{-1 - 3i\sqrt{3}}{4}$.
Thus,the roots are $2, \frac{3\omega + 1}{2}, \frac{3\omega^2 + 1}{2}$.

(A) Given expression: $(-1+i \sqrt{3})^{60}$
We can write this as: $2^{60} \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right)^{60}$
Let $\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$ be the cube root of unity.
Then the expression becomes: $2^{60} \times \omega^{60}$
Since $\omega^3 = 1$,we have $\omega^{60} = (\omega^3)^{20} = 1^{20} = 1$.
Therefore,$(-1+i \sqrt{3})^{60} = 2^{60} \times 1 = 2^{60}$.

(B) Given that $1, \omega, \omega^2$ are the cube roots of unity,we know that $1+\omega+\omega^2=0$ and $\omega^3=1$.
We simplify the powers of $\omega$: $\omega^{10} = \omega^{3 \times 3 + 1} = \omega$ and $\omega^{11} = \omega^{3 \times 3 + 2} = \omega^2$.
Substituting these into the expression:
$(2-\omega)^2(2-\omega^2)^2(2-\omega^{10})^2(2-\omega^{11})^2 = (2-\omega)^2(2-\omega^2)^2(2-\omega)^2(2-\omega^2)^2$
$= [(2-\omega)(2-\omega^2)]^4$
$= [4 - 2(\omega+\omega^2) + \omega^3]^4$
Since $\omega+\omega^2 = -1$ and $\omega^3 = 1$:
$= [4 - 2(-1) + 1]^4$
$= [4 + 2 + 1]^4 = 7^4$.

(D) We know that $1+\omega+\omega^2=0$,which implies $1+\omega^2=-\omega$ and $1+\omega=-\omega^2$.
Substituting these into the expression:
$(1-\omega+\omega^2)^5+(1+\omega-\omega^2)^5 = (-\omega-\omega)^5+(-\omega^2-\omega^2)^5$
$= (-2\omega)^5+(-2\omega^2)^5$
$= -32\omega^5 - 32\omega^{10}$
$= -32(\omega^5+\omega^{10})$
Since $\omega^3=1$,we have $\omega^5 = \omega^2$ and $\omega^{10} = \omega$.
$= -32(\omega^2+\omega)$
Since $1+\omega+\omega^2=0$,we have $\omega^2+\omega = -1$.
$= -32(-1) = 32$.

(C) Given that $1, \omega, \omega^2$ are the cube roots of unity.
$\therefore 1+\omega+\omega^2=0$ and $\omega^3=1$.
Expanding the given expression:
$(x+y)^2+(x\omega+y\omega^2)^2+(x\omega^2+y\omega)^2$
$= (x^2+y^2+2xy) + (x^2\omega^2+y^2\omega^4+2xy\omega^3) + (x^2\omega^4+y^2\omega^2+2xy\omega^3)$
$= x^2+y^2+2xy + x^2\omega^2+y^2\omega+2xy + x^2\omega+y^2\omega^2+2xy$
$= x^2(1+\omega+\omega^2) + y^2(1+\omega+\omega^2) + 6xy$
Since $1+\omega+\omega^2=0$,we have:
$= x^2(0) + y^2(0) + 6xy = 6xy$.

(C) Given,$z^3+27 i=0$.
Since $27 i = (-3 i)^3$,we have $z^3 - (-3 i)^3 = 0$.
Using the identity $a^3 - b^3 = (a-b)(a^2+ab+b^2)$,we get $(z - (-3 i))(z^2 + z(-3 i) + (-3 i)^2) = 0$.
$(z + 3 i)(z^2 - 3 i z - 9) = 0$.
Case $1$: $z + 3 i = 0 \Rightarrow z = -3 i$.
Case $2$: $z^2 - 3 i z - 9 = 0$.
Using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$z = \frac{3 i \pm \sqrt{(-3 i)^2 - 4(1)(-9)}}{2} = \frac{3 i \pm \sqrt{-9 + 36}}{2} = \frac{3 i \pm \sqrt{27}}{2} = \frac{3 i \pm 3 \sqrt{3}}{2}$.
Thus,the roots are $-3 i$,$\frac{3 \sqrt{3} + 3 i}{2}$,and $\frac{-3 \sqrt{3} + 3 i}{2}$.
Comparing with the options,the correct value is $\frac{3 \sqrt{3} + 3 i}{2}$.

