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

(C) The given equation is $x^3 + 27 = 0$,which can be written as $x^3 = -27$.
The roots are $x = -3, -3\omega, -3\omega^2$,where $\omega$ is the complex cube root of unity.
Let $\alpha = -3$,$\beta = -3\omega$,and $\gamma = -3\omega^2$.
Then $\frac{\beta}{\alpha} = \frac{-3\omega}{-3} = \omega$ and $\frac{\gamma}{\alpha} = \frac{-3\omega^2}{-3} = \omega^2$.
The roots of the required quadratic equation are $\omega^2$ and $(\omega^2)^2 = \omega^4 = \omega$.
The sum of the roots is $\omega^2 + \omega = -1$.
The product of the roots is $\omega^2 \cdot \omega = \omega^3 = 1$.
The 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) Let $S = \sum_{k=1}^{10} \left( \sin \frac{2k\pi}{11} + i\cos \frac{2k\pi}{11} \right)$.
We can factor out $i$ from the expression:
$S = i \sum_{k=1}^{10} \left( \cos \frac{2k\pi}{11} - i \sin \frac{2k\pi}{11} \right)$.
Using Euler's formula $e^{i\theta} = \cos \theta + i \sin \theta$,we have $\cos \theta - i \sin \theta = e^{-i\theta}$.
Thus,$S = i \sum_{k=1}^{10} e^{-i \frac{2k\pi}{11}}$.
Let $\omega = e^{-i \frac{2\pi}{11}}$. Then $S = i \sum_{k=1}^{10} \omega^k$.
This is a geometric series with $10$ terms: $\sum_{k=1}^{10} \omega^k = \omega + \omega^2 + \dots + \omega^{10}$.
We know that the sum of all $11$th roots of unity is $\sum_{k=0}^{10} \omega^k = 0$.
Therefore,$\sum_{k=1}^{10} \omega^k = -\omega^0 = -1$.
Substituting this back into the expression for $S$:
$S = i(-1) = -i$.

(D) Given $z^2 + z + 1 = 0$,the roots are $z = \omega$ or $z = \omega^2$,where $\omega$ is a cube root of unity.
Since $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$,we have $\omega + \frac{1}{\omega} = \omega + \omega^2 = -1$.
Calculating each term:
$1$. $\left( z + \frac{1}{z} \right)^2 = (-1)^2 = 1$
$2$. $\left( z^2 + \frac{1}{z^2} \right)^2 = (\omega^2 + \omega)^2 = (-1)^2 = 1$
$3$. $\left( z^3 + \frac{1}{z^3} \right)^2 = (1 + 1)^2 = 2^2 = 4$
$4$. $\left( z^4 + \frac{1}{z^4} \right)^2 = (\omega + \omega^2)^2 = (-1)^2 = 1$
$5$. $\left( z^5 + \frac{1}{z^5} \right)^2 = (\omega^2 + \omega)^2 = (-1)^2 = 1$
$6$. $\left( z^6 + \frac{1}{z^6} \right)^2 = (1 + 1)^2 = 2^2 = 4$
Sum $= 1 + 1 + 4 + 1 + 1 + 4 = 12$.

(C) We know that for the cube roots of unity,$1 + \omega + \omega^2 = 0$,which implies $1 + \omega = -\omega^2$.
Substituting this into the given expression:
$(1 + \omega)^7 = (-\omega^2)^7$
$= -\omega^{14}$
Since $\omega^3 = 1$,we have $\omega^{14} = \omega^{12} \cdot \omega^2 = (\omega^3)^4 \cdot \omega^2 = 1^4 \cdot \omega^2 = \omega^2$.
Thus,$(1 + \omega)^7 = -\omega^2$.
Using the identity $1 + \omega + \omega^2 = 0$,we get $-\omega^2 = 1 + \omega$.
Comparing $1 + \omega$ with $A + B\omega$,we get $A = 1$ and $B = 1$.

(B) The roots of the equation $x^2 - x + 1 = 0$ are given by $x = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm i\sqrt{3}}{2}$.
These are $-\omega$ and $-\omega^2$,where $\omega$ is a complex cube root of unity.
Let $\alpha = -\omega$ and $\beta = -\omega^2$.
Then $\alpha^{101} + \beta^{107} = (-\omega)^{101} + (-\omega^2)^{107}$.
$= -\omega^{101} - \omega^{214}$.
Since $\omega^3 = 1$,we have $\omega^{101} = \omega^{3 \times 33 + 2} = \omega^2$ and $\omega^{214} = \omega^{3 \times 71 + 1} = \omega$.
Thus,$\alpha^{101} + \beta^{107} = -(\omega^2 + \omega)$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.
Therefore,$\alpha^{101} + \beta^{107} = -(-1) = 1$.

(D) Let $z = \cos \frac{{2\pi }}{7} + i\sin \frac{{2\pi }}{7}$. By De Moivre's theorem,${z^k} = \cos \frac{{2\pi k}}{7} + i\sin \frac{{2\pi k}}{7}$.
The given sum is $S = \sum\limits_{k = 1}^6 {\left( {\sin \frac{{2\pi k}}{7} - i\cos \frac{{2\pi k}}{7}} \right)}$.
We can rewrite the term inside the sum as: $\sin \frac{{2\pi k}}{7} - i\cos \frac{{2\pi k}}{7} = -i \left( \cos \frac{{2\pi k}}{7} + i\sin \frac{{2\pi k}}{7} \right) = -i z^k$.
Thus,$S = -i \sum\limits_{k = 1}^6 z^k = -i (z + z^2 + z^3 + z^4 + z^5 + z^6)$.
Since $z$ is a $7^{th}$ root of unity ($z^7 = 1$ and $z \neq 1$),the sum of all $7^{th}$ roots of unity is $1 + z + z^2 + z^3 + z^4 + z^5 + z^6 = 0$.
Therefore,$\sum\limits_{k = 1}^6 z^k = -1$.
Substituting this into the expression for $S$,we get $S = -i(-1) = i$.

(A) The first equation can be factored as $(z + 1)(z^2 + z + 1) = 0$.
Its roots are $-1, \omega, \text{ and } \omega^2$.
Let $f(z) = z^{1985} + z^{100} + 1$.
Testing $z = -1$: $f(-1) = (-1)^{1985} + (-1)^{100} + 1 = -1 + 1 + 1 = 1 \neq 0$.
Therefore,$-1$ is not a root of $f(z) = 0$.
Testing $z = \omega$: $f(\omega) = \omega^{1985} + \omega^{100} + 1$.
Since $\omega^3 = 1$,we have $f(\omega) = (\omega^3)^{661} \cdot \omega^2 + (\omega^3)^{33} \cdot \omega + 1 = \omega^2 + \omega + 1 = 0$.
Therefore,$\omega$ is a root of $f(z) = 0$.
Testing $z = \omega^2$: $f(\omega^2) = (\omega^2)^{1985} + (\omega^2)^{100} + 1 = \omega^{3970} + \omega^{200} + 1$.
$f(\omega^2) = (\omega^3)^{1323} \cdot \omega + (\omega^3)^{66} \cdot \omega^2 + 1 = \omega + \omega^2 + 1 = 0$.
Therefore,$\omega^2$ is also a root of $f(z) = 0$.
Thus,the common roots are $\omega \text{ and } \omega^2$.

