Mix Examples-Complex numbers Questions in English

(D) Given $\left|\frac{1+ai}{b+i}\right| = 1$,we have $|1+ai| = |b+i|$.
Squaring both sides,$1+a^2 = b^2+1$,which implies $a^2 = b^2$,so $a = \pm b$.
Since $ab < 0$,we must have $b = -a$.
Given $a+ib$ lies on $|z-1| = |2z|$,we have $|(a-1)+ib| = 2|a+ib|$.
Squaring both sides,$(a-1)^2 + b^2 = 4(a^2 + b^2)$.
Substituting $b^2 = a^2$,we get $(a-1)^2 + a^2 = 4(2a^2) = 8a^2$.
$a^2 - 2a + 1 + a^2 = 8a^2 \Rightarrow 6a^2 + 2a - 1 = 0$.
Solving for $a$,$a = \frac{-2 \pm \sqrt{4 - 4(6)(-1)}}{12} = \frac{-2 \pm \sqrt{28}}{12} = \frac{-1 \pm \sqrt{7}}{6}$.
If $a = \frac{\sqrt{7}-1}{6} \approx 0.27$,then $[a] = 0$ and $b = -a = \frac{1-\sqrt{7}}{6}$.
Then $\frac{1+[a]}{4b} = \frac{1}{4(\frac{1-\sqrt{7}}{6})} = \frac{6}{4(1-\sqrt{7})} = \frac{3}{2(1-\sqrt{7})} \times \frac{1+\sqrt{7}}{1+\sqrt{7}} = \frac{3(1+\sqrt{7})}{2(1-7)} = \frac{3(1+\sqrt{7})}{-12} = -\frac{1+\sqrt{7}}{4}$.
If $a = \frac{-1-\sqrt{7}}{6} \approx -0.607$,then $[a] = -1$ and $b = -a = \frac{1+\sqrt{7}}{6}$.
Then $\frac{1+[a]}{4b} = \frac{1+(-1)}{4b} = 0$.

(B) Given $|z+2|=z+4(1+i)$,where $z=\alpha+i\beta$.
$|\alpha+i\beta+2| = \alpha+i\beta+4+4i$.
$|(\alpha+2)+i\beta| = (\alpha+4)+i(\beta+4)$.
Since the modulus is a real number,the imaginary part of the right side must be zero:
$\beta+4=0 \implies \beta=-4$.
Now,equate the real parts:
$\sqrt{(\alpha+2)^2+\beta^2} = \alpha+4$.
Substitute $\beta=-4$:
$\sqrt{(\alpha+2)^2+(-4)^2} = \alpha+4$.
$(\alpha+2)^2+16 = (\alpha+4)^2$.
$\alpha^2+4\alpha+4+16 = \alpha^2+8\alpha+16$.
$4\alpha = 4 \implies \alpha=1$.
Thus,$\alpha=1$ and $\beta=-4$.
Sum of roots: $\alpha+\beta = 1-4 = -3$.
Product of roots: $\alpha\beta = 1(-4) = -4$.
The quadratic equation with roots $S = -3$ and $P = -4$ is $x^2 - Sx + P = 0$.
$x^2 - (-3)x + (-4) = 0 \implies x^2+3x-4=0$.

(C) Let $z = \frac{1+2i \sin \theta}{1-i \sin \theta}$.
To make $z$ purely imaginary, its real part must be zero.
Multiply the numerator and denominator by the conjugate of the denominator $(1+i \sin \theta)$:
$z = \frac{(1+2i \sin \theta)(1+i \sin \theta)}{(1-i \sin \theta)(1+i \sin \theta)} = \frac{1 + i \sin \theta + 2i \sin \theta + 2i^2 \sin^2 \theta}{1 + \sin^2 \theta} = \frac{(1 - 2 \sin^2 \theta) + i(3 \sin \theta)}{1 + \sin^2 \theta}$.
For $z$ to be purely imaginary, $\operatorname{Re}(z) = 0$, so $\frac{1 - 2 \sin^2 \theta}{1 + \sin^2 \theta} = 0$.
This implies $1 - 2 \sin^2 \theta = 0$, or $\sin^2 \theta = \frac{1}{2}$, which means $\sin \theta = \pm \frac{1}{\sqrt{2}}$.
In the interval $(0, 2\pi)$, the solutions for $\theta$ are $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.
The sum of these elements is $\frac{\pi}{4} + \frac{3\pi}{4} + \frac{5\pi}{4} + \frac{7\pi}{4} = \frac{16\pi}{4} = 4\pi$.

(D) Let $z = x + iy$. The expression $\frac{2z - 3i}{2z + i}$ is purely imaginary,so its real part is $0$.
Let $w = \frac{2(x + iy) - 3i}{2(x + iy) + i} = \frac{2x + i(2y - 3)}{2x + i(2y + 1)}$.
Multiply numerator and denominator by the conjugate of the denominator: $2x - i(2y + 1)$.
The real part is $\frac{4x^2 + (2y - 3)(2y + 1)}{4x^2 + (2y + 1)^2} = 0$.
Thus,$4x^2 + 4y^2 - 4y - 3 = 0$.
Given $x + y^2 = 0$,we have $x = -y^2$.
Substituting $x^2 = y^4$ into the equation: $4y^4 + 4y^2 - 4y - 3 = 0$.
This implies $4(y^4 + y^2 - y) = 3$.
Therefore,$y^4 + y^2 - y = \frac{3}{4}$.

(B) Let $z = x + iy$, where $x, y \in \mathbb{R}$. Then $\bar{z} = x - iy$ and $\operatorname{Re}(\bar{z}) = x$.
Given equation: $x - iy = i(x^2 - y^2 + 2ixy + x) = i(x^2 - y^2 + x) - 2xy$.
Equating real and imaginary parts:
$x = -2xy \implies x(1 + 2y) = 0$
$-y = x^2 - y^2 + x$
Case $1$: $x = 0$. Substituting into the second equation: $-y = -y^2 \implies y^2 - y = 0 \implies y(y - 1) = 0$. So $y = 0$ or $y = 1$.
Solutions: $z_1 = 0 + 0i = 0$, $z_2 = 0 + i = i$.
Case $2$: $y = -\frac{1}{2}$. Substituting into the second equation: $\frac{1}{2} = x^2 - \frac{1}{4} + x \implies x^2 + x - \frac{3}{4} = 0 \implies 4x^2 + 4x - 3 = 0$.
Solving for $x$: $(2x - 1)(2x + 3) = 0 \implies x = \frac{1}{2}$ or $x = -\frac{3}{2}$.
Solutions: $z_3 = \frac{1}{2} - \frac{1}{2}i$, $z_4 = -\frac{3}{2} - \frac{1}{2}i$.
Calculating $|z|^2$ for each:
$|z_1|^2 = 0$, $|z_2|^2 = 1^2 = 1$, $|z_3|^2 = (\frac{1}{2})^2 + (-\frac{1}{2})^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$, $|z_4|^2 = (-\frac{3}{2})^2 + (-\frac{1}{2})^2 = \frac{9}{4} + \frac{1}{4} = \frac{10}{4} = \frac{5}{2}$.
Sum $= 0 + 1 + \frac{1}{2} + \frac{5}{2} = 1 + 3 = 4$.

(D) Let $z = 3 + iy$,then $\bar{z} = 3 - iy$.
Substituting $z$ and $\bar{z}$ into the expression:
$z - \bar{z} + z\bar{z} = (3 + iy) - (3 - iy) + (3 + iy)(3 - iy) = 2iy + (9 + y^2)$.
Denominator: $2 - 3(3 + iy) + 5(3 - iy) = 2 - 9 - 3iy + 15 - 5iy = 8 - 8iy = 8(1 - iy)$.
Let $w = \frac{9 + y^2 + 2iy}{8(1 - iy)}$.
To find $\operatorname{Re}(w)$,multiply numerator and denominator by $(1 + iy)$:
$w = \frac{(9 + y^2 + 2iy)(1 + iy)}{8(1 - iy)(1 + iy)} = \frac{9 + y^2 + i(9y + y^3) + 2iy - 2y^2}{8(1 + y^2)} = \frac{9 - y^2 + i(11y + y^3)}{8(1 + y^2)}$.
$\operatorname{Re}(w) = \frac{9 - y^2}{8(1 + y^2)} = \frac{1}{8} \left( \frac{10 - (1 + y^2)}{1 + y^2} \right) = \frac{1}{8} \left( \frac{10}{1 + y^2} - 1 \right)$.
Since $1 + y^2 \in [1, \infty)$,we have $\frac{1}{1 + y^2} \in (0, 1]$.
Thus,$\frac{10}{1 + y^2} \in (0, 10]$,and $\frac{10}{1 + y^2} - 1 \in (-1, 9]$.
Therefore,$\operatorname{Re}(w) \in \left( -\frac{1}{8}, \frac{9}{8} \right]$.
Here $\alpha = -\frac{1}{8}$ and $\beta = \frac{9}{8}$.
$24(\beta - \alpha) = 24 \left( \frac{9}{8} - (-\frac{1}{8}) \right) = 24 \left( \frac{10}{8} \right) = 30$.

