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

(A) The cube roots of unity are the solutions to the equation $z^3 = 1$.
These roots are given by $1$,$\omega = \frac{-1 + i\sqrt{3}}{2}$,and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$.
Comparing these with the given options,the correct value is $\frac{-1 + i\sqrt{3}}{2}$.

(C) We know that $1 + \omega + \omega^2 = 0$,which implies $1 + \omega = -\omega^2$.
Substituting this into the given equation:
$(1 + \omega)^7 = A + B\omega$
$(-\omega^2)^7 = A + B\omega$
$-\omega^{14} = A + B\omega$
Since $\omega^3 = 1$,we have $\omega^{14} = (\omega^3)^4 \cdot \omega^2 = 1^4 \cdot \omega^2 = \omega^2$.
So,$-\omega^2 = A + B\omega$.
Using $1 + \omega + \omega^2 = 0$,we get $\omega^2 = -1 - \omega$.
Therefore,$-(-1 - \omega) = A + B\omega$.
$1 + \omega = A + B\omega$.
Comparing the coefficients,we get $A = 1$ and $B = 1$.

(B) The equation for the $n^{th}$ roots of unity is given by $x^n = 1$.
We can write $1$ in polar form as $\cos(2r\pi) + i\sin(2r\pi) = e^{i2r\pi}$ for $r = 0, 1, 2, \dots, n-1$.
Thus,the roots are $x = e^{i(2r\pi/n)}$.
Substituting values of $r$,the roots are $1, e^{i(2\pi/n)}, e^{i(4\pi/n)}, \dots, e^{i(2(n-1)\pi/n)}$.
These terms form a geometric progression $(G.P.)$ with the first term $a = 1$ and common ratio $r = e^{i(2\pi/n)}$.

(A) We know that for the cube roots of unity,$1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
Given expression: $(3 + \omega^2 + \omega^4)^6$.
Since $\omega^4 = \omega^3 \times \omega = 1 \times \omega = \omega$,the expression becomes $(3 + \omega^2 + \omega)^6$.
Using the identity $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.
Substituting this into the expression: $(3 + (-1))^6 = (2)^6$.
$(2)^6 = 64$.

(A) The expression $1 + \omega + \omega^2 + ... + \omega^{n-1}$ is a geometric progression with $n$ terms,where the first term $a = 1$ and the common ratio is $\omega$.
The sum of a geometric progression is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.
Substituting $a = 1$ and $r = \omega$,we get $S_n = \frac{\omega^n - 1}{\omega - 1}$.
Since $\omega$ is an $n^{th}$ root of unity,$\omega^n = 1$.
Therefore,$S_n = \frac{1 - 1}{\omega - 1} = \frac{0}{\omega - 1} = 0$ (given $\omega \neq 1$).

(D) The $n^{th}$ roots of unity are given by the formula:
${z_k} = {e^{\frac{i 2\pi (k - 1)}{n}}}$ for $k = 1, 2, ....., n$.
Taking the modulus of both sides:
$|{z_k}| = |{e^{\frac{i 2\pi (k - 1)}{n}}}| = 1$ for all $k = 1, 2, ....., n$.
Since the modulus of every $n^{th}$ root of unity is $1$,it follows that:
$|{z_k}| = |{z_{k + 1}}| = 1$ for all $k = 1, 2, ....., n-1$.
Therefore,option $D$ is correct.

(A) Given $a + b + c = 0$,we have $c = -(a + b)$.
Substitute $c$ into the expression:
$E = (a + b\omega + c\omega^2)^3 + (a + b\omega^2 + c\omega)^3$
Using $1 + \omega + \omega^2 = 0$,we know $\omega^2 = -1 - \omega$ and $\omega = -1 - \omega^2$.
Alternatively,using the property that for $a+b+c=0$,the expression simplifies to $27abc$.
Let $a=1, b=1, c=-2$. Then $a+b+c=0$.
Expression $= (1 + \omega - 2\omega^2)^3 + (1 + \omega^2 - 2\omega)^3$
$= (1 + \omega - 2(-1 - \omega))^3 + (1 + \omega^2 - 2(-1 - \omega^2))^3$
$= (1 + \omega + 2 + 2\omega)^3 + (1 + \omega^2 + 2 + 2\omega^2)^3$
$= (3 + 3\omega)^3 + (3 + 3\omega^2)^3$
$= 27(1 + \omega)^3 + 27(1 + \omega^2)^3$
$= 27(-\omega^2)^3 + 27(-\omega)^3$
$= 27(-\omega^6) + 27(-\omega^3)$
$= 27(-1) + 27(-1) = -54$.
Checking option $A$: $27abc = 27(1)(1)(-2) = -54$.
Thus,the expression is equal to $27abc$.