(D) Given the equation $(x - 2)^3 + 27 = 0$.
This can be rewritten as $(x - 2)^3 = -27$.
Taking the cube root on both sides,we get $x - 2 = (-27)^{1/3}$.
Since $(-27)^{1/3} = -3(1)^{1/3}$,and the cube roots of unity are $1, \omega, \omega^2$,we have:
$x - 2 = -3(1), -3(\omega), -3(\omega^2)$.
$x - 2 = -3, -3\omega, -3\omega^2$.
Adding $2$ to each term,we get:
$x = 2 - 3, 2 - 3\omega, 2 - 3\omega^2$.
$x = -1, 2 - 3\omega, 2 - 3\omega^2$.

(B) Given $S = 1 + 3\alpha + 5\alpha^2 + \dots + (2n - 1)\alpha^{n-1}$ $(i)$
Multiply by $\alpha$: $\alpha S = \alpha + 3\alpha^2 + 5\alpha^3 + \dots + (2n - 1)\alpha^n$ $(ii)$
Subtracting $(ii)$ from $(i)$:
$(1 - \alpha)S = 1 + 2\alpha + 2\alpha^2 + \dots + 2\alpha^{n-1} - (2n - 1)\alpha^n$
Since $\alpha^n = 1$ for any $n^{th}$ root of unity:
$(1 - \alpha)S = 1 + 2(\alpha + \alpha^2 + \dots + \alpha^{n-1}) - (2n - 1)$
Using the sum of geometric progression $\alpha + \alpha^2 + \dots + \alpha^{n-1} = \frac{\alpha(1 - \alpha^{n-1})}{1 - \alpha} = \frac{\alpha - 1}{1 - \alpha} = -1$:
$(1 - \alpha)S = 1 + 2(-1) - 2n + 1 = 1 - 2 - 2n + 1 = -2n$
Therefore,$S = -\frac{2n}{1 - \alpha}$.

(A) Let $S = \sum_{n=1}^{2017} (n-\omega)(n-\omega^2)$.
Since $\omega^2+\omega+1=0$,we have $(n-\omega)(n-\omega^2) = n^2 - n(\omega+\omega^2) + \omega^3 = n^2 + n + 1$.
Thus,$S = \sum_{n=1}^{2017} (n^2 + n + 1) = \sum_{n=1}^{2017} n^2 + \sum_{n=1}^{2017} n + \sum_{n=1}^{2017} 1$.
Using standard summation formulas: $S = \frac{2017(2018)(4035)}{6} + \frac{2017(2018)}{2} + 2017$.
Dividing by $2017$: $\frac{S}{2017} = \frac{2018 \cdot 4035}{6} + \frac{2018}{2} + 1 = 1009 \cdot 1345 + 1009 + 1 = 1009(1345+1) + 1 = 1009(1346) + 1$.
Since $1009(1346)$ is even,$\frac{S}{2017} = \text{even} + 1 = \text{odd}$.
Therefore,$\cos\left( \frac{S\pi}{2017} \right) = \cos(\text{odd} \cdot \pi) = -1$.

(A) The given quadratic equation is $x^2 + \omega x + \omega^2 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:
$x = \frac{-\omega \pm \sqrt{\omega^2 - 4\omega^2}}{2} = \frac{-\omega \pm \sqrt{-3\omega^2}}{2} = \frac{-\omega \pm \sqrt{3}i\omega}{2} = \omega \left( \frac{-1 \pm \sqrt{3}i}{2} \right)$.
Since $\frac{-1 \pm \sqrt{3}i}{2}$ are the complex cube roots of unity (denoted as $\omega$ and $\omega^2$),the roots are $\alpha = \omega^2$ and $\beta = \omega^3 = 1$ (or vice versa).
Given $z = \alpha^9 + i\beta^9$,we substitute the values:
$z = (\omega^2)^9 + i(1)^9 = \omega^{18} + i$.
Since $\omega^3 = 1$,$\omega^{18} = (\omega^3)^6 = 1^6 = 1$.
Thus,$z = 1 + i$.
The modulus $|z| = |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}$.

(B) We know that $1 + \omega^r + \omega^{2r}$ is a geometric series sum.
If $r$ is a multiple of $3$,then $\omega^r = 1$ and $\omega^{2r} = 1$,so $1 + \omega^r + \omega^{2r} = 1 + 1 + 1 = 3$.
If $r$ is not a multiple of $3$,then $1 + \omega^r + \omega^{2r} = \frac{1 - (\omega^r)^3}{1 - \omega^r} = \frac{1 - (\omega^3)^r}{1 - \omega^r} = \frac{1 - 1}{1 - \omega^r} = 0$.
For $r = 1, 2, 3, 4, 5$,the terms are:
$r=1: 1 + \omega + \omega^2 = 0$
$r=2: 1 + \omega^2 + \omega^4 = 1 + \omega^2 + \omega = 0$
$r=3: 1 + \omega^3 + \omega^6 = 1 + 1 + 1 = 3$
$r=4: 1 + \omega^4 + \omega^8 = 1 + \omega + \omega^2 = 0$
$r=5: 1 + \omega^5 + \omega^{10} = 1 + \omega^2 + \omega = 0$
Sum $= 0 + 0 + 3 + 0 + 0 = 3$.

(A) The $n$-th roots of unity are given by $z_k = e^{i \frac{2k\pi}{n}}$ for $k = 0, 1, \dots, n-1$.
For $n=10$,the roots are $\omega^k = e^{i \frac{2k\pi}{10}}$ where $k \in \{0, 1, \dots, 9\}$.
Let $z_1 = \omega^r$ and $z_2 = \omega^s$ be two such roots,where $r, s \in \{0, 1, \dots, 9\}$.
The product is $z_1 z_2 = \omega^r \cdot \omega^s = \omega^{r+s} = e^{i \frac{2(r+s)\pi}{10}}$.
Since $\omega^{10} = 1$,we have $\omega^{r+s} = \omega^{(r+s) \pmod{10}}$.
Since $(r+s) \pmod{10}$ is always an integer in the set $\{0, 1, \dots, 9\}$,the product $z_1 z_2$ is always one of the tenth roots of unity.

(A) Given the quadratic equation $x^2 - \sqrt{2}x + 1 = 0$.
Using the quadratic formula,the roots are $x = \frac{\sqrt{2} \pm \sqrt{2 - 4}}{2} = \frac{\sqrt{2} \pm i\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \pm i\frac{1}{\sqrt{2}}$.
These can be written in polar form as $\alpha = e^{i\pi/4}$ and $\beta = e^{-i\pi/4}$.
Then,$\alpha^{50} + \beta^{50} = (e^{i\pi/4})^{50} + (e^{-i\pi/4})^{50} = e^{i25\pi/2} + e^{-i25\pi/2}$.
Since $e^{i25\pi/2} = e^{i(12\pi + \pi/2)} = e^{i\pi/2} = i$ and $e^{-i25\pi/2} = -i$.
Therefore,$\alpha^{50} + \beta^{50} = i + (-i) = 0$.