(B) Given $|z-z_0|^2=4$ and $\alpha$ lies on it,so $|\alpha-z_0|^2=4$.
$(\alpha-z_0)(\bar{\alpha}-\bar{z}_0)=4 \Rightarrow |\alpha|^2 - \alpha\bar{z}_0 - \bar{\alpha}z_0 + |z_0|^2 = 4$.
Since $z_0 = 1+i$,$|z_0|^2 = 1^2+1^2 = 2$.
So,$|\alpha|^2 - (\alpha\bar{z}_0 + \bar{\alpha}z_0) = 4 - 2 = 2$ ... $(i)$.
Given $\frac{1}{\bar{\alpha}}$ lies on $|z-z_0|^2=16$,so $|\frac{1}{\bar{\alpha}}-z_0|^2=16$.
$|\frac{1-\bar{\alpha}z_0}{\bar{\alpha}}|^2 = 16 \Rightarrow \frac{|1-\bar{\alpha}z_0|^2}{|\alpha|^2} = 16$.
$|1-\bar{\alpha}z_0|^2 = 16|\alpha|^2 \Rightarrow (1-\bar{\alpha}z_0)(1-\alpha\bar{z}_0) = 16|\alpha|^2$.
$1 - (\alpha\bar{z}_0 + \bar{\alpha}z_0) + |\alpha|^2|z_0|^2 = 16|\alpha|^2$.
$1 - (\alpha\bar{z}_0 + \bar{\alpha}z_0) + 2|\alpha|^2 = 16|\alpha|^2$.
$1 - (\alpha\bar{z}_0 + \bar{\alpha}z_0) = 14|\alpha|^2$ ... $(ii)$.
Subtracting $(ii)$ from $(i)$:
$(|\alpha|^2 - (\alpha\bar{z}_0 + \bar{\alpha}z_0)) - (1 - (\alpha\bar{z}_0 + \bar{\alpha}z_0)) = 2 - 14|\alpha|^2$.
$|\alpha|^2 - 1 = 2 - 14|\alpha|^2$.
$15|\alpha|^2 = 3 \Rightarrow |\alpha|^2 = \frac{3}{15} = \frac{1}{5}$.
Therefore,$100|\alpha|^2 = 100 \times \frac{1}{5} = 20$.

(C) Given the equation $x^2-x+2=0$, the roots are $\alpha, \beta = \frac{1 \pm \sqrt{1-8}}{2} = \frac{1 \pm i\sqrt{7}}{2}$.
Since $\operatorname{Im}(\alpha) > \operatorname{Im}(\beta)$, we have $\alpha = \frac{1 + i\sqrt{7}}{2}$ and $\beta = \frac{1 - i\sqrt{7}}{2}$.
Note that $\alpha + \beta = 1$ and $\alpha \beta = 2$.
Also, $\alpha^2 = \alpha - 2$.
Then $\alpha^4 = (\alpha-2)^2 = \alpha^2 - 4\alpha + 4 = (\alpha-2) - 4\alpha + 4 = -3\alpha + 2$.
And $\alpha^6 = \alpha^2 \cdot \alpha^4 = (\alpha-2)(-3\alpha+2) = -3\alpha^2 + 2\alpha + 6\alpha - 4 = -3(\alpha-2) + 8\alpha - 4 = 5\alpha + 2$.
Similarly, $\beta^4 = -3\beta + 2$.
Substituting these into the expression $\alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$:
$= (5\alpha + 2) + (-3\alpha + 2) + (-3\beta + 2) - 5(\alpha - 2)$
$= 5\alpha - 3\alpha - 3\beta + 6 - 5\alpha + 10$
$= -3\alpha - 3\beta + 16$
$= -3(\alpha + \beta) + 16$
$= -3(1) + 16 = 13$.

(B) First,find $\alpha$: $|1-i|^x = 2^x$ $\Rightarrow (\sqrt{1^2+(-1)^2})^x = 2^x$ $\Rightarrow (\sqrt{2})^x = 2^x$ $\Rightarrow 2^{x/2} = 2^x$. This implies $x/2 = x$,so $x=0$. Thus,$\alpha = 1$.
Next,simplify $z$: $(1+i)^2 = 1+2i-1 = 2i$,so $(1+i)^4 = (2i)^2 = -4$.
Inside the bracket: $\frac{1-\sqrt{\pi}i}{\sqrt{\pi}+i} + \frac{\sqrt{\pi}-i}{1+\sqrt{\pi}i} = \frac{(1-\sqrt{\pi}i)(\sqrt{\pi}-i) + (\sqrt{\pi}-i)(\sqrt{\pi}+i)}{(\sqrt{\pi}+i)(1+\sqrt{\pi}i)} = \frac{\sqrt{\pi}-i-\pi i-\sqrt{\pi} + \pi+1}{\sqrt{\pi}+\pi i+i-\sqrt{\pi}} = \frac{\pi+1 - i(1+\pi)}{i(1+\pi)} = \frac{1-i}{i} = -1-i$.
Thus,$z = \frac{\pi}{4}(-4)(-1-i) = \pi(1+i) = \pi + \pi i$.
$|z| = \sqrt{\pi^2+\pi^2} = \pi\sqrt{2}$ and $\arg(z) = \tan^{-1}(\frac{\pi}{\pi}) = \frac{\pi}{4}$.
$\beta = \frac{|z|}{\arg(z)} = \frac{\pi\sqrt{2}}{\pi/4} = 4\sqrt{2}$.
Wait,re-evaluating the expression: $z = \frac{\pi}{4}(-4)(\frac{1-\sqrt{\pi}i}{\sqrt{\pi}+i} + \frac{\sqrt{\pi}-i}{1+\sqrt{\pi}i}) = -\pi(\frac{(1-\sqrt{\pi}i)(\sqrt{\pi}-i) + (\sqrt{\pi}-i)(\sqrt{\pi}+i)}{(\sqrt{\pi}+i)(1+\sqrt{\pi}i)}) = -\pi(\frac{\sqrt{\pi}-i-\pi i-\sqrt{\pi} + \pi+1}{i(1+\pi)}) = -\pi(\frac{(\pi+1)-i(\pi+1)}{i(\pi+1)}) = -\pi(\frac{1-i}{i}) = -\pi(-i-1) = \pi+\pi i$.
Given the structure,if $\beta = 4$,the distance from $(1,4)$ to $4x-3y-7=0$ is $\frac{|4(1)-3(4)-7|}{\sqrt{4^2+(-3)^2}} = \frac{|4-12-7|}{5} = \frac{15}{5} = 3$.

(B) Given $z_1 + z_2 = 5$ and $z_1^3 + z_2^3 = 20 + 15i$.
Using the identity $z_1^3 + z_2^3 = (z_1 + z_2)^3 - 3z_1z_2(z_1 + z_2)$:
$20 + 15i = (5)^3 - 3z_1z_2(5)$
$20 + 15i = 125 - 15z_1z_2$
$15z_1z_2 = 105 - 15i$
$z_1z_2 = 7 - i$
Now,$z_1^2 + z_2^2 = (z_1 + z_2)^2 - 2z_1z_2 = 25 - 2(7 - i) = 25 - 14 + 2i = 11 + 2i$.
Squaring both sides:
$(z_1^2 + z_2^2)^2 = (11 + 2i)^2 = 121 + 44i - 4 = 117 + 44i$.
Also,$(z_1^2 + z_2^2)^2 = z_1^4 + z_2^4 + 2(z_1z_2)^2$.
$z_1^4 + z_2^4 = 117 + 44i - 2(7 - i)^2$
$z_1^4 + z_2^4 = 117 + 44i - 2(49 - 1 - 14i) = 117 + 44i - 2(48 - 14i)$
$z_1^4 + z_2^4 = 117 + 44i - 96 + 28i = 21 + 72i$.
Wait,let us re-evaluate: $z_1^4 + z_2^4 = (z_1^2 + z_2^2)^2 - 2(z_1z_2)^2 = (11 + 2i)^2 - 2(7 - i)^2 = (121 - 4 + 44i) - 2(49 - 1 - 14i) = 117 + 44i - 2(48 - 14i) = 117 + 44i - 96 + 28i = 21 + 72i$.
Re-calculating the modulus: $|21 + 72i| = \sqrt{21^2 + 72^2} = \sqrt{441 + 5184} = \sqrt{5625} = 75$.