(A) The roots of the equation $z^4 = 1$ are the fourth roots of unity,which are $z_1 = 1, z_2 = i, z_3 = -1, z_4 = -i$.
We need to calculate the sum $\sum_{i=1}^4 z_i^3 = z_1^3 + z_2^3 + z_3^3 + z_4^3$.
Substituting the values: $1^3 + (i)^3 + (-1)^3 + (-i)^3$.
$= 1 - i - 1 + i = 0$.
Therefore,the sum is $0$.

(B) Since $\alpha$ is an imaginary cube root of unity,let $\alpha = \omega$.
Given expression: $\alpha^{3n + 1} + \alpha^{3n + 3} + \alpha^{3n + 5}$.
Using the properties of cube roots of unity,$\omega^3 = 1$ and $\omega^{3n} = 1$ for any $n \in N$.
Substituting these values:
$\omega^{3n} \cdot \omega^1 + \omega^{3n} \cdot \omega^3 + \omega^{3n} \cdot \omega^5 = 1 \cdot \omega + 1 \cdot 1 + 1 \cdot \omega^2$.
$= \omega + 1 + \omega^2$.
Since $1 + \omega + \omega^2 = 0$,the value is $0$.

(D) We know that the cube roots of unity are $1, \omega, \omega^2$,where $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$.
Given expression is $\omega^{20} + (\omega^2)^{20} = \omega^{20} + \omega^{40}$.
Since $\omega^3 = 1$,we have $\omega^{20} = (\omega^3)^6 \cdot \omega^2 = 1^6 \cdot \omega^2 = \omega^2$.
Similarly,$\omega^{40} = (\omega^3)^{13} \cdot \omega = 1^{13} \cdot \omega = \omega$.
Thus,the expression becomes $\omega^2 + \omega$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 + \omega = -1$.

(C) Given that $\omega$ is a complex cube root of unity,we have $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
We simplify the expression: $(3 + 5\omega + 3\omega^2)^2 + (3 + 3\omega + 5\omega^2)^2$.
Using $3 + 3\omega + 3\omega^2 = 3(1 + \omega + \omega^2) = 3(0) = 0$,we can rewrite the terms:
First term: $(3 + 3\omega + 3\omega^2 + 2\omega)^2 = (0 + 2\omega)^2 = 4\omega^2$.
Second term: $(3 + 3\omega + 3\omega^2 + 2\omega^2)^2 = (0 + 2\omega^2)^2 = 4\omega^4$.
Since $\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$,the sum is $4\omega^2 + 4\omega$.
Using $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.
Therefore,$4\omega^2 + 4\omega = 4(\omega^2 + \omega) = 4(-1) = -4$.

(C) Given that $\omega$ is an imaginary cube root of unity,we know that $\omega^3 = 1$ 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 $\omega + \omega^2 = -1$.
So,the expression becomes $\sin \left[ -1 \cdot \pi - \frac{\pi}{4} \right] = \sin \left( -\pi - \frac{\pi}{4} \right)$.
Using the property $\sin(-\theta) = -\sin(\theta)$:
$-\sin \left( \pi + \frac{\pi}{4} \right) = -(-\sin \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

(A) Let $z_1 = \frac{\sqrt{3} + i}{2} = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$.
Then $z_1^6 = (e^{i\pi/6})^6 = e^{i\pi} = -1$.
Let $z_2 = \frac{i - \sqrt{3}}{2} = \frac{-\sqrt{3} + i}{2} = \cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right) = e^{i5\pi/6}$.
Then $z_2^6 = (e^{i5\pi/6})^6 = e^{i5\pi} = \cos(5\pi) + i\sin(5\pi) = -1$.
Therefore,$z_1^6 + z_2^6 = -1 + (-1) = -2$.

(D) We know that $1 + \omega + \omega^2 = 0$,which implies $1 + \omega = -\omega^2$.
Substituting this into the expression:
$(1 + \omega - \omega^2)^7 = (-\omega^2 - \omega^2)^7$
$= (-2\omega^2)^7$
$= (-2)^7 \times (\omega^2)^7$
$= -128 \times \omega^{14}$
Since $\omega^3 = 1$,we have $\omega^{14} = (\omega^3)^4 \times \omega^2 = 1^4 \times \omega^2 = \omega^2$.
Therefore,the expression equals $-128\omega^2$.