(B) Let $z_1 = \cos \theta + i \sin \theta$,$z_2 = \cos \phi + i \sin \phi$,and $z_3 = \cos \psi + i \sin \psi$.
Given equations are $\cos \theta + 2\cos \phi + 3\cos \psi = 0$ and $\sin \theta + 2\sin \phi + 3\sin \psi = 0$.
Combining these,we get $z_1 + 2z_2 + 3z_3 = 0$.
Using the algebraic identity $x^3 + y^3 + z^3 = 3xyz$ if $x + y + z = 0$,where $x = z_1$,$y = 2z_2$,and $z = 3z_3$:
$z_1^3 + (2z_2)^3 + (3z_3)^3 = 3(z_1)(2z_2)(3z_3) = 18 z_1 z_2 z_3$.
By De Moivre's Theorem,$z_1^3 = \cos 3\theta + i \sin 3\theta$,$(2z_2)^3 = 8(\cos 3\phi + i \sin 3\phi)$,and $(3z_3)^3 = 27(\cos 3\psi + i \sin 3\psi)$.
Thus,$(\cos 3\theta + 8\cos 3\phi + 27\cos 3\psi) + i(\sin 3\theta + 8\sin 3\phi + 27\sin 3\psi) = 18(\cos(\theta + \phi + \psi) + i \sin(\theta + \phi + \psi))$.
Equating the real parts,we get $\cos 3\theta + 8\cos 3\phi + 27\cos 3\psi = 18\cos(\theta + \phi + \psi)$.

(B) The given quadratic equation is $x^2 - 2x + 4 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{2 \pm \sqrt{4 - 16}}{2} = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm i\sqrt{3}$.
We can write the roots in polar form:
$\alpha = 1 + i\sqrt{3} = 2 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)$
$\beta = 1 - i\sqrt{3} = 2 \left( \cos \frac{\pi}{3} - i \sin \frac{\pi}{3} \right)$
Using De Moivre's Theorem,$\alpha^n = 2^n \left( \cos \frac{n\pi}{3} + i \sin \frac{n\pi}{3} \right)$ and $\beta^n = 2^n \left( \cos \frac{n\pi}{3} - i \sin \frac{n\pi}{3} \right)$.
Adding these,we get $\alpha^n + \beta^n = 2^n \left( \cos \frac{n\pi}{3} + i \sin \frac{n\pi}{3} + \cos \frac{n\pi}{3} - i \sin \frac{n\pi}{3} \right)$.
$\alpha^n + \beta^n = 2^n \left( 2 \cos \frac{n\pi}{3} \right) = 2^{n+1} \cos \frac{n\pi}{3}$.

(D) Let $z = \frac{1 + i\sqrt{3}}{1 - i\sqrt{3}}$.
Multiplying the numerator and denominator by the conjugate of the denominator $(1 + i\sqrt{3})$:
$z = \frac{(1 + i\sqrt{3})(1 + i\sqrt{3})}{(1 - i\sqrt{3})(1 + i\sqrt{3})} = \frac{1 + 2i\sqrt{3} + (i\sqrt{3})^2}{1^2 + (\sqrt{3})^2} = \frac{1 + 2i\sqrt{3} - 3}{1 + 3} = \frac{-2 + 2i\sqrt{3}}{4} = \frac{-1 + i\sqrt{3}}{2}$.
We recognize this as $\omega$,where $\omega = \frac{-1 + i\sqrt{3}}{2}$ is a complex cube root of unity.
Thus,$z = \omega$.
The equation becomes $\omega^n = 1$.
The smallest positive integer $n$ for which $\omega^n = 1$ is $n = 3$.

(A) The given equation is $x^2 + 2x + 2 = 0$.
Completing the square,we get $(x+1)^2 + 1 = 0$,so $(x+1)^2 = -1$.
Thus,$x+1 = \pm i$,which gives $x = -1 \pm i$.
In polar form,$x = \sqrt{2} \left( \cos \frac{3\pi}{4} \pm i \sin \frac{3\pi}{4} \right) = \sqrt{2} e^{\pm i(3\pi/4)}$.
Using De Moivre's Theorem,$\alpha^{15} + \beta^{15} = (\sqrt{2})^{15} e^{i(45\pi/4)} + (\sqrt{2})^{15} e^{-i(45\pi/4)}$.
$\alpha^{15} + \beta^{15} = 2^{7} \sqrt{2} \times 2 \cos \left( \frac{45\pi}{4} \right)$.
Since $\frac{45\pi}{4} = 11\pi + \frac{\pi}{4}$,$\cos \left( \frac{45\pi}{4} \right) = \cos \left( 11\pi + \frac{\pi}{4} \right) = -\cos \left( \frac{\pi}{4} \right) = -\frac{1}{\sqrt{2}}$.
Therefore,$\alpha^{15} + \beta^{15} = 2^8 \sqrt{2} \times \left( -\frac{1}{\sqrt{2}} \right) = -2^8 = -256$.

(D) We know that $\frac{\sqrt{3}}{2} + \frac{i}{2} = e^{i\pi/6}$ and $\frac{\sqrt{3}}{2} - \frac{i}{2} = e^{-i\pi/6}$.
Thus,$z = (e^{i\pi/6})^5 + (e^{-i\pi/6})^5 = e^{i5\pi/6} + e^{-i5\pi/6}$.
Using Euler's formula,$e^{i\theta} + e^{-i\theta} = 2\cos(\theta)$,we get $z = 2\cos(5\pi/6)$.
Since $\cos(5\pi/6) = -\frac{\sqrt{3}}{2}$,we have $z = 2(-\frac{\sqrt{3}}{2}) = -\sqrt{3}$.
Here,$R(z) = -\sqrt{3}$ and $I(z) = 0$.
Therefore,$I(z) = 0$ is the correct statement.

(A) The given equation is $x^2 - 2x + 2 = 0$.
Using the quadratic formula,$x = \frac{2 \pm \sqrt{4 - 8}}{2} = 1 \pm i$.
Let $\alpha = 1 + i$ and $\beta = 1 - i$.
Then $\frac{\alpha}{\beta} = \frac{1 + i}{1 - i} = \frac{(1 + i)^2}{(1 - i)(1 + i)} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i$.
We need to find the least natural number $n$ such that $(\frac{\alpha}{\beta})^n = 1$,which means $i^n = 1$.
The powers of $i$ are $i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1$.
Thus,the least value of $n$ is $4$.

(A) Given $\alpha = \frac{-1 + i \sqrt{3}}{2} = \omega$,where $\omega$ is the cube root of unity,satisfying $\omega^{3} = 1$ and $1 + \omega + \omega^{2} = 0$.
For $a = (1 + \alpha) \sum_{k=0}^{100} \alpha^{2k} = (1 + \omega) (1 + \omega^{2} + \omega^{4} + \dots + \omega^{200})$.
Since $\omega^{3} = 1$,the sequence $1, \omega^{2}, \omega^{4}, \dots$ repeats every three terms as $1, \omega^{2}, \omega$. The sum of three consecutive terms is $1 + \omega^{2} + \omega = 0$.
There are $101$ terms in the sum. The sum of the first $99$ terms is $0$. The remaining two terms are the $100^{th}$ and $101^{st}$ terms: $1 + \omega^{2}$.
Thus,$a = (1 + \omega)(1 + \omega^{2}) = 1 + \omega^{2} + \omega + \omega^{3} = 0 + 1 = 1$.
For $b = \sum_{k=0}^{100} \alpha^{3k} = \sum_{k=0}^{100} (\omega^{3})^{k} = \sum_{k=0}^{100} (1)^{k} = 101$.
The quadratic equation with roots $a = 1$ and $b = 101$ is given by $(x - a)(x - b) = 0$,which is $(x - 1)(x - 101) = x^{2} - 102x + 101 = 0$.