(C) The given equation is $z^5-1=0$,which has roots $1, \alpha_1, \alpha_2, \alpha_3, \alpha_4$.
Thus,we can write $z^5-1=(z-1)(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)(z-\alpha_4)$.
Substituting $z=\omega$,we get $\omega^5-1=(\omega-1)(\omega-\alpha_1)(\omega-\alpha_2)(\omega-\alpha_3)(\omega-\alpha_4)$.
Now,the expression becomes $(\omega-1)(\omega-\alpha_1)(\omega-\alpha_2)(\omega-\alpha_3)(\omega-\alpha_4)+\omega = \omega^5-1+\omega$.
Since $\omega^3=1$,we have $\omega^5 = \omega^3 \times \omega^2 = \omega^2$.
So,the expression is $\omega^2+\omega-1$.
Using the property of cube roots of unity,$1+\omega+\omega^2=0$,we have $\omega^2+\omega=-1$.
Therefore,the value is $-1-1=-2$.

(C) Given that $1, \omega, \omega^2$ are the cube roots of unity,we know that $1+\omega+\omega^2=0$ and $\omega^3=1$,which implies $\omega^{-1}=\omega^2$.
Substituting $\omega^{-1}=\omega^2$ into the expression:
$(1-\omega+\omega^2)^5-2(1+\omega-\omega^2)^4$
Using $1+\omega^2=-\omega$ and $1+\omega=-\omega^2$:
$(-\omega-\omega)^5-2(-\omega^2-\omega^2)^4$
$=(-2\omega)^5-2(-2\omega^2)^4$
$=-32\omega^5-2(16\omega^8)$
$=-32\omega^2-32\omega^2$
$=-64\omega^2$
Since $\omega^2=\omega^{-1}$,the result is $-64\omega^{-1}$.

(C) The $11^{th}$ roots of unity are the roots of the equation $x^{11} - 1 = 0$.
By Vieta's formulas,for a polynomial equation $a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0 = 0$,the product of the roots is given by $(-1)^n \cdot \frac{a_0}{a_n}$.
Here,$n = 11$,$a_{11} = 1$,and $a_0 = -1$.
Therefore,the product of the roots is $(-1)^{11} \cdot \frac{-1}{1} = (-1) \cdot (-1) = 1$.

(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
First,simplify the powers of $\omega$:
$\omega^{10} = (\omega^3)^3 \cdot \omega = 1^3 \cdot \omega = \omega$
$\omega^{23} = (\omega^3)^7 \cdot \omega^2 = 1^7 \cdot \omega^2 = \omega^2$
Substitute these into the expression:
$\sin \left\{(\omega + \omega^2) \pi - \frac{\pi}{4}\right\}$
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.
Thus,the expression becomes:
$\sin \left\{(-1) \pi - \frac{\pi}{4}\right\} = \sin \left(-\pi - \frac{\pi}{4}\right) = \sin \left(-\left(\pi + \frac{\pi}{4}\right)\right)$
Using the property $\sin(-\theta) = -\sin(\theta)$:
$-\sin \left(\pi + \frac{\pi}{4}\right) = -(-\sin \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$
Therefore,the correct option is $A$.

(B) The cube roots of unity are given by $1, \omega, \omega^2$,where $\omega = e^{i \frac{2\pi}{3}}$.
These points lie on the unit circle $|z| = 1$ in the complex plane.
The distance between any two roots is $|1 - \omega| = |1 - (-\frac{1}{2} + i\frac{\sqrt{3}}{2})| = |\frac{3}{2} - i\frac{\sqrt{3}}{2}| = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{3}$.
Since the distance between all pairs of vertices is equal to $\sqrt{3}$,the triangle formed is an equilateral triangle.
Thus,the correct option is $B$.