(A) Let $z_1 = -1 + i\sqrt{3} = 2\omega$ and $z_2 = -1 - i\sqrt{3} = 2\omega^2$,where $\omega$ is the cube root of unity.
Then the expression is $E = \frac{(2\omega)^{15}}{(1 - i)^{20}} + \frac{(2\omega^2)^{15}}{(1 + i)^{20}}$.
Since $\omega^3 = 1$,$\omega^{15} = 1$ and $\omega^{30} = 1$.
$E = \frac{2^{15}}{(1 - i)^{20}} + \frac{2^{15}}{(1 + i)^{20}} = 2^{15} \left[ \frac{(1 + i)^{20} + (1 - i)^{20}}{(1 - i^2)^{20}} \right]$.
Since $1 - i^2 = 2$,the denominator is $2^{20}$.
$E = \frac{2^{15}}{2^{20}} [(1 + i)^{20} + (1 - i)^{20}]$.
Note that $(1 + i)^2 = 1 + i^2 + 2i = 2i$ and $(1 - i)^2 = 1 + i^2 - 2i = -2i$.
$E = \frac{1}{2^5} [(2i)^{10} + (-2i)^{10}] = \frac{1}{32} [2^{10} i^{10} + 2^{10} i^{10}] = \frac{2 \times 2^{10} \times (-1)}{32} = \frac{-2^{11}}{2^5} = -2^6 = -64$.

(A) The exponent is a geometric series with first term $a = \frac{1}{2}$ and common ratio $r = \frac{3}{4}$.
Sum of the infinite geometric series $S = \frac{a}{1 - r} = \frac{1/2}{1 - 3/4} = \frac{1/2}{1/4} = 2$.
Thus,the expression becomes $\omega + \omega^2$.
Since $\omega$ is a complex cube root of unity,$1 + \omega + \omega^2 = 0$.
Therefore,$\omega + \omega^2 = -1$.

(C) Given expression: $(3 + \omega + 3\omega^2)^4$.
We know that $1 + \omega + \omega^2 = 0$,which implies $\omega + \omega^2 = -1$.
Also,$3 + 3\omega^2 = 3(1 + \omega^2) = 3(-\omega) = -3\omega$.
Substituting this into the expression:
$(3 + 3\omega^2 + \omega)^4 = (-3\omega + \omega)^4$.
$= (-2\omega)^4$.
$= 16\omega^4$.
Since $\omega^3 = 1$,we have $\omega^4 = \omega^3 \times \omega = 1 \times \omega = \omega$.
Therefore,the value is $16\omega$.

(C) We know that $1 + \omega + \omega^2 = 0$.
Therefore,$1 + \omega^2 = -\omega$ and $1 + \omega = -\omega^2$.
Substituting these into the expression:
$(1 - \omega + \omega^2)(1 - \omega^2 + \omega)^6 = (-\omega - \omega)(-\omega^2 - \omega^2)^6$
$= (-2\omega)(-2\omega^2)^6$
$= (-2\omega)(64\omega^{12})$
Since $\omega^3 = 1$,$\omega^{12} = (\omega^3)^4 = 1^4 = 1$.
$= (-2\omega)(64 \times 1) = -128\omega$.

(D) The cube roots of unity are $1, \omega, \omega^2$.
Their product is $1 \times \omega \times \omega^2 = \omega^3$.
Since $\omega$ is a cube root of unity,$\omega^3 = 1$.
Therefore,the product is $1$.

(C) Given $z = \frac{\sqrt{3} + i}{-2} = -\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)$.
We know that $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$.
Multiplying $z$ by $i$,we get $iz = -\left(i\frac{\sqrt{3}}{2} - \frac{1}{2}\right) = -\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -\omega$.
Thus,$z = \frac{-\omega}{i} = i\omega$.
Now,$z^{69} = (i\omega)^{69} = i^{69} \cdot \omega^{69}$.
Since $i^{69} = i^{68} \cdot i = (i^4)^{17} \cdot i = 1^{17} \cdot i = i$ and $\omega^{69} = (\omega^3)^{23} = 1^{23} = 1$.
Therefore,$z^{69} = i \cdot 1 = i$.