(A) First,simplify the base expressions:
$\frac{1+i}{1-i} = \frac{(1+i)^2}{1^2 - i^2} = \frac{1+2i-1}{2} = i$.
$\frac{1+i}{i-1} = \frac{1+i}{-(1-i)} = -\frac{1+i}{1-i} = -i$.
Given the equations:
$(i)^{m/2} = 1$ and $(-i)^{n/3} = 1$.
For $(i)^{m/2} = 1$,the exponent $m/2$ must be a multiple of $4$. Thus,$m/2 = 4k_1 \Rightarrow m = 8k_1$. The least value of $m$ is $8$.
For $(-i)^{n/3} = 1$,we know $(-i)^4 = 1$. Thus,$n/3 = 4k_2 \Rightarrow n = 12k_2$. The least value of $n$ is $12$.
The greatest common divisor of $8$ and $12$ is $4$.

(D) Given $\alpha = \frac{-1+i \sqrt{3}}{2} = \omega,$ where $\omega$ is the complex cube root of unity.
We have $\omega^2 + \omega + 1 = 0$ and $\omega^3 = 1$.
Expanding $(2+\omega)^4$ using the binomial theorem:
$(2+\omega)^4 = 2^4 + 4(2^3)(\omega) + 6(2^2)(\omega^2) + 4(2)(\omega^3) + \omega^4$
$= 16 + 32\omega + 24\omega^2 + 8(1) + \omega$
$= 24 + 33\omega + 24\omega^2$
Since $\omega^2 = -1 - \omega$,we substitute this into the expression:
$= 24 + 33\omega + 24(-1 - \omega)$
$= 24 + 33\omega - 24 - 24\omega$
$= 9\omega$
Comparing this with $a + b\omega$,we get $a = 0$ and $b = 9$.
Therefore,$a + b = 0 + 9 = 9$.

(D) Given $P(x) = f(x^3) + xg(x^3)$.
Since $P(x)$ is divisible by $x^2 + x + 1$,it must vanish at the roots of $x^2 + x + 1 = 0$. Let $\omega$ be a complex cube root of unity,then $\omega^2 + \omega + 1 = 0$ and $\omega^3 = 1$.
The roots are $\omega$ and $\omega^2$. Thus,$P(\omega) = 0$ and $P(\omega^2) = 0$.
$P(\omega) = f(\omega^3) + \omega g(\omega^3) = f(1) + \omega g(1) = 0$ (Equation $1$)
$P(\omega^2) = f((\omega^2)^3) + \omega^2 g((\omega^2)^3) = f(\omega^6) + \omega^2 g(\omega^6) = f(1) + \omega^2 g(1) = 0$ (Equation $2$)
Subtracting Equation $2$ from Equation $1$:
$(f(1) + \omega g(1)) - (f(1) + \omega^2 g(1)) = 0$
$(\omega - \omega^2) g(1) = 0$
Since $\omega \neq \omega^2$,we must have $g(1) = 0$.
Substituting $g(1) = 0$ into Equation $1$:
$f(1) + \omega(0) = 0 \Rightarrow f(1) = 0$.
We need to find $P(1)$:
$P(1) = f(1^3) + 1 \cdot g(1^3) = f(1) + g(1) = 0 + 0 = 0$.

(C) Given the equation $x^{3}-2x^{2}+2x-1=0$.
By inspection,$x=1$ satisfies the equation as $1-2+2-1=0$.
Thus,$(x-1)$ is a factor of the polynomial.
Performing division,we get $(x-1)(x^{2}-x+1)=0$.
The roots are $x=1$ and the roots of $x^{2}-x+1=0$,which are $x = \frac{1 \pm i\sqrt{3}}{2}$.
These roots are $-\omega^{2}$ and $-\omega$,where $\omega$ is the complex cube root of unity.
The sum of the $162^{\text{th}}$ powers is $S = (1)^{162} + (-\omega^{2})^{162} + (-\omega)^{162}$.
$S = 1 + \omega^{324} + \omega^{162}$.
Since $\omega^{3} = 1$,we have $\omega^{324} = (\omega^{3})^{108} = 1$ and $\omega^{162} = (\omega^{3})^{54} = 1$.
Therefore,$S = 1 + 1 + 1 = 3$.

(B) Given $k = \frac{(-1+i \sqrt{3})^{21}}{(1-i)^{24}}+\frac{(1+i \sqrt{3})^{21}}{(1+i)^{24}}$.
We know that $-1+i \sqrt{3} = 2(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}) = 2e^{i \frac{2\pi}{3}}$ and $1+i \sqrt{3} = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = 2e^{i \frac{\pi}{3}}$.
Also $1-i = \sqrt{2}(\cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4})) = \sqrt{2}e^{-i \frac{\pi}{4}}$ and $1+i = \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) = \sqrt{2}e^{i \frac{\pi}{4}}$.
Substituting these values:
$k = \frac{(2e^{i \frac{2\pi}{3}})^{21}}{(\sqrt{2}e^{-i \frac{\pi}{4}})^{24}} + \frac{(2e^{i \frac{\pi}{3}})^{21}}{(\sqrt{2}e^{i \frac{\pi}{4}})^{24}} = \frac{2^{21} e^{i 14\pi}}{2^{12} e^{-i 6\pi}} + \frac{2^{21} e^{i 7\pi}}{2^{12} e^{i 6\pi}}$
$k = 2^9 (e^{i 20\pi} + e^{i \pi}) = 512(1 - 1) = 0$.
Thus,$n = [|k|] = 0$.
The expression becomes $\sum_{j=0}^{5} (j+5)^2 - \sum_{j=0}^{5} (j+5) = \sum_{j=0}^{5} ((j+5)^2 - (j+5)) = \sum_{j=0}^{5} (j^2 + 10j + 25 - j - 5) = \sum_{j=0}^{5} (j^2 + 9j + 20)$.
$= \sum_{j=0}^{5} j^2 + 9 \sum_{j=0}^{5} j + \sum_{j=0}^{5} 20 = \frac{5(6)(11)}{6} + 9 \frac{5(6)}{2} + 20(6) = 55 + 135 + 120 = 310$.

(D) Given $z = \frac{1 - i \sqrt{3}}{2} = e^{-i \frac{\pi}{3}}$.
Then $z^r + \frac{1}{z^r} = e^{-i \frac{r\pi}{3}} + e^{i \frac{r\pi}{3}} = 2 \cos\left(\frac{r\pi}{3}\right)$.
We need to evaluate $S = 21 + \sum_{r=1}^{21} \left(z^r + \frac{1}{z^r}\right)^3 = 21 + \sum_{r=1}^{21} \left(2 \cos\left(\frac{r\pi}{3}\right)\right)^3$.
Using the identity $8 \cos^3 \theta = 2(4 \cos^3 \theta) = 2(\cos 3\theta + 3 \cos \theta)$,we have:
$S = 21 + \sum_{r=1}^{21} 2 \left(\cos(r\pi) + 3 \cos\left(\frac{r\pi}{3}\right)\right) = 21 + 2 \sum_{r=1}^{21} \cos(r\pi) + 6 \sum_{r=1}^{21} \cos\left(\frac{r\pi}{3}\right)$.
For $\sum_{r=1}^{21} \cos(r\pi) = (-1) + 1 + (-1) + \dots + (-1) = -1$ (since there are $21$ terms).
For $\sum_{r=1}^{21} \cos\left(\frac{r\pi}{3}\right)$,the sum of $\cos\left(\frac{r\pi}{3}\right)$ over one period ($r=1$ to $6$) is $0$. Since $21 = 3 \times 6 + 3$,the sum is $\sum_{r=1}^{3} \cos\left(\frac{r\pi}{3}\right) = \cos\frac{\pi}{3} + \cos\frac{2\pi}{3} + \cos\pi = \frac{1}{2} - \frac{1}{2} - 1 = -1$.
Thus,$S = 21 + 2(-1) + 6(-1) = 21 - 2 - 6 = 13$.