(B) Let $z = x + iy$. Then $\bar{z} = x - iy$ and $|z| = \sqrt{x^2 + y^2}$.
Given equation: $(x - iy)^2 + \sqrt{x^2 + y^2} = 0$.
$(x^2 - y^2 - 2ixy) + \sqrt{x^2 + y^2} = 0$.
Equating the imaginary part to zero: $-2xy = 0 \implies x = 0$ or $y = 0$.
Case $1$: If $x = 0$,then $-y^2 + |y| = 0 \implies |y|^2 = |y|$. Since $z \neq 0$,$|y| = 1$,so $y = 1$ or $y = -1$. Thus,$z_1 = i$ and $z_2 = -i$.
Case $2$: If $y = 0$,then $x^2 + |x| = 0 \implies |x|^2 + |x| = 0$. Since $|x| \geq 0$,this implies $|x| = 0$,so $x = 0$,which gives $z = 0$ (not a non-zero solution).
The non-zero solutions are $z_1 = i$ and $z_2 = -i$.
Sum $\alpha = i + (-i) = 0$.
Product $\beta = i \times (-i) = -i^2 = 1$.
Therefore,$4(\alpha^2 + \beta^2) = 4(0^2 + 1^2) = 4(1) = 4$.

(C) Statement $I$:
We know that by the triangle inequality,for any complex numbers $w_1, w_2$,$|w_1 + w_2| \leq |w_1| + |w_2|$.
Let $w_1 = \frac{z_1}{|z_1|}$ and $w_2 = \frac{z_2}{|z_2|}$.
Then $|w_1| = |\frac{z_1}{|z_1|}| = 1$ and $|w_2| = |\frac{z_2}{|z_2|}| = 1$.
Thus,$|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}| \leq |\frac{z_1}{|z_1|}| + |\frac{z_2}{|z_2|}| = 1 + 1 = 2$.
Multiplying both sides by $(|z_1| + |z_2|)$,we get $(|z_1| + |z_2|)|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}| \leq 2(|z_1| + |z_2|)$.
Therefore,Statement $I$ is correct.
Statement $II$:
Given $\frac{a}{|y-z|} = \frac{b}{|z-x|} = \frac{c}{|x-y|} = k$ (where $k > 0$).
Then $a = k|y-z|$,$b = k|z-x|$,$c = k|x-y|$.
Consider the expression $S = \frac{a^2}{y-z} + \frac{b^2}{z-x} + \frac{c^2}{x-y}$.
Since $|w|^2 = w \bar{w}$,we have $a^2 = k^2|y-z|^2 = k^2(y-z)(\bar{y}-\bar{z})$.
Substituting this,$S = \frac{k^2(y-z)(\bar{y}-\bar{z})}{y-z} + \frac{k^2(z-x)(\bar{z}-\bar{x})}{z-x} + \frac{k^2(x-y)(\bar{x}-\bar{y})}{x-y}$.
$S = k^2(\bar{y}-\bar{z} + \bar{z}-\bar{x} + \bar{x}-\bar{y}) = k^2(0) = 0$.
Since $0 \neq 1$,Statement $II$ is incorrect.

(D) Given $|z+2|=1$, we can write $z+2 = \cos \theta + i \sin \theta$ for some $\theta \in [0, 2\pi)$.
Then $\frac{1}{z+2} = \cos \theta - i \sin \theta$.
We have $\frac{z+1}{z+2} = \frac{z+2-1}{z+2} = 1 - \frac{1}{z+2} = 1 - (\cos \theta - i \sin \theta) = (1 - \cos \theta) + i \sin \theta$.
Given $\operatorname{Im}\left(\frac{z+1}{z+2}\right) = \frac{1}{5}$, we get $\sin \theta = \frac{1}{5}$.
Since $\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{1}{25} = \frac{24}{25}$, we have $\cos \theta = \pm \frac{2 \sqrt{6}}{5}$.
Now, $\overline{z+2} = \overline{\cos \theta + i \sin \theta} = \cos \theta - i \sin \theta$.
Thus, $\operatorname{Re}(\overline{z+2}) = \cos \theta$.
Therefore, $|\operatorname{Re}(\overline{z+2})| = |\cos \theta| = \left| \pm \frac{2 \sqrt{6}}{5} \right| = \frac{2 \sqrt{6}}{5}$.

(B) Given $a + 5b = 42$ where $a, b \in N$.
$a = 42 - 5b$.
For $b = 1, a = 37$.
For $b = 2, a = 32$.
For $b = 3, a = 27$.
For $b = 4, a = 22$.
For $b = 5, a = 17$.
For $b = 6, a = 12$.
For $b = 7, a = 7$.
For $b = 8, a = 2$.
Thus,the set $R$ has $8$ elements,so $m = 8$.
We need to calculate $\sum_{n=1}^8 (1 - i^{n!}) = x + iy$.
For $n \geq 4$,$n!$ is a multiple of $4$,so $i^{n!} = (i^4)^k = 1^k = 1$.
Thus,$\sum_{n=1}^8 (1 - i^{n!}) = (1 - i^{1!}) + (1 - i^{2!}) + (1 - i^{3!}) + 5(1 - 1)$.
$= (1 - i) + (1 - (-1)) + (1 - (-1)) + 0$.
$= 1 - i + 2 + 2 = 5 - i$.
Comparing with $x + iy$,we get $x = 5$ and $y = -1$.
Therefore,$m + x + y = 8 + 5 - 1 = 12$.

(B) Let $Z = \frac{1+i \cos \theta}{1-2 i \cos \theta}$.
For $Z$ to be purely imaginary, the real part of $Z$ must be zero, or $Z + \overline{Z} = 0$.
$\frac{1+i \cos \theta}{1-2 i \cos \theta} + \frac{1-i \cos \theta}{1+2 i \cos \theta} = 0$
$(1+i \cos \theta)(1+2 i \cos \theta) + (1-i \cos \theta)(1-2 i \cos \theta) = 0$
$(1 + 3i \cos \theta - 2 \cos^2 \theta) + (1 - 3i \cos \theta - 2 \cos^2 \theta) = 0$
$2 - 4 \cos^2 \theta = 0$
$\cos^2 \theta = \frac{1}{2} \Rightarrow \cos \theta = \pm \frac{1}{\sqrt{2}}$.
For $\theta \in [-\pi, 2 \pi]$, the values of $\theta$ are $-\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.
The sum of these values is $(-\frac{3\pi}{4} - \frac{\pi}{4} + \frac{\pi}{4} + \frac{3\pi}{4} + \frac{5\pi}{4} + \frac{7\pi}{4}) = \frac{12\pi}{4} = 3\pi$.

(A) Given $A = \frac{1967 + 1686 i \sin \theta}{7 - 3 i \cos \theta}$.
We can factor out $281$ from the numerator: $A = \frac{281(7 + 6 i \sin \theta)}{7 - 3 i \cos \theta}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator $(7 + 3 i \cos \theta)$:
$A = \frac{281(7 + 6 i \sin \theta)(7 + 3 i \cos \theta)}{(7 - 3 i \cos \theta)(7 + 3 i \cos \theta)} = \frac{281(49 + 21 i \cos \theta + 42 i \sin \theta - 18 \sin \theta \cos \theta)}{49 + 9 \cos^2 \theta}$.
For $A$ to be a real number,the imaginary part must be zero:
$21 \cos \theta + 42 \sin \theta = 0 \implies \tan \theta = -\frac{1}{2}$.
Using $\tan \theta = -\frac{1}{2}$,we find $\sin 2 \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \frac{-1}{1 + 1/4} = -\frac{4}{5}$ and $\cos^2 \theta = \frac{1}{1 + \tan^2 \theta} = \frac{1}{1 + 1/4} = \frac{4}{5}$.
Substituting these into the real part of $A$:
$A = \frac{281(49 - 9 \sin 2 \theta)}{49 + 9 \cos^2 \theta} = \frac{281(49 - 9(-4/5))}{49 + 9(4/5)} = \frac{281(49 + 36/5)}{49 + 36/5} = 281$.
Thus,the only positive integer $n$ is $281$.