(C) Given ${\omega _n} = \cos \left( {\frac{{2\pi }}{n}} \right) + i\sin \left( {\frac{{2\pi }}{n}} \right)$.
For $n=3$,${\omega _3} = \cos \left( {\frac{{2\pi }}{3}} \right) + i\sin \left( {\frac{{2\pi }}{3}} \right) = -\frac{1}{2} + i\frac{{\sqrt{3}}}{2} = \omega$,where $\omega$ is the complex cube root of unity.
Similarly,${\omega _3}^2 = \cos \left( {\frac{{4\pi }}{3}} \right) + i\sin \left( {\frac{{4\pi }}{3}} \right) = -\frac{1}{2} - i\frac{{\sqrt{3}}}{2} = {\omega ^2}$.
Now,the expression is $(x + y\omega + z{\omega ^2})(x + y{\omega ^2} + z\omega)$.
Expanding this product:
$= x(x + y{\omega ^2} + z\omega) + y\omega(x + y{\omega ^2} + z\omega) + z{\omega ^2}(x + y{\omega ^2} + z\omega)$
$= {x^2} + xy{\omega ^2} + xz\omega + xy\omega + {y^2}{\omega ^3} + yz{\omega ^2} + xz{\omega ^2} + yz{\omega ^4} + {z^2}{\omega ^3}$
Since ${\omega ^3} = 1$ and ${\omega ^4} = \omega$:
$= {x^2} + {y^2} + {z^2} + xy({\omega ^2} + \omega) + yz({\omega ^2} + \omega) + xz({\omega ^2} + \omega)$
Since $1 + \omega + {\omega ^2} = 0$,we have $\omega + {\omega ^2} = -1$.
$= {x^2} + {y^2} + {z^2} + xy(-1) + yz(-1) + xz(-1)$
$= {x^2} + {y^2} + {z^2} - xy - yz - zx$.

(D) Given $z + z^{-1} = 1$,we have $z + \frac{1}{z} = 1$,which implies $z^2 - z + 1 = 0$.
Using the quadratic formula,$z = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm i\sqrt{3}}{2} = -\omega$ or $-\omega^2$,where $\omega$ is a complex cube root of unity.
Case $1$: If $z = -\omega$,then $z^{100} + z^{-100} = (-\omega)^{100} + (-\omega)^{-100} = \omega^{100} + \frac{1}{\omega^{100}}$.
Since $\omega^3 = 1$,$\omega^{100} = (\omega^3)^{33} \cdot \omega = \omega$.
Thus,$z^{100} + z^{-100} = \omega + \frac{1}{\omega} = \omega + \omega^2 = -1$.
Case $2$: If $z = -\omega^2$,then $z^{100} + z^{-100} = (-\omega^2)^{100} + (-\omega^2)^{-100} = \omega^{200} + \frac{1}{\omega^{200}}$.
Since $\omega^3 = 1$,$\omega^{200} = (\omega^3)^{66} \cdot \omega^2 = \omega^2$.
Thus,$z^{100} + z^{-100} = \omega^2 + \frac{1}{\omega^2} = \omega^2 + \omega = -1$.
In both cases,the result is $-1$.

(C) Let $z = \frac{1 + i\sqrt{3}}{1 - i\sqrt{3}}$.
Multiply the numerator and denominator by the conjugate of the denominator $(1 + i\sqrt{3})$:
$z = \frac{(1 + i\sqrt{3})(1 + i\sqrt{3})}{(1 - i\sqrt{3})(1 + i\sqrt{3})} = \frac{1 + 2i\sqrt{3} + (i\sqrt{3})^2}{1^2 + (\sqrt{3})^2} = \frac{1 + 2i\sqrt{3} - 3}{1 + 3} = \frac{-2 + 2i\sqrt{3}}{4} = \frac{-1 + i\sqrt{3}}{2}$.
This is the value of $\omega$,the complex cube root of unity.
Thus,$z^n = \omega^n$.
For $\omega^n$ to be an integer,$n$ must be a multiple of $3$ because $\omega^3 = 1$.
The smallest positive integer $n$ is $3$.

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

(A) Given the expression: $(a + b)(a + b\omega)(a + b\omega^2)$
Since $\omega$ is a non-real cube root of unity,we know that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
First,multiply the last two terms: $(a + b\omega)(a + b\omega^2) = a^2 + ab\omega^2 + ab\omega + b^2\omega^3 = a^2 + ab(\omega + \omega^2) + b^2(1)$.
Since $\omega + \omega^2 = -1$,this simplifies to: $a^2 - ab + b^2$.
Now,multiply by the first term: $(a + b)(a^2 - ab + b^2) = a^3 + b^3$.