(A) Given $(\sqrt{3}+i)^{100}=2^{99}(p+iq)$.
Express $\sqrt{3}+i$ in polar form: $2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$.
Using De Moivre's theorem: $(2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}))^{100} = 2^{100}(\cos \frac{100\pi}{6} + i \sin \frac{100\pi}{6}) = 2^{100}(\cos \frac{50\pi}{3} + i \sin \frac{50\pi}{3})$.
Since $\frac{50\pi}{3} = 16\pi + \frac{2\pi}{3}$,we have $\cos \frac{50\pi}{3} = \cos \frac{2\pi}{3} = -\frac{1}{2}$ and $\sin \frac{50\pi}{3} = \sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}$.
Thus,$2^{100}(-\frac{1}{2} + i \frac{\sqrt{3}}{2}) = 2^{99}(p+iq)$.
Dividing by $2^{99}$,we get $2(-\frac{1}{2} + i \frac{\sqrt{3}}{2}) = p+iq$,so $p = -1$ and $q = \sqrt{3}$.
The quadratic equation with roots $p$ and $q$ is $x^2 - (p+q)x + pq = 0$.
$p+q = \sqrt{3}-1$ and $pq = -\sqrt{3}$.
Therefore,the equation is $x^2 - (\sqrt{3}-1)x - \sqrt{3} = 0$.

(C) Given $z^{2} + z + 1 = 0$,the roots are $z = \omega$ or $z = \omega^{2}$,where $\omega$ is a complex cube root of unity.
Since $\omega^{3} = 1$,we have $\frac{1}{\omega} = \omega^{2}$ and $\frac{1}{\omega^{2}} = \omega$.
Let $S = \sum_{n=1}^{15} \left( z^{n} + (-1)^{n} \frac{1}{z^{n}} \right)^{2} = \sum_{n=1}^{15} \left( z^{2n} + \frac{1}{z^{2n}} + 2(-1)^{n} \right)$.
For $z = \omega$,$z^{2n} + \frac{1}{z^{2n}} = \omega^{2n} + \omega^{-2n} = \omega^{2n} + \omega^{n} = 2 \cos\left(\frac{2n\pi}{3}\right)$.
Summing the terms:
$\sum_{n=1}^{15} z^{2n} = \sum_{n=1}^{15} \omega^{2n} = 0$ (since sum of $3$ consecutive powers of $\omega$ is $0$ and $15$ is a multiple of $3$).
$\sum_{n=1}^{15} \frac{1}{z^{2n}} = \sum_{n=1}^{15} \omega^{-2n} = 0$.
$\sum_{n=1}^{15} 2(-1)^{n} = 2(-1 + 1 - 1 + 1 ... - 1) = 2(-1) = -2$.
Thus,$|S| = |0 + 0 - 2| = |-2| = 2$.

(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 $\omega, \omega^{2}$ and the roots of $x^{2}-x+1=0$ are $-\omega, -\omega^{2}$,where $\omega = e^{i2\pi/3}$.
In all cases,$\alpha^{6}=1$ because $\alpha^{2}$ is a root of $y^{2}+y+1=0$,so $(\alpha^{2})^{3}=1$.
We calculate the powers:
$\alpha^{1011} = (\alpha^{6})^{168} \cdot \alpha^{3} = 1^{168} \cdot \alpha^{3} = \alpha^{3}$.
Since $\alpha^{2}+\alpha^{4}+1=0$,we have $\alpha^{2}+\alpha^{4} = -1$. Also $\alpha^{6}=1 \Rightarrow \alpha^{3} = \pm 1$.
For $x^{4}+x^{2}+1=0$,if $\alpha^{2} = \omega$,then $\alpha^{3} = \pm \sqrt{\omega} = \pm e^{i\pi/3}$.
Actually,$\alpha^{6}=1$ and $\alpha^{2} \neq 1$. Thus $\alpha^{3} = \pm 1$.
Specifically,$\alpha^{1011} = (\alpha^{3})^{337} = (\pm 1)^{337} = \pm 1$.
$\alpha^{2022} = (\alpha^{6})^{337} = 1^{337} = 1$.
$\alpha^{3033} = (\alpha^{6})^{505} \cdot \alpha^{3} = 1 \cdot \alpha^{3} = \pm 1$.
Thus,$\alpha^{1011}+\alpha^{2022}-\alpha^{3033} = \pm 1 + 1 - (\pm 1) = 1$.

(B) The given equation is $x^{4}+x^{3}+x^{2}+x+1=0$.
This is a geometric series sum,which can be written as $\frac{x^{5}-1}{x-1} = 0$ for $x \neq 1$.
The roots $\alpha, \beta, \gamma, \delta$ are the $5^{\text{th}}$ roots of unity excluding $1$,i.e.,$\omega, \omega^{2}, \omega^{3}, \omega^{4}$ where $\omega = e^{i \frac{2\pi}{5}}$.
Since $\omega^{5} = 1$,we have $\omega^{2021} = (\omega^{5})^{404} \cdot \omega = \omega$.
Similarly,$\beta^{2021} = \omega^{2}$,$\gamma^{2021} = \omega^{3}$,and $\delta^{2021} = \omega^{4}$.
Thus,the sum is $\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021} = \omega + \omega^{2} + \omega^{3} + \omega^{4}$.
From the equation $x^{4}+x^{3}+x^{2}+x+1=0$,the sum of the roots is $\alpha+\beta+\gamma+\delta = -\frac{1}{1} = -1$.

(A) Let the vertices of the regular $10$-gon be $z_k = e^{i \frac{2k\pi}{10}}$ for $k = 0, 1, \ldots, 9$. Fix the vertex at $z_0 = 1$. The lengths of the chords are $l_k = |1 - z_k| = |1 - e^{i \frac{2k\pi}{10}}| = 2 \sin \frac{k\pi}{10}$ for $k = 1, 2, \ldots, 9$.
We need to calculate the product $P = \prod_{k=1}^{9} 2 \sin \frac{k\pi}{10}$.
Using the identity $\prod_{k=1}^{n-1} \sin \frac{k\pi}{n} = \frac{n}{2^{n-1}}$,we have:
$P = 2^9 \prod_{k=1}^{9} \sin \frac{k\pi}{10} = 2^9 \times \frac{10}{2^{10-1}} = 2^9 \times \frac{10}{2^9} = 10$.

(D) The roots of unity are given by the solutions to the equation $z^n = 1$ for all positive integers $n \in \mathbb{N}$.
These roots are of the form $z_k = e^{i \frac{2k\pi}{n}}$ for $k = 0, 1, 2, \dots, n-1$.
As $n$ increases,the set of all roots of unity $\bigcup_{n=1}^{\infty} \{e^{i \frac{2k\pi}{n}} : k=0, 1, \dots, n-1\}$ becomes dense on the unit circle $|z|=1$.
Since any arc of positive length on the unit circle contains infinitely many points from this dense set,there are infinitely many roots of unity on any such arc.
Therefore,the correct option is $D$.