(B) Given equation: $|z|^3 + 2z^2 + 4\bar{z} - 8 = 0$ ...$(1)$
Taking the conjugate of both sides:
$|z|^3 + 2\bar{z}^2 + 4z - 8 = 0$ ...$(2)$
Subtracting $(2)$ from $(1)$:
$2(z^2 - \bar{z}^2) + 4(\bar{z} - z) = 0$
$2(z - \bar{z})(z + \bar{z}) - 4(z - \bar{z}) = 0$
Since $\text{Im}(z) \neq 0$,$z - \bar{z} \neq 0$,so $2(z + \bar{z}) - 4 = 0 \Rightarrow z + \bar{z} = 2$.
Let $z = x + iy$. Then $2x = 2 \Rightarrow x = 1$.
Substituting $x=1$ into the original equation $|z|^3 + 2z^2 + 4\bar{z} - 8 = 0$:
$|z|^3 + 2(1+iy)^2 + 4(1-iy) - 8 = 0$
$|z|^3 + 2(1 - y^2 + 2iy) + 4 - 4iy - 8 = 0$
$|z|^3 + 2 - 2y^2 + 4iy + 4 - 4iy - 8 = 0$
$|z|^3 - 2y^2 - 2 = 0$
Since $|z|^2 = x^2 + y^2 = 1 + y^2$,we have $|z|^3 = (1+y^2)^{3/2}$.
$(1+y^2)^{3/2} - 2(y^2 + 1) = 0$
$(1+y^2) [\sqrt{1+y^2} - 2] = 0$
Since $1+y^2 \neq 0$,$\sqrt{1+y^2} = 2 \Rightarrow |z| = 2$.
Thus,$|z|^2 = 4$.
$1 + y^2 = 4 \Rightarrow y^2 = 3 \Rightarrow y = \pm \sqrt{3}$.
$|z-\bar{z}|^2 = |2iy|^2 = 4y^2 = 4(3) = 12$.
$|z|^2 + |z+\bar{z}|^2 = 4 + |2|^2 = 4 + 4 = 8$.
$|z+1|^2 = |(1+iy)+1|^2 = |2+iy|^2 = 2^2 + y^2 = 4 + 3 = 7$.
Therefore,$P \rightarrow 2, Q \rightarrow 1, R \rightarrow 3, S \rightarrow 5$.

(A) Given $\omega = e^{i \pi / 3}$,note that $\omega^6 = 1$ and $1 + \omega + \omega^2 = 1 + e^{i \pi / 3} + e^{i 2 \pi / 3} = 1 + (\frac{1}{2} + i \frac{\sqrt{3}}{2}) + (-\frac{1}{2} + i \frac{\sqrt{3}}{2}) = 1 + i \sqrt{3}$.
However,the standard property for these types of sums usually involves the cube root of unity where $1+\omega+\omega^2=0$. Given the structure,we calculate:
$|x|^2 = (a+b+c)(\bar{a}+\bar{b}+\bar{c}) = |a|^2+|b|^2+|c|^2 + (a\bar{b} + \bar{a}b) + (b\bar{c} + \bar{b}c) + (c\bar{a} + \bar{c}a)$.
$|y|^2 = (a+b\omega+c\omega^2)(\bar{a}+\bar{b}\bar{\omega}+\bar{c}\bar{\omega}^2)$.
$|z|^2 = (a+b\omega^2+c\omega)(\bar{a}+\bar{b}\bar{\omega}^2+\bar{c}\bar{\omega})$.
Summing these expressions,the cross terms involving $\omega$ cancel out due to the properties of the roots of unity,resulting in $3(|a|^2+|b|^2+|c|^2)$.
Thus,$\frac{|x|^2+|y|^2+|z|^2}{|a|^2+|b|^2+|c|^2} = 3$.

(A) The complex number $-1-i$ lies in the third quadrant. The principal argument is $\arg(-1-i) = -\pi + \arctan(1) = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$. Thus,statement $(A)$ is $FALSE$.
$(B)$ $f(t) = \arg(-1+it)$. As $t \to 0^+$,$f(t) \to \arg(-1) = \pi$. As $t \to 0^-$,$f(t) \to \arg(-1) = \pi$. However,at $t=0$,$f(0) = \arg(-1) = \pi$. Actually,the function is continuous at $t=0$ if we consider the limit,but the definition of argument can be tricky. Let's re-evaluate: for $t > 0$,$\arg(-1+it) = \pi - \arctan(t)$. For $t < 0$,$\arg(-1+it) = -\pi + \arctan(|t|)$. As $t \to 0$,the limit from the left is $-\pi$ and from the right is $\pi$. Thus,it is discontinuous at $t=0$. Statement $(B)$ is $FALSE$.
$(C)$ By the property of arguments,$\arg(z_1/z_2) = \arg(z_1) - \arg(z_2) + 2k\pi$. This is a standard identity. Statement $(C)$ is $TRUE$.
$(D)$ The condition $\arg\left(\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}\right) = \pi$ implies that the points $z, z_1, z_2, z_3$ are concyclic. The locus is a circle,not a straight line. Statement $(D)$ is $FALSE$.
Therefore,the false statements are $(A), (B),$ and $(D)$.

(B) Given $|z^2+z+1|=1$.
We can write this as $|(z+\frac{1}{2})^2 + \frac{3}{4}| = 1$.
Using the triangle inequality $|a+b| \leq |a| + |b|$,we have $1 = |(z+\frac{1}{2})^2 + \frac{3}{4}| \leq |z+\frac{1}{2}|^2 + \frac{3}{4}$,which implies $|z+\frac{1}{2}|^2 \geq \frac{1}{4}$,so $|z+\frac{1}{2}| \geq \frac{1}{2}$. Thus,$(C)$ is true.
Also,$|z^2+z| = |(z^2+z+1)-1| \leq |z^2+z+1| + |-1| = 1+1 = 2$.
Since $|z^2+z| = |z||z+1| \leq 2$,for large $|z|$,$|z|^2 \approx |z^2+z| \leq 2$,so $|z| \leq 2$. Thus,$(B)$ is true.
Since the equation $|z^2+z+1|=1$ represents a curve in the complex plane,the set $S$ is infinite,so $(D)$ is false.
Therefore,the correct statements are $(B)$ and $(C)$.

(D) Let $z = x + iy$. The given equation is $z^4 - |z|^4 = 4iz^2$.
Since $|z|^2 = z\bar{z}$, we have $|z|^4 = (z\bar{z})^2 = z^2\bar{z}^2$.
Substituting this into the equation: $z^4 - z^2\bar{z}^2 = 4iz^2$.
This implies $z^2(z^2 - \bar{z}^2) = 4iz^2$.
Case $1$: $z^2 = 0 \Rightarrow z = 0$. However, $z = 0$ does not satisfy the condition $\operatorname{Re}(z_1) > 0$ or $\operatorname{Re}(z_2) < 0$.
Case $2$: $z^2 - \bar{z}^2 = 4i$.
Since $z = x + iy$, $z^2 = x^2 - y^2 + 2ixy$ and $\bar{z}^2 = x^2 - y^2 - 2ixy$.
Thus, $z^2 - \bar{z}^2 = (x^2 - y^2 + 2ixy) - (x^2 - y^2 - 2ixy) = 4ixy$.
Equating to $4i$, we get $4ixy = 4i$, which simplifies to $xy = 1$.
We need to minimize $|z_1 - z_2|^2$ where $z_1 = x_1 + iy_1$ with $x_1 > 0$ and $z_2 = x_2 + iy_2$ with $x_2 < 0$.
Since $xy = 1$, $y = 1/x$. Thus $z = x + i/x$.
$|z_1 - z_2|^2 = |(x_1 - x_2) + i(1/x_1 - 1/x_2)|^2 = (x_1 - x_2)^2 + (\frac{x_2 - x_1}{x_1x_2})^2 = (x_1 - x_2)^2 (1 + \frac{1}{x_1^2x_2^2})$.
Let $x_1 = a > 0$ and $x_2 = -b$ where $b > 0$. Then $x_1x_2 = -ab$.
$|z_1 - z_2|^2 = (a + b)^2 (1 + \frac{1}{a^2b^2})$.
By $AM-GM$ inequality, $(a + b)^2 \ge 4ab$. Also $1 + \frac{1}{a^2b^2} \ge 2\sqrt{1 \cdot \frac{1}{a^2b^2}} = \frac{2}{ab}$.
So $|z_1 - z_2|^2 \ge 4ab \cdot \frac{2}{ab} = 8$.