(B) The given complex number is $z = \frac{1}{2} + i\frac{\sqrt{3}}{2}$.
In polar form,this is $z = \cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right) = cis\left(\frac{\pi}{3}\right)$.
By De Moivre's Theorem,the $n$-th roots of $cis(\theta)$ are given by $cis\left(\frac{\theta + 2k\pi}{n}\right)$ for $k = 0, 1, 2, 3$.
For the fourth root $(n=4)$ and $k=0$,we get $cis\left(\frac{\pi/3}{4}\right) = cis\left(\frac{\pi}{12}\right)$.
Thus,$cis\left(\frac{\pi}{12}\right)$ is a fourth root of the given number.

(D) Let $(8)^{1/3} = x$.
Then $x^3 = 8$,which implies $x^3 - 8 = 0$.
Using the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$,we get $(x - 2)(x^2 + 2x + 4) = 0$.
This gives $x = 2$ or $x^2 + 2x + 4 = 0$.
Solving the quadratic equation $x^2 + 2x + 4 = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{-2 \pm \sqrt{4 - 16}}{2} = \frac{-2 \pm \sqrt{-12}}{2} = \frac{-2 \pm 2i\sqrt{3}}{2} = -1 \pm i\sqrt{3}$.
Thus,the cube roots of $8$ are $2, -1 + i\sqrt{3}, -1 - i\sqrt{3}$.
Therefore,all the given options are correct.

(C) Given that $\omega = \frac{-1 + \sqrt{3}i}{2}$ is the complex cube root of unity,we know that $1 + \omega + \omega^2 = 0$,which implies $1 + \omega^2 = -\omega$.
Substituting this into the expression:
$(3 + \omega + 3\omega^2)^4 = [3(1 + \omega^2) + \omega]^4$
$= [3(-\omega) + \omega]^4$
$= [-3\omega + \omega]^4$
$= [-2\omega]^4$
$= 16\omega^4$
Since $\omega^3 = 1$,we have $\omega^4 = \omega^3 \times \omega = 1 \times \omega = \omega$.
Therefore,$16\omega^4 = 16\omega$.

(A) We know that $1 + \omega + \omega^2 = 0$,which implies $1 + \omega^2 = -\omega$.
Substituting this into the expression:
$(1 - 2\omega + \omega^2)^6 = (1 + \omega^2 - 2\omega)^6$
$= (-\omega - 2\omega)^6$
$= (-3\omega)^6$
$= (-3)^6 \times \omega^6$
Since $\omega^3 = 1$,we have $\omega^6 = (\omega^3)^2 = 1^2 = 1$.
Therefore,$(-3)^6 \times 1 = 729$.

(D) Given expression: $\omega^{99} + \omega^{100} + \omega^{101}$
Factor out $\omega^{99}$: $\omega^{99}(1 + \omega + \omega^{2})$
Since $\omega$ is a complex cube root of unity,we know that $1 + \omega + \omega^{2} = 0$ and $\omega^{3} = 1$.
Substituting these values: $\omega^{99}(0) = 1^{33} \times 0 = 0$.
Therefore,the value is $0$.

(C) Let $y = |a + b\omega + c\omega^2|$.
For $y$ to be minimum,$y^2$ must be minimum.
$y^2 = |a + b\omega + c\omega^2|^2 = (a + b\omega + c\omega^2)(a + b\bar{\omega} + c\bar{\omega}^2)$.
Using $\omega^2 = \bar{\omega}$ and $\omega = \bar{\omega}^2$,we get $y^2 = a^2 + b^2 + c^2 - ab - bc - ca$.
This can be written as $y^2 = \frac{1}{2}[(a - b)^2 + (b - c)^2 + (c - a)^2]$.
Since $a, b, c$ are integers and not all equal,the minimum value occurs when two variables are equal and the third differs by $1$ (e.g.,$a=0, b=0, c=1$).
Substituting these values,$y^2 = \frac{1}{2}[(0-0)^2 + (0-1)^2 + (1-0)^2] = \frac{1}{2}[0 + 1 + 1] = 1$.
Thus,the minimum value of $y$ is $\sqrt{1} = 1$.

(D) We know that $1 + \omega + \omega^2 = 0$,which implies $1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$.
Substituting these values into the expression:
$\omega^2(1 + \omega)^3 - (1 + \omega^2)\omega = \omega^2(-\omega^2)^3 - (-\omega)\omega$
$= \omega^2(-\omega^6) + \omega^2$
$= -\omega^8 + \omega^2$
Since $\omega^3 = 1$,we have $\omega^8 = \omega^6 \times \omega^2 = 1 \times \omega^2 = \omega^2$.
Thus,$-\omega^2 + \omega^2 = 0$.