(C) We know that $\cos(n\theta) + i \sin(n\theta) = (\cos \theta + i \sin \theta)^n$.
For $\theta = 1^{\circ}$ and $n = 90$,we have $\cos 90^{\circ} + i \sin 90^{\circ} = (\cos 1^{\circ} + i \sin 1^{\circ})^{90} = i$.
This implies that $\cos 1^{\circ} + i \sin 1^{\circ}$ is a root of the polynomial $z^{90} - i = 0$.
Since $a + ib$ is a root of $z^{90} - i = 0$,it is an algebraic number.
Any complex number $z = a + ib$ is algebraic if and only if both its real part $a$ and imaginary part $b$ are algebraic.
Since $\cos 1^{\circ} + i \sin 1^{\circ}$ is a root of $z^{90} = i$,then $z^{180} = -1$,or $z^{180} + 1 = 0$.
Thus,$\cos 1^{\circ} + i \sin 1^{\circ}$ is an algebraic number.
It is a known result that if $\alpha$ is an algebraic number,then $\cos \alpha$ and $\sin \alpha$ are algebraic for rational multiples of $\pi$.
Since $1^{\circ} = \frac{\pi}{180}$ radians,both $\cos 1^{\circ}$ and $\sin 1^{\circ}$ are algebraic numbers.

(A) The vertices $z_1, z_2, \ldots, z_7$ are the roots of the equation $z^7 - 1 = 0$.
By Vieta's formulas for the polynomial $P(z) = z^7 + 0z^6 + 0z^5 + 0z^4 + 0z^3 + 0z^2 + 0z - 1 = 0$,the sum of the roots taken two at a time is given by the coefficient of $z^5$ divided by the coefficient of $z^7$.
Thus,$w = \sum_{1 \leq i < j \leq 7} z_i z_j = 0$.
Therefore,$|w| = |0| = 0$.

(C) We have the expression $\left|e^{\sum_{k=0}^n {^nC_k} \omega^k}\right|$,where $\omega$ is a cube root of unity.
Using the binomial theorem,$\sum_{k=0}^n {^nC_k} \omega^k = (1+\omega)^n$.
Since $1+\omega+\omega^2=0$,we have $1+\omega = -\omega^2$.
Thus,the exponent is $(-\omega^2)^n = (-1)^n \omega^{2n}$.
We need to evaluate $\left|e^{(-1)^n \omega^{2n}}\right|$.
Let $z = (-1)^n \omega^{2n}$. Then $|e^z| = |e^{\text{Re}(z) + i \text{Im}(z)}| = e^{\text{Re}(z)}$.
Case $1$: If $n$ is a multiple of $3$,say $n=3m$,then $\omega^{2n} = (\omega^3)^{2m} = 1^{2m} = 1$. So $z = (-1)^{3m} = (-1)^m$. The value is $e^{\pm 1}$.
Case $2$: If $n = 3m+1$,then $\omega^{2n} = \omega^{6m+2} = \omega^2$. So $z = (-1)^{3m+1} \omega^2 = (-1)^{3m+1} (-\frac{1}{2} - i\frac{\sqrt{3}}{2})$. The real part is $\pm \frac{1}{2}$. The value is $e^{\pm 1/2}$.
Case $3$: If $n = 3m+2$,then $\omega^{2n} = \omega^{6m+4} = \omega$. So $z = (-1)^{3m+2} \omega = (-1)^{3m+2} (-\frac{1}{2} + i\frac{\sqrt{3}}{2})$. The real part is $\pm \frac{1}{2}$. The value is $e^{\pm 1/2}$.
Combining these,the possible values for the real part are $1, -1, 1/2, -1/2$.
Thus,the possible values of the magnitude are $e^1, e^{-1}, e^{1/2}, e^{-1/2}$.
There are $4$ possible values.

(B) We are given the expression $|a + b\omega + c\omega^2|$ where $a, b, c \in \{1, -1\}$.
Since $\omega$ is a cube root of unity,we know that $1 + \omega + \omega^2 = 0$,which implies $\omega + \omega^2 = -1$.
We test the possible combinations of $a, b, c \in \{1, -1\}$:
If $a=1, b=-1, c=-1$,then $|1 - \omega - \omega^2| = |1 - (\omega + \omega^2)| = |1 - (-1)| = |1 + 1| = 2$.
If $a=1, b=1, c=-1$,then $|1 + \omega - \omega^2| = |1 + \omega - (-1 - \omega)| = |2 + 2\omega| = 2|1 + \omega| = 2|-\omega^2| = 2$.
If $a=1, b=1, c=1$,then $|1 + \omega + \omega^2| = |0| = 0$.
Thus,the maximum possible value is $2$.

(B) Given that $1, \omega, \omega^2$ are the cube roots of unity,we know that $1+\omega+\omega^2=0$.
The roots of the polynomial are given as:
$z_1 = 2\omega$
$z_2 = 2\omega^2$
$z_3 = 3+4\omega$
$z_4 = 3+4\omega^2$
$z_5 = 5-(\omega+\omega^2) = 5-(-1) = 6$
Since the polynomial has real coefficients,if a complex number $z$ is a root,its conjugate $\bar{z}$ must also be a root.
$1$. For $z_1 = 2\omega$,the conjugate is $\bar{z_1} = 2\bar{\omega} = 2\omega^2$,which is $z_2$.
$2$. For $z_3 = 3+4\omega$,the conjugate is $\bar{z_3} = 3+4\bar{\omega} = 3+4\omega^2$,which is $z_4$.
$3$. For $z_5 = 6$,the conjugate is $\bar{z_5} = 6$,which is $z_5$ itself.
All these roots are already present in the given set. Thus,the set of roots is $\{2\omega, 2\omega^2, 3+4\omega, 3+4\omega^2, 6\}$.
There are $5$ distinct roots. Therefore,the minimum degree of the polynomial is $5$.

(B) Given that $2x^2+2x+1$ divides $(x+1)^n-r$, the roots of $2x^2+2x+1=0$ must satisfy $(x+1)^n-r=0$.
Solving $2x^2+2x+1=0$ using the quadratic formula:
$x = \frac{-2 \pm \sqrt{4-8}}{4} = \frac{-1 \pm i}{2}$.
Substituting $x = \frac{-1+i}{2}$ into $(x+1)^n = r$:
$(\frac{-1+i}{2} + 1)^n = r \Rightarrow (\frac{1+i}{2})^n = r$.
Expressing $\frac{1+i}{2}$ in polar form: $\frac{1+i}{2} = \frac{1}{\sqrt{2}} e^{i\pi/4}$.
So, $(\frac{1}{\sqrt{2}} e^{i\pi/4})^n = r \Rightarrow \frac{1}{2^{n/2}} e^{in\pi/4} = r$.
Since $r$ is a real number, the imaginary part must be zero, so $\sin(\frac{n\pi}{4}) = 0$, which implies $n$ must be a multiple of $4$.
For $n=4000$, $r = \frac{1}{2^{4000/2}} \cos(\frac{4000\pi}{4}) = \frac{1}{2^{2000}} \cos(1000\pi) = \frac{1}{2^{2000}} = \frac{1}{4^{1000}}$.
Thus, $(n, r) = (4000, \frac{1}{4^{1000}})$.