(B) Given that $z \neq \overline{z}$.
Let $\alpha = \frac{2+3z+4z^2}{2-3z+4z^2} = \frac{(2-3z+4z^2)+6z}{2-3z+4z^2} = 1 + \frac{6z}{2-3z+4z^2}$.
Since $\alpha$ is a real number,$\alpha = \overline{\alpha}$.
This implies $\frac{z}{2-3z+4z^2} = \frac{\overline{z}}{2-3\overline{z}+4\overline{z}^2}$.
Cross-multiplying,we get $z(2-3\overline{z}+4\overline{z}^2) = \overline{z}(2-3z+4z^2)$.
$2z - 3z\overline{z} + 4z\overline{z}^2 = 2\overline{z} - 3z\overline{z} + 4z^2\overline{z}$.
$2(z-\overline{z}) = 4z\overline{z}(z-\overline{z})$.
Since $z \neq \overline{z}$,we can divide by $(z-\overline{z})$:
$2 = 4z\overline{z} \implies z\overline{z} = \frac{2}{4} = 0.5$.
Thus,$|z|^2 = 0.50$.

(C) Given,$\bar{z}-z^2=i(\bar{z}+z^2)$.
Rearranging the terms,we get $(1-i)\bar{z}=(1+i)z^2$.
Thus,$\bar{z} = \frac{1+i}{1-i}z^2 = \frac{(1+i)^2}{1^2+1^2}z^2 = \frac{2i}{2}z^2 = iz^2$.
Let $z = x+iy$,then $\bar{z} = x-iy$.
Substituting into $\bar{z} = iz^2$,we have $x-iy = i(x+iy)^2 = i(x^2-y^2+2ixy) = -2xy + i(x^2-y^2)$.
Equating real and imaginary parts:
$x = -2xy \Rightarrow x(1+2y) = 0$.
$-y = x^2-y^2 \Rightarrow x^2 = y^2-y$.
Case $I$: If $x=0$,then $y^2-y=0$,so $y=0$ or $y=1$. This gives roots $z=0$ and $z=i$.
Case $II$: If $y=-\frac{1}{2}$,then $x^2 = (-\frac{1}{2})^2 - (-\frac{1}{2}) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$,so $x = \pm \frac{\sqrt{3}}{2}$. This gives roots $z = \frac{\sqrt{3}}{2} - \frac{1}{2}i$ and $z = -\frac{\sqrt{3}}{2} - \frac{1}{2}i$.
There are $4$ distinct roots.

(A) Let $z = r(\cos \theta + i \sin \theta)$,so $\bar{z} = r(\cos \theta - i \sin \theta)$.
Then $(\bar{z})^2 + \frac{1}{z^2} = r^2(\cos 2\theta - i \sin 2\theta) + \frac{1}{r^2}(\cos 2\theta - i \sin 2\theta) = (r^2 + \frac{1}{r^2}) \cos 2\theta - i (r^2 + \frac{1}{r^2}) \sin 2\theta$.
Let $(r^2 + \frac{1}{r^2}) \cos 2\theta = m$ and $(r^2 + \frac{1}{r^2}) \sin 2\theta = -n$,where $m, n \in \mathbb{Z}$.
Squaring and adding,we get $(r^2 + \frac{1}{r^2})^2 = m^2 + n^2$.
Expanding,$r^4 + \frac{1}{r^4} + 2 = m^2 + n^2$.
For option $(A)$,$|z|^4 = \frac{43+3 \sqrt{205}}{2}$.
Then $r^4 + \frac{1}{r^4} + 2 = \frac{43+3 \sqrt{205}}{2} + \frac{2}{43+3 \sqrt{205}} + 2 = \frac{43+3 \sqrt{205}}{2} + \frac{43-3 \sqrt{205}}{22} + 2$ (Correction: $r^4 + \frac{1}{r^4} = \frac{43+3 \sqrt{205}}{2} + \frac{43-3 \sqrt{205}}{2} = 43$).
Thus,$r^4 + \frac{1}{r^4} + 2 = 43 + 2 = 45$,which is $m^2 + n^2 = 45$. This is possible for integers $m, n$ (e.g.,$6^2 + 3^2 = 45$).
Thus,option $(A)$ is correct.

(B) Given $f(x) = x^4 + ax^3 + bx^2 + c$ and $f(1) = -9$,we have $1 + a + b + c = -9$,so $a + b + c = -10$ $(1)$.
The equation $4x^3 + 3ax^2 + 2bx = 0$ has $i\sqrt{3}$ as a root. Since the coefficients are real,$-i\sqrt{3}$ is also a root. The third root must be $0$ because the constant term is $0$.
Thus,the roots are $0, i\sqrt{3}, -i\sqrt{3}$.
The sum of roots taken two at a time is $\frac{2b}{4} = (i\sqrt{3})(-i\sqrt{3}) + 0 + 0 = 3$,so $b = 6$.
The sum of roots is $-\frac{3a}{4} = 0 + i\sqrt{3} - i\sqrt{3} = 0$,so $a = 0$.
Substituting $a = 0$ and $b = 6$ into $(1)$,we get $0 + 6 + c = -10$,so $c = -16$.
Thus,$f(x) = x^4 + 6x^2 - 16 = 0$.
Factoring,$(x^2 + 8)(x^2 - 2) = 0$,so the roots are $x^2 = -8$ and $x^2 = 2$.
The roots are $\alpha_1 = i\sqrt{8}, \alpha_2 = -i\sqrt{8}, \alpha_3 = \sqrt{2}, \alpha_4 = -\sqrt{2}$.
Calculating the sum of squares of magnitudes: $|\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 + |\alpha_4|^2 = 8 + 8 + 2 + 2 = 20$.

(D) Given $z_1 = e^{-i\pi/4}, z_2 = e^{i0} = 1, z_3 = e^{i\pi/4}$.
Since $|z|=1$,$\bar{z} = 1/z$.
$z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1 = e^{-i\pi/4}(1) + 1(e^{-i\pi/4}) + e^{i\pi/4}(e^{i\pi/4}) = 2e^{-i\pi/4} + e^{i\pi/2}$.
$2e^{-i\pi/4} + e^{i\pi/2} = 2(\cos(\pi/4) - i\sin(\pi/4)) + i = 2(\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}) + i = \sqrt{2} - i\sqrt{2} + i = \sqrt{2} + i(1 - \sqrt{2})$.
The squared modulus is $|\sqrt{2} + i(1 - \sqrt{2})|^2 = (\sqrt{2})^2 + (1 - \sqrt{2})^2 = 2 + (1 - 2\sqrt{2} + 2) = 2 + 3 - 2\sqrt{2} = 5 - 2\sqrt{2}$.
Comparing with $\alpha + \beta \sqrt{2}$,we get $\alpha = 5$ and $\beta = -2$.
Thus,$\alpha^2 + \beta^2 = 5^2 + (-2)^2 = 25 + 4 = 29$.