(C) Given $x = \alpha + \beta$,$y = \alpha \omega + \beta \omega^2$,and $z = \alpha \omega^2 + \beta \omega$.
We know that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
First,calculate the product $yz$:
$yz = (\alpha \omega + \beta \omega^2)(\alpha \omega^2 + \beta \omega)$
$yz = \alpha^2 \omega^3 + \alpha \beta \omega^2 + \alpha \beta \omega^4 + \beta^2 \omega^3$
Since $\omega^3 = 1$ and $\omega^4 = \omega$,we get:
$yz = \alpha^2(1) + \alpha \beta \omega^2 + \alpha \beta \omega + \beta^2(1)$
$yz = \alpha^2 + \alpha \beta(\omega^2 + \omega) + \beta^2$
Since $\omega^2 + \omega = -1$,we have:
$yz = \alpha^2 - \alpha \beta + \beta^2$
Now,calculate $xyz = x(yz)$:
$xyz = (\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)$
Using the algebraic identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$,we get:
$xyz = \alpha^3 + \beta^3$.

(D) Given $(1 + x)^n = C_0 + C_1x + C_2x^2 + ..... + C_nx^n$.
Put $x = i$ on both sides:
$(1 + i)^n = (C_0 - C_2 + C_4 - .....) + i(C_1 - C_3 + C_5 - .....) \dots (i)$
Express $1 + i$ in polar form: $1 + i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)$.
Using De Moivre's theorem:
$(1 + i)^n = (\sqrt{2})^n \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)^n = 2^{n/2} \left( \cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4} \right) \dots (ii)$
Equating the real parts from $(i)$ and $(ii)$,we get:
$C_0 - C_2 + C_4 - C_6 + ..... = 2^{n/2} \cos \frac{n\pi}{4}$.

(C) Given $x = \cos \theta + i\sin \theta = e^{i\theta}$ and $y = \cos \phi + i\sin \phi = e^{i\phi}$.
Applying De Moivre's theorem,we have ${x^m} = e^{im\theta}$ and ${y^n} = e^{in\phi}$.
Therefore,${x^m}{y^n} = e^{i(m\theta + n\phi)}$ and ${x^{-m}}{y^{-n}} = e^{-i(m\theta + n\phi)}$.
Using Euler's formula $e^{i\alpha} + e^{-i\alpha} = 2\cos \alpha$,we get:
${x^m}{y^n} + {x^{-m}}{y^{-n}} = e^{i(m\theta + n\phi)} + e^{-i(m\theta + n\phi)} = 2\cos (m\theta + n\phi)$.

(D) We have $\sum\limits_{r = 1}^8 {\left( {\sin \frac{{2r\pi }}{9} + i\cos \frac{{2r\pi }}{9}} \right)} = \sum\limits_{r = 1}^8 {i\left( {\cos \frac{{2r\pi }}{9} - i\sin \frac{{2r\pi }}{9}} \right)}$.
Using Euler's formula,this becomes $i\sum\limits_{r = 1}^8 {e^{-i\frac{2r\pi}{9}}} = i\sum\limits_{r = 1}^8 {\alpha^r}$,where $\alpha = e^{-i\frac{2\pi}{9}}$.
This is a geometric progression with $8$ terms,so the sum is $i\alpha \frac{1 - \alpha^8}{1 - \alpha} = i \frac{\alpha - \alpha^9}{1 - \alpha}$.
Since $\alpha^9 = e^{-i2\pi} = 1$,the expression simplifies to $i \frac{\alpha - 1}{1 - \alpha} = i(-1) = -i$.

(C) Given $\cos \alpha + \cos \beta + \cos \gamma = 0$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$.
Let $a = \cos \alpha + i\sin \alpha$,$b = \cos \beta + i\sin \beta$,and $c = \cos \gamma + i\sin \gamma$.
Then $a + b + c = (\cos \alpha + \cos \beta + \cos \gamma) + i(\sin \alpha + \sin \beta + \sin \gamma) = 0 + i0 = 0$.
Using the identity if $a + b + c = 0$,then $a^3 + b^3 + c^3 = 3abc$.
Substituting the values,we get $(\cos \alpha + i\sin \alpha)^3 + (\cos \beta + i\sin \beta)^3 + (\cos \gamma + i\sin \gamma)^3 = 3(\cos \alpha + i\sin \alpha)(\cos \beta + i\sin \beta)(\cos \gamma + i\sin \gamma)$.
By De Moivre's Theorem,$(\cos 3\alpha + i\sin 3\alpha) + (\cos 3\beta + i\sin 3\beta) + (\cos 3\gamma + i\sin 3\gamma) = 3[\cos(\alpha + \beta + \gamma) + i\sin(\alpha + \beta + \gamma)]$.
Equating the real parts,we get $\cos 3\alpha + \cos 3\beta + \cos 3\gamma = 3\cos(\alpha + \beta + \gamma)$.