(B) Given $(1-\sqrt{3}i)^{200} = 2^{199}(p + iq)$.
Converting to polar form: $1-\sqrt{3}i = 2(\cos(\frac{-\pi}{3}) + i\sin(\frac{-\pi}{3}))$.
So,$(1-\sqrt{3}i)^{200} = 2^{200}(\cos(\frac{-200\pi}{3}) + i\sin(\frac{-200\pi}{3}))$.
Since $\frac{-200\pi}{3} = -66\pi - \frac{2\pi}{3}$,we have $\cos(\frac{-200\pi}{3}) = \cos(\frac{-2\pi}{3}) = -\frac{1}{2}$ and $\sin(\frac{-200\pi}{3}) = \sin(\frac{-2\pi}{3}) = -\frac{\sqrt{3}}{2}$.
Thus,$2^{200}(-\frac{1}{2} - i\frac{\sqrt{3}}{2}) = 2^{199}(p + iq)$.
$2 \times 2^{199}(-\frac{1}{2} - i\frac{\sqrt{3}}{2}) = 2^{199}(p + iq)$.
$p + iq = -1 - i\sqrt{3}$,so $p = -1$ and $q = -\sqrt{3}$.
Let $\alpha = p + q + q^2 = -1 - \sqrt{3} + 3 = 2 - \sqrt{3}$.
Let $\beta = p - q + q^2 = -1 + \sqrt{3} + 3 = 2 + \sqrt{3}$.
Sum of roots $\alpha + \beta = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$.
Product of roots $\alpha \beta = (2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$.
The quadratic equation is $x^2 - (\alpha + \beta)x + \alpha\beta = 0$,which is $x^2 - 4x + 1 = 0$.

(D) Given the quadratic equation $x^2-\sqrt{2} x+2=0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $x = \frac{\sqrt{2} \pm \sqrt{2-8}}{2} = \frac{\sqrt{2} \pm \sqrt{-6}}{2} = \frac{\sqrt{2} \pm i\sqrt{6}}{2}$.
We can write the roots in polar form: $\alpha = \sqrt{2} \left( \frac{1}{2} + i\frac{\sqrt{3}}{2} \right) = \sqrt{2} e^{i\pi/3}$ and $\beta = \sqrt{2} e^{-i\pi/3}$.
Now,calculate $\alpha^{14} + \beta^{14}$:
$\alpha^{14} = (\sqrt{2})^{14} e^{i14\pi/3} = 2^7 e^{i(4\pi + 2\pi/3)} = 128 e^{i2\pi/3}$.
$\beta^{14} = (\sqrt{2})^{14} e^{-i14\pi/3} = 128 e^{-i2\pi/3}$.
$\alpha^{14} + \beta^{14} = 128 (e^{i2\pi/3} + e^{-i2\pi/3}) = 128 \times 2 \cos(2\pi/3)$.
Since $\cos(2\pi/3) = -1/2$,we have $\alpha^{14} + \beta^{14} = 256 \times (-1/2) = -128$.

(B) Given $x^2+x+1=0$,the roots are $\omega$ and $\omega^2$. Let $\alpha = \omega$.
Since $1+\omega+\omega^2=0$,we have $1+\omega = -\omega^2$.
Then $(1+\alpha)^7 = (1+\omega)^7 = (-\omega^2)^7 = -\omega^{14} = -\omega^2$.
Using $1+\omega+\omega^2=0$,we have $-\omega^2 = 1+\omega$.
Comparing $1+\omega$ with $A+B\alpha+C\alpha^2 = A+B\omega+C\omega^2$,we get $A=1, B=1, C=0$.
Thus,$5(3A-2B-C) = 5(3(1)-2(1)-0) = 5(3-2) = 5(1) = 5$.

(A) The given equation is $x^2-\sqrt{6}x+3=0$.
Using the quadratic formula,$x = \frac{\sqrt{6} \pm \sqrt{6-12}}{2} = \frac{\sqrt{6} \pm i\sqrt{6}}{2} = \sqrt{\frac{3}{2}}(1 \pm i)$.
Since $\operatorname{Im}(\alpha) > \operatorname{Im}(\beta)$,we have $\alpha = \sqrt{\frac{3}{2}}(1+i) = \sqrt{3} e^{i\pi/4}$ and $\beta = \sqrt{\frac{3}{2}}(1-i) = \sqrt{3} e^{-i\pi/4}$.
We need to evaluate $\frac{\alpha^{99}}{\beta} + \alpha^{98} = \alpha^{98} \left( \frac{\alpha}{\beta} + 1 \right) = \alpha^{98} \left( \frac{\alpha+\beta}{\beta} \right)$.
Note that $\alpha+\beta = \sqrt{6}$ and $\alpha\beta = 3$.
So,$\frac{\alpha^{99}}{\beta} + \alpha^{98} = \alpha^{98} \left( \frac{\sqrt{6}}{\beta} \right) = \alpha^{98} \left( \frac{\sqrt{6}\alpha}{\alpha\beta} \right) = \alpha^{99} \frac{\sqrt{6}}{3} = \alpha^{99} \sqrt{2}$.
Substituting $\alpha = \sqrt{3} e^{i\pi/4}$,we get $\alpha^{99} = (\sqrt{3})^{99} e^{i99\pi/4} = 3^{49} \sqrt{3} \cdot e^{i(24\pi + 3\pi/4)} = 3^{49} \sqrt{3} \left( -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \right)$.
Multiplying by $\sqrt{2}$,we get $3^{49} \sqrt{3} \left( -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \right) \cdot \sqrt{2} = 3^{49} \sqrt{3} (-1+i) = 3^{49} (-\sqrt{3} + i\sqrt{3})$.
Wait,re-evaluating: $\alpha^{99} \sqrt{2} = (\sqrt{3} e^{i\pi/4})^{99} \sqrt{2} = 3^{49} \sqrt{3} (\cos(99\pi/4) + i\sin(99\pi/4)) \sqrt{2} = 3^{49} \sqrt{3} (-\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}) \sqrt{2} = 3^{49} (- \sqrt{3} + i\sqrt{3})$.
Comparing with $3^n(a+ib)$,we have $n=49, a=-\sqrt{3}, b=\sqrt{3}$. Since $a, b$ must be integers,let's re-check: $\frac{\alpha^{99}}{\beta} + \alpha^{98} = \alpha^{98}(\frac{\alpha+\beta}{\beta}) = \alpha^{98}(\frac{\sqrt{6}}{\beta}) = \frac{\alpha^{98} \sqrt{6}}{\beta} \cdot \frac{\alpha}{\alpha} = \frac{\alpha^{99} \sqrt{6}}{3} = \alpha^{99} \sqrt{2}$.
Actually,$\alpha^2 = \frac{3}{2}(1+i)^2 = \frac{3}{2}(2i) = 3i$. Thus $\alpha^{98} = (\alpha^2)^{49} = (3i)^{49} = 3^{49} i^{49} = 3^{49} i$.
Then $\alpha^{99} = 3^{49} i \cdot \alpha = 3^{49} i \cdot \sqrt{\frac{3}{2}}(1+i) = 3^{49} \sqrt{\frac{3}{2}} (i-1)$.
So $\alpha^{99} \sqrt{2} = 3^{49} \sqrt{3} (i-1) = 3^{49} (-\sqrt{3} + i\sqrt{3})$.
Given $a, b$ are integers not divisible by $3$,the expression simplifies to $3^{49}(-1+i)$ if $\alpha = \sqrt{3}e^{i\pi/4}$. The result is $n=49, a=-1, b=1$. Thus $n+a+b = 49-1+1 = 49$.