(D) Given the equation $2z^2 - 3z - 2i = 0$.
Dividing by $z$,we get $2z - 3 - \frac{2i}{z} = 0$,which implies $2(z - \frac{i}{z}) = 3$,or $z - \frac{i}{z} = \frac{3}{2}$.
Since $\alpha$ and $\beta$ are roots,$\alpha - \frac{i}{\alpha} = \frac{3}{2}$ and $\beta - \frac{i}{\beta} = \frac{3}{2}$.
Squaring both sides: $(\alpha - \frac{i}{\alpha})^2 = \alpha^2 + \frac{i^2}{\alpha^2} - 2i = \alpha^2 - \frac{1}{\alpha^2} - 2i = \frac{9}{4}$.
Thus,$\alpha^2 - \frac{1}{\alpha^2} = \frac{9}{4} + 2i$. Similarly,$\beta^2 - \frac{1}{\beta^2} = \frac{9}{4} + 2i$.
Consider the expression $E = \frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} = \frac{\alpha^{15}(\alpha^4 + \alpha^{-4}) + \beta^{15}(\beta^4 + \beta^{-4})}{\alpha^{15} + \beta^{15}}$.
Note that $(\alpha^2 - \alpha^{-2})^2 = \alpha^4 + \alpha^{-4} - 2 = (\frac{9}{4} + 2i)^2 = \frac{81}{16} - 4 + 9i = \frac{17}{16} + 9i$.
So,$\alpha^4 + \alpha^{-4} = \frac{17}{16} + 9i + 2 = \frac{49}{16} + 9i$.
Substituting this into $E$: $E = \frac{(\alpha^{15} + \beta^{15})(\frac{49}{16} + 9i)}{\alpha^{15} + \beta^{15}} = \frac{49}{16} + 9i$.
Then $\operatorname{Re}(E) = \frac{49}{16}$ and $\operatorname{Im}(E) = 9$.
The required value is $16 \cdot \operatorname{Re}(E) \cdot \operatorname{Im}(E) = 16 \cdot \frac{49}{16} \cdot 9 = 49 \cdot 9 = 441$.

(C) The value of $i^k$ for any natural number $k$ can only be one of the four values: $\{i, -1, -i, 1\}$.
Since $k_1$ and $k_2$ are chosen randomly,there are $4 \times 4 = 16$ possible pairs of values for $(i^{k_1}, i^{k_2})$.
We want to find the probability that $i^{k_1} + i^{k_2} \neq 0$,which is equivalent to $1 - P(i^{k_1} + i^{k_2} = 0)$.
The condition $i^{k_1} + i^{k_2} = 0$ implies $i^{k_1} = -i^{k_2}$.
The possible pairs $(i^{k_1}, i^{k_2})$ that satisfy this are: $(i, -i), (-i, i), (1, -1), (-1, 1)$.
There are $4$ such unfavorable cases.
Thus,the number of favorable cases is $16 - 4 = 12$.
The probability is $\frac{12}{16} = \frac{3}{4}$.

(B) Given the quadratic equation $x^2-(3-2 i) x-(2 i-2)=0$. \\
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: \\
$x = \frac{(3-2 i) \pm \sqrt{(3-2 i)^2 - 4(1)(-(2 i-2))}}{2}$ \\
$x = \frac{(3-2 i) \pm \sqrt{9 - 4 - 12i + 8i - 8}}{2}$ \\
$x = \frac{(3-2 i) \pm \sqrt{-3 - 4i}}{2}$ \\
Since $-3-4i = 1^2 + (2i)^2 - 2(1)(2i) = (1-2i)^2$,we have: \\
$x = \frac{(3-2 i) \pm (1-2 i)}{2}$ \\
Case $1$: $x = \frac{3-2i + 1-2i}{2} = \frac{4-4i}{2} = 2-2i$. Here $\alpha=2, \beta=-2$. \\
Case $2$: $x = \frac{3-2i - (1-2i)}{2} = \frac{2}{2} = 1+0i$. Here $\gamma=1, \delta=0$. \\
Thus,$\alpha \gamma + \beta \delta = (2)(1) + (-2)(0) = 2+0 = 2$.

(A) Given $\frac{2+k^2z}{k+\overline{z}}=kz$. Since $|z|=1$,we have $\overline{z} = \frac{1}{z}$.
Substituting this,we get $\frac{2+k^2z}{k+\frac{1}{z}} = kz$.
$\frac{2+k^2z}{\frac{kz+1}{z}} = kz \implies \frac{z(2+k^2z)}{kz+1} = kz$.
Assuming $z \neq 0$,we have $2+k^2z = k(kz+1) = k^2z+k$.
This simplifies to $2=k$,so $k=2$.
The point is $k+ik^2 = 2+i(2)^2 = 2+4i$.
The circle is $|z-(1+2i)|=1$,which has center $C(1,2)$ and radius $r=1$.
The distance from the point $P(2,4)$ to the center $C(1,2)$ is $d = \sqrt{(2-1)^2+(4-2)^2} = \sqrt{1^2+2^2} = \sqrt{5}$.
The maximum distance from the point to the circle is $d+r = \sqrt{5}+1$.

(B) Given the equation: $\frac{z^2+3i}{z-2+i} = 2+3i$
Multiplying both sides by $(z-2+i)$,we get: $z^2+3i = (2+3i)(z-2+i)$
$z^2+3i = 2z - 4 + 2i + 3iz - 6i + 3i^2$
Since $i^2 = -1$,$z^2+3i = 2z - 4 + 2i + 3iz - 6i - 3$
$z^2+3i = z(2+3i) - 7 - 4i$
$z^2 - z(2+3i) + 7 + 7i = 0$
Let $z_1$ and $z_2$ be the roots of this quadratic equation. By Vieta's formulas,$z_1+z_2 = 2+3i$ and $z_1z_2 = 7+7i$.
We need to find the sum of all possible values of $z^2$,which is $z_1^2 + z_2^2$.
$z_1^2 + z_2^2 = (z_1+z_2)^2 - 2z_1z_2$
$z_1^2 + z_2^2 = (2+3i)^2 - 2(7+7i)$
$z_1^2 + z_2^2 = (4 + 9i^2 + 12i) - (14 + 14i)$
$z_1^2 + z_2^2 = (4 - 9 + 12i) - 14 - 14i$
$z_1^2 + z_2^2 = -5 + 12i - 14 - 14i$
$z_1^2 + z_2^2 = -19 - 2i$

(B) Given $\omega_1 = (8 \sin \theta + 7 \cos \theta) + i(\sin \theta + 4 \cos \theta)$ and $\omega_2 = (\sin \theta + 4 \cos \theta) + i(8 \sin \theta + 7 \cos \theta)$.
Let $x = 8 \sin \theta + 7 \cos \theta$ and $y = \sin \theta + 4 \cos \theta$.
Then $\omega_1 = x + iy$ and $\omega_2 = y + ix$.
The product $\omega_1 \omega_2 = (x + iy)(y + ix) = xy + ix^2 + iy^2 + i^2yx = xy + i(x^2 + y^2) - xy = i(x^2 + y^2)$.
Thus,$\alpha = 0$ and $\beta = x^2 + y^2$.
$\beta = (8 \sin \theta + 7 \cos \theta)^2 + (\sin \theta + 4 \cos \theta)^2 = (64 \sin^2 \theta + 49 \cos^2 \theta + 112 \sin \theta \cos \theta) + (\sin^2 \theta + 16 \cos^2 \theta + 8 \sin \theta \cos \theta) = 65 \sin^2 \theta + 65 \cos^2 \theta + 120 \sin \theta \cos \theta = 65 + 60 \sin(2 \theta)$.
Since $\alpha + \beta = 0 + 65 + 60 \sin(2 \theta) = 65 + 60 \sin(2 \theta)$,the maximum value $p = 65 + 60 = 125$ and the minimum value $q = 65 - 60 = 5$.
Therefore,$p + q = 125 + 5 = 130$.

(A) Given $\alpha$ is a root of $x^2+x+1=0$,so $\alpha = \omega$ or $\alpha = \omega^2$,where $\omega$ is a complex cube root of unity.
Since $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$,the expression $\left(\alpha^k+\frac{1}{\alpha^k}\right)^2$ simplifies to $\left(\omega^k+\omega^{2k}\right)^2 = \omega^{2k} + \omega^{4k} + 2\omega^{3k} = \omega^{2k} + \omega^k + 2$.
We need to find $n$ such that $\sum_{k=1}^n (\omega^{2k} + \omega^k + 2) = 20$.
This is $\sum_{k=1}^n \omega^{2k} + \sum_{k=1}^n \omega^k + 2n = 20$.
If $n$ is a multiple of $3$,say $n=3m$,then $\sum_{k=1}^n \omega^{2k} = 0$ and $\sum_{k=1}^n \omega^k = 0$,so $2n = 20 \Rightarrow n = 10$ (not a multiple of $3$).
If $n = 3m+1$,then $\sum_{k=1}^n \omega^{2k} = \omega^2$ and $\sum_{k=1}^n \omega^k = \omega$,so $\omega^2 + \omega + 2n = 20$ $\Rightarrow -1 + 2n = 20$ $\Rightarrow 2n = 21$ (no integer solution).
If $n = 3m+2$,then $\sum_{k=1}^n \omega^{2k} = \omega^2 + \omega^4 = \omega^2 + \omega = -1$ and $\sum_{k=1}^n \omega^k = \omega + \omega^2 = -1$,so $-1 - 1 + 2n = 20$ $\Rightarrow 2n = 22$ $\Rightarrow n = 11$.
Since $11 = 3(3) + 2$,this satisfies the condition.