(B) Given the equation $(x - 1)^3 + 8 = 0$.
This can be written as $(x - 1)^3 = -8$.
Taking the cube root on both sides,we get $x - 1 = (-8)^{1/3}$.
Since the cube roots of unity are $1, \omega, \omega^2$,the cube roots of $-8$ are $-2, -2\omega, -2\omega^2$.
Therefore,$x - 1 = -2, -2\omega, -2\omega^2$.
Adding $1$ to all sides,we get $x = 1 - 2, 1 - 2\omega, 1 - 2\omega^2$.
Thus,the roots are $-1, 1 - 2\omega, 1 - 2\omega^2$.

(C) Since $1, \omega, \omega^2, \dots, \omega^{n-1}$ are the $n^{th}$ roots of unity,they are the roots of the equation $x^n - 1 = 0$.
Therefore,we can write the polynomial as:
$x^n - 1 = (x - 1)(x - \omega)(x - \omega^2) \dots (x - \omega^{n-1})$
Dividing both sides by $(x - 1)$,we get:
$\frac{x^n - 1}{x - 1} = (x - \omega)(x - \omega^2) \dots (x - \omega^{n-1})$
Using the geometric series formula,the left side simplifies to:
$x^{n-1} + x^{n-2} + \dots + x + 1 = (x - \omega)(x - \omega^2) \dots (x - \omega^{n-1})$
Now,taking the limit as $x \to 1$ or simply substituting $x = 1$ on both sides:
$1^{n-1} + 1^{n-2} + \dots + 1 + 1 = (1 - \omega)(1 - \omega^2) \dots (1 - \omega^{n-1})$
Since there are $n$ terms on the left side,the sum is $n$.
Thus,$(1 - \omega)(1 - \omega^2) \dots (1 - \omega^{n-1}) = n$.

(C) Let $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ be the complex cube root of unity,where $\omega^3 = 1$.
Given expression is $4 + 5\omega^{334} + 3\omega^{365}$.
Since $\omega^3 = 1$,we have $\omega^{334} = \omega^{333} \cdot \omega = (\omega^3)^{111} \cdot \omega = 1^{111} \cdot \omega = \omega$.
Similarly,$\omega^{365} = \omega^{363} \cdot \omega^2 = (\omega^3)^{121} \cdot \omega^2 = 1^{121} \cdot \omega^2 = \omega^2$.
Substituting these values into the expression:
$4 + 5\omega + 3\omega^2$.
Using the property $1 + \omega + \omega^2 = 0$,we have $\omega^2 = -1 - \omega$.
Substituting this:
$4 + 5\omega + 3(-1 - \omega) = 4 + 5\omega - 3 - 3\omega = 1 + 2\omega$.
Substituting $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$:
$1 + 2\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 1 - 1 + i\sqrt{3} = i\sqrt{3}$.

(D) The $n^{th}$ roots of unity are given by ${z_r} = \cos \frac{2r\pi}{n} + i\sin \frac{2r\pi}{n}$ for $r = 0, 1, \dots, n-1$.
Let ${z_1} = \cos \frac{2r_1\pi}{n} + i\sin \frac{2r_1\pi}{n}$ and ${z_2} = \cos \frac{2r_2\pi}{n} + i\sin \frac{2r_2\pi}{n}$.
The angle subtended by the segment joining ${z_1}$ and ${z_2}$ at the origin is given by the argument of the ratio $\frac{z_1}{z_2}$.
$\text{arg}\left(\frac{z_1}{z_2}\right) = \text{arg}(z_1) - \text{arg}(z_2) = \frac{2(r_1 - r_2)\pi}{n}$.
Since the angle is a right angle,we have $\frac{2(r_1 - r_2)\pi}{n} = \pm \frac{\pi}{2}$.
This implies $\frac{2(r_1 - r_2)}{n} = \pm \frac{1}{2}$,which simplifies to $n = \pm 4(r_1 - r_2)$.
Since $r_1$ and $r_2$ are integers,$n$ must be a multiple of $4$,i.e.,$n = 4k$ for some integer $k$.