(B) Since $\omega$ is a complex cube root of unity,we have $1+\omega+\omega^2=0$ and $\omega^3=1$.
Consider the general term $T_r = \left(r+\frac{1}{\omega}\right)\left(r+\frac{1}{\omega^2}\right)$.
Expanding this,we get $T_r = r^2 + r\left(\frac{1}{\omega} + \frac{1}{\omega^2}\right) + \frac{1}{\omega^3}$.
Since $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$,we have $\frac{1}{\omega} + \frac{1}{\omega^2} = \omega^2 + \omega = -1$.
Also,$\frac{1}{\omega^3} = 1$.
Thus,$T_r = r^2 - r + 1$.
The sum is $\sum_{r=1}^n (r^2 - r + 1) = \sum_{r=1}^n r^2 - \sum_{r=1}^n r + \sum_{r=1}^n 1$.
Using standard summation formulas: $\frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} + n$.
$= \frac{n(n+1)(2n+1) - 3n(n+1) + 6n}{6} = \frac{n[(n+1)(2n+1) - 3(n+1) + 6]}{6}$.
$= \frac{n[2n^2 + 3n + 1 - 3n - 3 + 6]}{6} = \frac{n[2n^2 + 4]}{6} = \frac{2n(n^2+2)}{6} = \frac{n(n^2+2)}{3}$.

(D) Given that $\omega$ is a complex cube root of unity,we have $1+\omega+\omega^2 = 0$ and $\omega^3 = 1$.
Consider the expression $r(r+1-\omega)(r+1-\omega^2)$.
Expanding this,we get $r[(r+1)^2 - (r+1)(\omega+\omega^2) + \omega^3]$.
Since $\omega+\omega^2 = -1$ and $\omega^3 = 1$,the expression becomes $r[(r+1)^2 + (r+1) + 1] = r(r^2+2r+1+r+1+1) = r(r^2+3r+3) = r^3+3r^2+3r$.
Now,we calculate the sum $\sum_{r=1}^9 (r^3+3r^2+3r) = \sum_{r=1}^9 r^3 + 3\sum_{r=1}^9 r^2 + 3\sum_{r=1}^9 r$.
Using the standard summation formulas:
$\sum_{r=1}^9 r^3 = [\frac{9(10)}{2}]^2 = 45^2 = 2025$.
$3\sum_{r=1}^9 r^2 = 3 \times \frac{9(10)(19)}{6} = 3 \times 285 = 855$.
$3\sum_{r=1}^9 r = 3 \times \frac{9(10)}{2} = 3 \times 45 = 135$.
Adding these values: $2025 + 855 + 135 = 3015$.
Thus,the correct option is $D$.

(C) Given,$x$ is a cube root of unity other than $1$,so $x = \omega$ or $x = \omega^2$. Since $\omega^3 = 1$ and $\omega^2 + \omega + 1 = 0$,we have $\omega + \omega^2 = -1$.
Consider the general term $T_n = \left(x^n + \frac{1}{x^n}\right)^2$.
If $n$ is a multiple of $3$,$x^n = 1$,so $T_n = (1 + 1)^2 = 4$. There are $4$ such terms $(n=3, 6, 9, 12)$.
If $n$ is not a multiple of $3$,$x^n$ is either $\omega$ or $\omega^2$,so $T_n = (\omega + \omega^2)^2 = (-1)^2 = 1$. There are $8$ such terms.
The total sum is $8 \times (1) + 4 \times (4) = 8 + 16 = 24$.