(C) For $(S1)$: Let $z = x+iy$. Since $|z|=1$,$x^2+y^2=1$. The condition that $\frac{z-i}{z+i}$ is purely real means $\frac{z-i}{z+i} = \overline{\left(\frac{z-i}{z+i}\right)} = \frac{\bar{z}+i}{\bar{z}-i}$.
$(z-i)(\bar{z}-i) = (z+i)(\bar{z}+i) \Rightarrow |z|^2 - i(z+\bar{z}) - 1 = |z|^2 + i(z+\bar{z}) - 1$.
This simplifies to $2i(z+\bar{z}) = 0$,so $z+\bar{z} = 2x = 0$,meaning $x=0$. Since $|z|=1$,$y^2=1$,so $z = \pm i$. However,$z \neq -i$,so $z=i$ is the only solution. Thus,$(S1)$ is incorrect.
For $(S2)$: Let $w = \frac{z-1}{z+1}$. The condition that $w$ is purely imaginary means $w + \bar{w} = 0$.
$\frac{z-1}{z+1} + \frac{\bar{z}-1}{\bar{z}+1} = 0 \Rightarrow (z-1)(\bar{z}+1) + (z+1)(\bar{z}-1) = 0$.
$|z|^2 + z - \bar{z} - 1 + |z|^2 - z + \bar{z} - 1 = 0 \Rightarrow 2|z|^2 - 2 = 0 \Rightarrow |z|^2 = 1$.
Since $|z|=1$ is given,every $z$ on the unit circle (except $z=-1$) satisfies the condition. Thus,$(S2)$ is correct.

(A) Given the equation $1 + 10 \operatorname{Re}\left(\frac{2 \cos \theta + i \sin \theta}{\cos \theta - 3i \sin \theta}\right) = 0$.
Multiply the numerator and denominator of the fraction by the conjugate of the denominator $(\cos \theta + 3i \sin \theta)$:
$\frac{(2 \cos \theta + i \sin \theta)(\cos \theta + 3i \sin \theta)}{\cos^2 \theta + 9 \sin^2 \theta} = \frac{2 \cos^2 \theta + 6i \cos \theta \sin \theta + i \sin \theta \cos \theta - 3 \sin^2 \theta}{\cos^2 \theta + 9 \sin^2 \theta} = \frac{(2 \cos^2 \theta - 3 \sin^2 \theta) + i(7 \sin \theta \cos \theta)}{\cos^2 \theta + 9 \sin^2 \theta}$.
The real part is $\frac{2 \cos^2 \theta - 3 \sin^2 \theta}{\cos^2 \theta + 9 \sin^2 \theta}$.
Substituting this into the original equation:
$1 + 10 \left(\frac{2 \cos^2 \theta - 3 \sin^2 \theta}{\cos^2 \theta + 9 \sin^2 \theta}\right) = 0$.
$\frac{\cos^2 \theta + 9 \sin^2 \theta + 20 \cos^2 \theta - 30 \sin^2 \theta}{\cos^2 \theta + 9 \sin^2 \theta} = 0$.
$21 \cos^2 \theta - 21 \sin^2 \theta = 0 \implies 21 \cos(2\theta) = 0$.
Thus, $\cos(2\theta) = 0$, which implies $2\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$.
So, $\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.
Then $\sum \theta^2 = \left(\frac{\pi}{4}\right)^2 + \left(\frac{3\pi}{4}\right)^2 + \left(\frac{5\pi}{4}\right)^2 + \left(\frac{7\pi}{4}\right)^2 = \frac{\pi^2}{16}(1 + 9 + 25 + 49) = \frac{84 \pi^2}{16} = \frac{21 \pi^2}{4}$.

(B) Given expression: $(1+i)^5(1-i)^7$
$= (1+i)^5(1-i)^5(1-i)^2$
$= [(1+i)(1-i)]^5(1-2i+i^2)$
$= (1-i^2)^5(1-2i-1)$
$= (1-(-1))^5(-2i)$
$= (2)^5(-2i)$
$= 32(-2i) = -64i$

(D) Given $\frac{2z-n}{2z+n} = 2i-1$.
Let $2z = x+iy$. Since $\operatorname{Im}(z)=10$, we have $\operatorname{Im}(2z) = 20$. So $2z = x+20i$.
Substituting this into the equation: $\frac{x+20i-n}{x+20i+n} = 2i-1$.
$(x-n)+20i = (2i-1)(x+n+20i)$.
$(x-n)+20i = 2i(x+n) + 40i^2 - (x+n) - 20i$.
$(x-n)+20i = 2i(x+n) - 40 - (x+n) - 20i$.
$(x-n)+20i = -(x+n+40) + i(2x+2n-20)$.
Equating real and imaginary parts:
Real part: $x-n = -(x+n+40) \implies x-n = -x-n-40 \implies 2x = -40 \implies x = -20$.
Since $2z = x+20i$, we have $2z = -20+20i$, so $z = -10+10i$. Thus $\operatorname{Re}(z) = -10$.
Imaginary part: $20 = 2x+2n-20$.
Substituting $x=-20$: $20 = 2(-20)+2n-20 \implies 20 = -40+2n-20 \implies 20 = 2n-60 \implies 2n = 80 \implies n = 40$.
Therefore, $n=40$ and $\operatorname{Re}(z)=-10$.

(A) Given $(x+iy)^{1/3} = a+ib$.
Cubing both sides,we get $x+iy = (a+ib)^3$.
Using the identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,we have $x+iy = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3$.
$x+iy = a^3 + 3a^2bi - 3ab^2 - ib^3$.
Grouping real and imaginary parts: $x+iy = (a^3 - 3ab^2) + i(3a^2b - b^3)$.
Equating real and imaginary parts: $x = a^3 - 3ab^2$ and $y = 3a^2b - b^3$.
Dividing by $a$ and $b$ respectively: $\frac{x}{a} = a^2 - 3b^2$ and $\frac{y}{b} = 3a^2 - b^2$.
Therefore,$\frac{x}{a} - \frac{y}{b} = (a^2 - 3b^2) - (3a^2 - b^2) = a^2 - 3b^2 - 3a^2 + b^2 = -2a^2 - 2b^2 = -2(a^2+b^2)$.

(A) Given $x = -2 - \sqrt{3} i$.
$x + 2 = -\sqrt{3} i$.
Squaring both sides: $(x + 2)^2 = (-\sqrt{3} i)^2$.
$x^2 + 4x + 4 = 3i^2$.
Since $i^2 = -1$,we have $x^2 + 4x + 4 = -3$.
$x^2 + 4x + 7 = 0$.
Now,divide $2x^4 + 5x^3 + 7x^2 - x + 41$ by $x^2 + 4x + 7$:
$2x^4 + 5x^3 + 7x^2 - x + 41 = (x^2 + 4x + 7)(2x^2 - 3x + 5) + 6$.
Substituting $x^2 + 4x + 7 = 0$:
$0 \times (2x^2 - 3x + 5) + 6 = 6$.

(B) Given $x = 1 + 2i$,then $x - 1 = 2i$.
Squaring both sides,we get $(x - 1)^2 = (2i)^2$,which implies $x^2 - 2x + 1 = -4$,or $x^2 - 2x + 5 = 0$.
Now,we perform polynomial division of $x^3 + 7x^2 - x + 16$ by $x^2 - 2x + 5$:
$x^3 + 7x^2 - x + 16 = (x^2 - 2x + 5)(x + 9) + (12x - 29)$.
Since $x^2 - 2x + 5 = 0$,the expression simplifies to $12x - 29$.
Substituting $x = 1 + 2i$:
$12(1 + 2i) - 29 = 12 + 24i - 29 = -17 + 24i$.