(B) Given equations are $z^{1985}+z^{100}+1=0$ and $z^3+2z^2+2z+1=0$.
First,factorize $z^3+2z^2+2z+1=0$:
$(z^3+1) + (2z^2+2z) = 0$
$(z+1)(z^2-z+1) + 2z(z+1) = 0$
$(z+1)(z^2-z+1+2z) = 0$
$(z+1)(z^2+z+1) = 0$
The roots are $z = -1$,$z = \omega$,and $z = \omega^2$,where $\omega$ is the complex cube root of unity.
Now,check these roots in the first equation $f(z) = z^{1985}+z^{100}+1=0$:
$1$. For $z = -1$:
$(-1)^{1985} + (-1)^{100} + 1 = -1 + 1 + 1 = 1 \neq 0$.
$2$. For $z = \omega$:
$\omega^{1985} + \omega^{100} + 1 = \omega^{1985 \pmod 3} + \omega^{100 \pmod 3} + 1 = \omega^2 + \omega + 1 = 0$.
$3$. For $z = \omega^2$:
$(\omega^2)^{1985} + (\omega^2)^{100} + 1 = \omega^{3970} + \omega^{200} + 1 = \omega^1 + \omega^2 + 1 = 0$.
Thus,the common roots are $z = \omega$ and $z = \omega^2$.
The number of common roots is $2$.

(B) We know that $|a + b\omega + c\omega^2|^2 = (a + b\omega + c\omega^2)(\overline{a + b\omega + c\omega^2})$.
Since $\overline{\omega} = \omega^2$ and $\overline{\omega^2} = \omega$,this becomes $(a + b\omega + c\omega^2)(a + b\omega^2 + c\omega)$.
Expanding this,we get $a^2 + ab\omega^2 + ac\omega + ab\omega + b^2\omega^3 + bc\omega^2 + ac\omega^2 + bc\omega^4 + c^2\omega^3$.
Using $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$,this simplifies to $a^2 + b^2 + c^2 - ab - bc - ca$.
This expression can be written as $\frac{1}{2}[(a - b)^2 + (b - c)^2 + (c - a)^2]$.
For distinct non-zero integers $a, b, c$,the smallest possible values for the differences are $1$ and $2$ (e.g.,$a=1, b=2, c=3$).
Substituting these,we get $\frac{1}{2}[(1 - 2)^2 + (2 - 3)^2 + (3 - 1)^2] = \frac{1}{2}[1 + 1 + 4] = \frac{6}{2} = 3$.

(C) $(P)$ Since $z_k = e^{i(2k\pi/10)}$,we have $z_k \cdot z_j = e^{i(2(k+j)\pi/10)} = 1$ if $k+j = 10$. For any $k \in \{1, \ldots, 9\}$,we can choose $j = 10-k \in \{1, \ldots, 9\}$. Thus,the statement is True $(1)$.
$(Q)$ The equation $z_1 \cdot z = z_k$ is a linear equation in complex numbers,which always has a solution $z = z_k / z_1$. Thus,the statement is False $(2)$.
$(R)$ The values $z_1, z_2, \ldots, z_9$ are the roots of $\frac{z^{10}-1}{z-1} = 0$. Thus,$z^{10}-1 = (z-1)(z-z_1)(z-z_2)\ldots(z-z_9)$. Dividing by $(z-1)$,we get $1+z+z^2+\ldots+z^9 = (z-z_1)(z-z_2)\ldots(z-z_9)$. Setting $z=1$,we get $10 = (1-z_1)(1-z_2)\ldots(1-z_9)$. Taking the modulus,$|1-z_1||1-z_2|\ldots|1-z_9| = 10$. Thus,the expression equals $10/10 = 1$ $(3)$.
$(S)$ The sum of all roots of $z^{10}-1=0$ is $1 + z_1 + z_2 + \ldots + z_9 = 0$. Thus,$\sum_{k=1}^9 z_k = -1$. Taking the real part,$\sum_{k=1}^9 \cos(2k\pi/10) = -1$. Then $1 - (-1) = 2$ $(4)$.
Therefore,the correct matching is $P-1, Q-2, R-3, S-4$.

(A) The sum is a geometric progression: $S = \sum_{n=1}^{2025} (-\omega)^n = (-\omega) + (-\omega)^2 + \dots + (-\omega)^{2025}$.
This is a geometric series with first term $a = -\omega$,common ratio $r = -\omega$,and $n = 2025$ terms.
$S = \frac{a(1-r^n)}{1-r} = \frac{-\omega(1-(-\omega)^{2025})}{1-(-\omega)} = \frac{-\omega(1 - (-\omega^{2025}))}{1+\omega}$.
Since $\omega^3 = 1$ and $2025$ is a multiple of $3$,$\omega^{2025} = 1$.
$S = \frac{-\omega(1 - (-1))}{1+\omega} = \frac{-\omega(2)}{1+\omega}$.
Using $1+\omega = -\omega^2$,we get $S = \frac{-2\omega}{-\omega^2} = \frac{2}{\omega} = 2\omega^2$.
Since $\omega = e^{i2\pi/3}$,$\omega^2 = e^{i4\pi/3} = e^{-i2\pi/3}$.
Thus,$\alpha = \arg(2\omega^2) = \arg(e^{-i2\pi/3}) = -\frac{2\pi}{3}$.
Therefore,$\frac{3\alpha}{\pi} = \frac{3}{\pi} \times \left(-\frac{2\pi}{3}\right) = -2$.

(A) Given $z(2-i) = 3+i$.
$z = \frac{3+i}{2-i} = \frac{(3+i)(2+i)}{(2-i)(2+i)} = \frac{6+3i+2i+i^2}{4+1} = \frac{6+5i-1}{5} = \frac{5+5i}{5} = 1+i$.
Now,express $z$ in polar form: $z = 1+i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)$.
Using De Moivre's Theorem,$z^{38} = (\sqrt{2})^{38} \left( \cos \frac{38\pi}{4} + i \sin \frac{38\pi}{4} \right)$.
$z^{38} = 2^{19} \left( \cos \frac{19\pi}{2} + i \sin \frac{19\pi}{2} \right)$.
Since $\frac{19\pi}{2} = 9\pi + \frac{\pi}{2}$,we have $\cos \frac{19\pi}{2} = 0$ and $\sin \frac{19\pi}{2} = -1$.
$z^{38} = 2^{19} (0 + i(-1)) = -2^{19} i$.

(B) Using De Moivre's Theorem,we know that $(\cos \theta + i \sin \theta) = e^{i\theta}$.
Thus,$f(x) = e^{ix} \cdot e^{i3x} \cdot e^{i5x} \cdots e^{i(2n-1)x}$.
$f(x) = e^{i(1 + 3 + 5 + \cdots + (2n-1))x}$.
The sum of the first $n$ odd numbers is $n^2$,so $f(x) = e^{i(n^2)x}$.
Now,find the first derivative: $f'(x) = i n^2 e^{i(n^2)x} = i n^2 f(x)$.
Find the second derivative: $f''(x) = i n^2 f'(x) = i n^2 (i n^2 f(x)) = i^2 n^4 f(x)$.
Since $i^2 = -1$,we get $f''(x) = -n^4 f(x)$.