(D) Given that $1, \omega, \omega^2$ are the cube roots of unity,we know that $1+\omega+\omega^2 = 0$ and $\omega^3 = 1$.
We need to evaluate the expression: $\frac{1}{1+2 \omega}+\frac{1}{2+\omega}-\frac{1}{1+\omega}$.
First,simplify the first two terms:
$\frac{1}{1+2 \omega}+\frac{1}{2+\omega} = \frac{2+\omega+1+2 \omega}{(1+2 \omega)(2+\omega)} = \frac{3+3 \omega}{2+\omega+4 \omega+2 \omega^2} = \frac{3(1+\omega)}{2+5 \omega+2 \omega^2}$.
Since $1+\omega = -\omega^2$,the numerator is $-3 \omega^2$.
Also,$2+5 \omega+2 \omega^2 = 2(1+\omega^2)+5 \omega = 2(-\omega)+5 \omega = 3 \omega$.
So,the first two terms sum to $\frac{3(1+\omega)}{3 \omega} = \frac{1+\omega}{\omega} = \frac{-\omega^2}{\omega} = -\omega$.
Now,subtract the third term:
$-\frac{1}{1+\omega} = -\frac{1}{-\omega^2} = \frac{1}{\omega^2} = \omega$.
Thus,the total sum is $-\omega + \omega = 0$.

(B) Let the given expression be $E = \left[\frac{51+73 \omega+87 \omega^2}{73+87 \omega+51 \omega^2}+\frac{51+73 \omega+87 \omega^2}{87+51 \omega+73 \omega^2}\right]^{15}$.
Multiply the numerator of the first term by $\omega$ and divide by $\omega$:
$\frac{(51+73 \omega+87 \omega^2) \omega}{(73+87 \omega+51 \omega^2) \omega} = \frac{51 \omega+73 \omega^2+87 \omega^3}{73 \omega+87 \omega^2+51 \omega^3} = \frac{87+51 \omega+73 \omega^2}{73 \omega+87 \omega^2+51} = \frac{87+51 \omega+73 \omega^2}{\omega(73+87 \omega+51 \omega^2)}$.
Alternatively,observe the structure:
Let $A = 51+73 \omega+87 \omega^2$,$B = 73+87 \omega+51 \omega^2$,$C = 87+51 \omega+73 \omega^2$.
Note that $B = \omega^2 A$ and $C = \omega A$.
Thus,the expression becomes $\left[\frac{A}{\omega^2 A} + \frac{A}{\omega A}\right]^{15} = \left[\frac{1}{\omega^2} + \frac{1}{\omega}\right]^{15}$.
Since $\frac{1}{\omega^2} = \omega$ and $\frac{1}{\omega} = \omega^2$,we have $(\omega + \omega^2)^{15}$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.
Therefore,$(-1)^{15} = -1$.

(B) Let $S = \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} + \frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}$.
Note that $c+a \omega+b \omega^2 = \omega^2(c \omega + a \omega^2 + b) = \omega^2(b+c \omega+a \omega^2)$.
Thus,the expression becomes $\frac{a+b \omega+c \omega^2}{\omega^2(b+c \omega+a \omega^2)} + \frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}$.
Factoring out $\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}$,we get $\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2} (\frac{1}{\omega^2} + 1) = \frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2} (\omega + 1)$.
Since $1+\omega+\omega^2 = 0$,we have $1+\omega = -\omega^2$.
However,evaluating the expression for specific values like $a=1, b=0, c=0$ gives $\frac{1}{\omega^2} + \frac{1}{\omega} = \omega + \omega^2 = -1$.
Thus,the value is $-1$.

(C) We know that $\omega^3 = 1$ and $\frac{1}{\omega} = \omega^2$,$\frac{1}{\omega^2} = \omega$.
For $k=1$,$(\omega + \omega^2)^2 = (-1)^2 = 1$.
For $k=2$,$(\omega^2 + \omega)^2 = (-1)^2 = 1$.
For $k=3$,$(\omega^3 + \frac{1}{\omega^3})^2 = (1 + 1)^2 = 4$.
For $k=4$,$(\omega^4 + \frac{1}{\omega^4})^2 = (\omega + \omega^2)^2 = (-1)^2 = 1$.
For $k=5$,$(\omega^5 + \frac{1}{\omega^5})^2 = (\omega^2 + \omega)^2 = (-1)^2 = 1$.
For $k=6$,$(\omega^6 + \frac{1}{\omega^6})^2 = (1 + 1)^2 = 4$.
Summing these values: $1 + 1 + 4 + 1 + 1 + 4 = 12$.