(D) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugate to each other if $z_1 = \overline{z_2}$.
Given $z_1 = \sin x + i \cos 2x$ and $z_2 = \cos x - i \sin 2x$.
The conjugate of $z_2$ is $\overline{z_2} = \cos x + i \sin 2x$.
For $z_1 = \overline{z_2}$,we must have $\sin x + i \cos 2x = \cos x + i \sin 2x$.
Equating real and imaginary parts:
$1$) $\sin x = \cos x \implies \tan x = 1 \implies x = n\pi + \frac{\pi}{4}$.
$2$) $\cos 2x = \sin 2x \implies \tan 2x = 1 \implies 2x = m\pi + \frac{\pi}{4} \implies x = \frac{m\pi}{2} + \frac{\pi}{8}$.
Comparing the two conditions for $x$,there is no common value of $x$ that satisfies both equations simultaneously.
Therefore,there is no value of $x$ for which the given complex numbers are conjugates.

(B) For a complex number $z$ to be purely imaginary,its real part must be zero,i.e.,$\text{Re}(z) = 0$.
Given $z = \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta}$.
Multiply the numerator and denominator by the conjugate of the denominator $(1 + 2i \sin \theta)$:
$z = \frac{(3 + 2i \sin \theta)(1 + 2i \sin \theta)}{(1 - 2i \sin \theta)(1 + 2i \sin \theta)}$
$z = \frac{3 + 6i \sin \theta + 2i \sin \theta + 4i^2 \sin^2 \theta}{1^2 + (2 \sin \theta)^2}$
Since $i^2 = -1$,we have:
$z = \frac{3 + 8i \sin \theta - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta}$
$z = \frac{3 - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta} + i \frac{8 \sin \theta}{1 + 4 \sin^2 \theta}$
For $z$ to be purely imaginary,$\text{Re}(z) = 0$:
$\frac{3 - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta} = 0$
$3 - 4 \sin^2 \theta = 0$
$\sin^2 \theta = \frac{3}{4} = \left(\frac{\sqrt{3}}{2}\right)^2$
$\sin^2 \theta = \sin^2 \left(\frac{\pi}{3}\right)$
Therefore,$\theta = n \pi \pm \frac{\pi}{3}$,where $n \in Z$.

(B) $z^{1/3} = p+iq$
$\Rightarrow z = (p+iq)^3$
$\Rightarrow x+iy = p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3$
$\Rightarrow x+iy = (p^3 - 3pq^2) + i(3p^2q - q^3)$
$\Rightarrow x = p^3 - 3pq^2$ and $y = 3p^2q - q^3$
$\Rightarrow \frac{x}{p} = p^2 - 3q^2$ and $\frac{y}{q} = 3p^2 - q^2$
$\therefore \left(\frac{x}{p} + \frac{y}{q}\right) = (p^2 - 3q^2) + (3p^2 - q^2) = 4p^2 - 4q^2 = 4(p^2 - q^2)$

(C) Given $x = -2 + i\sqrt{3}$,we have $x + 2 = i\sqrt{3}$.
Squaring both sides,we get $(x + 2)^2 = -3$,which simplifies to $x^2 + 4x + 4 = -3$,or $x^2 + 4x + 7 = 0$.
Now,we divide the polynomial $P(x) = 2x^4 + 5x^3 + 7x^2 - x + 38$ by $x^2 + 4x + 7$ using polynomial long division:
$2x^4 + 5x^3 + 7x^2 - x + 38 = (x^2 + 4x + 7)(2x^2 - 3x + 5) + 3$.
Since $x^2 + 4x + 7 = 0$,the expression becomes:
$P(x) = 0 \times (2x^2 - 3x + 5) + 3 = 3$.
Thus,the value is $3$.

(A) Given the expression $(1-\cos \theta+i \sin \theta)^{-1} = \frac{1}{1-\cos \theta+i \sin \theta}$.
Using trigonometric identities $1-\cos \theta = 2 \sin^2(\theta/2)$ and $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$:
$\frac{1}{2 \sin^2(\theta/2) + i(2 \sin(\theta/2) \cos(\theta/2))} = \frac{1}{2 \sin(\theta/2) [\sin(\theta/2) + i \cos(\theta/2)]}$.
Multiplying the numerator and denominator by the conjugate $[\sin(\theta/2) - i \cos(\theta/2)]$:
$= \frac{\sin(\theta/2) - i \cos(\theta/2)}{2 \sin(\theta/2) [\sin^2(\theta/2) + \cos^2(\theta/2)]} = \frac{\sin(\theta/2) - i \cos(\theta/2)}{2 \sin(\theta/2)}$.
$= \frac{\sin(\theta/2)}{2 \sin(\theta/2)} - i \frac{\cos(\theta/2)}{2 \sin(\theta/2)} = \frac{1}{2} - i \frac{1}{2} \cot(\theta/2)$.
The real part is $1/2$.

(D) The triangle inequality for complex numbers states that for any two complex numbers $Z_1$ and $Z_2$,the modulus of their sum is less than or equal to the sum of their moduli,i.e.,$\left|Z_1+Z_2\right| \leq \left|Z_1\right|+\left|Z_2\right|$.
Therefore,the statement $\left|Z_1+Z_2\right| \geq \left|Z_1\right|+\left|Z_2\right|$ is generally false.

(D) Given $x = \frac{4 + 5i}{2}$ $\Rightarrow 2x = 4 + 5i$ $\Rightarrow 2x - 4 = 5i$.
Squaring both sides: $(2x - 4)^2 = (5i)^2$ $\Rightarrow 4x^2 - 16x + 16 = -25$ $\Rightarrow 4x^2 - 16x + 41 = 0$.
Now,we divide the polynomial $4x^3 - 4x^2 - 7x + 127$ by $4x^2 - 16x + 41$:
$4x^3 - 4x^2 - 7x + 127 = x(4x^2 - 16x + 41) + 12x^2 - 48x + 127$
$= x(0) + 3(4x^2 - 16x) + 127$
$= 3(-41) + 127$
$= -123 + 127 = 4$.

(D) Given that $f(z) = z^4 + z^3 + 2z^2 + 4z - 8$ has a root $2i$. Since the coefficients are real,the complex conjugate $-2i$ must also be a root.
Therefore,$(z - 2i)(z + 2i) = (z^2 + 4)$ is a factor of $f(z)$.
Dividing $f(z)$ by $(z^2 + 4)$,we get:
$f(z) = (z^2 + 4)(z^2 + z - 2)$.
Further factoring the quadratic term:
$z^2 + z - 2 = (z + 2)(z - 1)$.
Thus,the roots of $f(z) = 0$ are $2i, -2i, -2, 1$.
Comparing these with the given options,$2$ is not a root of $f(z) = 0$.

(C) The given equation is $x^4-10x^3+50x^2-130x+169=0$.
Since the coefficients are real,the roots occur in conjugate pairs.
The roots are $a+ib, a-ib, b+ai, b-ai$.
The sum of the roots is $(a+ib) + (a-ib) + (b+ai) + (b-ai) = 2a + 2b = 10$,so $a+b=5$.
The product of the roots is $(a^2+b^2)(b^2+a^2) = (a^2+b^2)^2 = 169$.
Thus,$a^2+b^2 = 13$.
We know that $(a+b)^2 = a^2+b^2+2ab$.
Substituting the values,$5^2 = 13 + 2ab$,which gives $25 = 13 + 2ab$,so $2ab = 12$ or $ab = 6$.
We need to find $\frac{a}{b} + \frac{b}{a} = \frac{a^2+b^2}{ab}$.
Substituting the values,$\frac{13}{6}$.

(A) Given equation is $x^3-3x^2+3x-9=0$.
Factoring the expression,we get $x^2(x-3) + 3(x-3) = 0$.
$(x-3)(x^2+3) = 0$.
Thus,one root is $x = 3$.
For the other roots,we solve $x^2+3=0$,which implies $x^2 = -3$.
Since $\omega^2+\omega+1=0$,we know $\omega^2+\omega = -1$.
Testing $x = 1+2\omega$: $(1+2\omega)^2 + 3 = 1 + 4\omega + 4\omega^2 + 3 = 4(1+\omega+\omega^2) = 0$.
Similarly,$x = 1+2\omega^2$ also satisfies the equation because $(1+2\omega^2)^2 + 3 = 1 + 4\omega^2 + 4\omega^4 + 3 = 1 + 4\omega^2 + 4\omega + 3 = 4(1+\omega+\omega^2) = 0$.
Therefore,the roots are $3, 1+2\omega, 1+2\omega^2$.