(A) The general term of the series is $T_r = (r + 1)(r\omega + 1)(r\omega^2 + 1)$ for $r = 1$ to $n$.
Using the property $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$,we have $(r\omega + 1)(r\omega^2 + 1) = r^2\omega^3 + r\omega + r\omega^2 + 1 = r^2 + r(\omega + \omega^2) + 1 = r^2 - r + 1$.
Thus,$T_r = (r + 1)(r^2 - r + 1) = r^3 + 1$.
The sum is $\sum_{r=1}^n (r^3 + 1) = \sum_{r=1}^n r^3 + \sum_{r=1}^n 1$.
Using the formula $\sum_{r=1}^n r^3 = [\frac{n(n + 1)}{2}]^2$,the sum is $[\frac{n(n + 1)}{2}]^2 + n$.

(D) Given the equation: $(1 + \omega^2)^m = (1 + \omega^4)^m$.
Since $\omega^3 = 1$,we have $\omega^4 = \omega$.
Substituting this into the equation: $(1 + \omega^2)^m = (1 + \omega)^m$.
Using the property $1 + \omega + \omega^2 = 0$,we know $1 + \omega^2 = -\omega$ and $1 + \omega = -\omega^2$.
So,$(-\omega)^m = (-\omega^2)^m$.
Dividing both sides by $(-\omega)^m$ (assuming $\omega \neq 0$): $1 = \frac{(-\omega^2)^m}{(-\omega)^m} = (\frac{-\omega^2}{-\omega})^m = (\omega)^m$.
For $(\omega)^m = 1$,the smallest positive integer $m$ must be $3$ because $\omega^3 = 1$.

(D) The roots of the equation $x^2 + x + 1 = 0$ are the complex cube roots of unity,$\omega$ and $\omega^2$.
Let $\alpha = \omega$ and $\beta = \omega^2$.
We need to find the equation whose roots are $\alpha^{19}$ and $\beta^7$.
$\alpha^{19} = \omega^{19} = (\omega^3)^6 \cdot \omega = 1^6 \cdot \omega = \omega$.
$\beta^7 = (\omega^2)^7 = \omega^{14} = (\omega^3)^4 \cdot \omega^2 = 1^4 \cdot \omega^2 = \omega^2$.
The roots of the new equation are $\omega$ and $\omega^2$,which are the same as the roots of the original equation $x^2 + x + 1 = 0$.

(D) Given the equation $x^3 + 8 = 0$,the roots are $\alpha, \beta, \gamma$.
Let $y = x^2$,which implies $x = y^{1/2}$.
Substituting $x$ into the original equation: $(y^{1/2})^3 + 8 = 0$.
$y^{3/2} = -8$.
Squaring both sides: $(y^{3/2})^2 = (-8)^2$.
$y^3 = 64$.
$y^3 - 64 = 0$.
Thus,the equation whose roots are $\alpha^2, \beta^2, \gamma^2$ is $x^3 - 64 = 0$.

(D) Given $x + \frac{1}{x} = 2 \cos \alpha$.
Let $x = \cos \alpha + i \sin \alpha = e^{i\alpha}$.
Then $\frac{1}{x} = \cos \alpha - i \sin \alpha = e^{-i\alpha}$.
Thus,$x + \frac{1}{x} = e^{i\alpha} + e^{-i\alpha} = 2 \cos \alpha$.
Now,$x^n + \frac{1}{x^n} = (e^{i\alpha})^n + (e^{-i\alpha})^n = e^{in\alpha} + e^{-in\alpha}$.
Using Euler's formula,$e^{in\alpha} + e^{-in\alpha} = 2 \cos n\alpha$.
Therefore,$x^n + \frac{1}{x^n} = 2 \cos n\alpha$.

(D) The given equation is $x^2 + x + 1 = 0$.
The roots of this equation are the complex cube roots of unity,$\omega$ and $\omega^2$.
Let $\alpha = \omega$ and $\beta = \omega^2$.
We need to find the equation whose roots are $\alpha^{19}$ and $\beta^7$.
$\alpha^{19} = \omega^{19} = (\omega^3)^6 \cdot \omega = 1^6 \cdot \omega = \omega$.
$\beta^7 = (\omega^2)^7 = \omega^{14} = (\omega^3)^4 \cdot \omega^2 = 1^4 \cdot \omega^2 = \omega^2$.
Since the new roots are $\omega$ and $\omega^2$,the required equation is the same as the original equation: $x^2 + x + 1 = 0$.

(C) Given that $x = a$,$y = b\omega$,and $z = c\omega^2$.
Substituting these values into the expression:
$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = \frac{a}{a} + \frac{b\omega}{b} + \frac{c\omega^2}{c}$
$= 1 + \omega + \omega^2$
Since $1 + \omega + \omega^2 = 0$ for the complex cube roots of unity,the result is $0$.