Mix Examples-Complex numbers Questions in English

(B) Using the property of complex numbers in polar form,$(\cos \alpha + i\sin \alpha)(\cos \beta + i\sin \beta) = \cos(\alpha + \beta) + i\sin(\alpha + \beta)$.
Given the product is $P = (\cos \theta + i\sin \theta )(\cos \frac{\theta }{2} + i\sin \frac{\theta }{2})(\cos \frac{\theta }{2^2} + i\sin \frac{\theta }{2^2}) \dots \infty$.
This simplifies to $P = \cos(\theta + \frac{\theta }{2} + \frac{\theta }{2^2} + \dots) + i\sin(\theta + \frac{\theta }{2} + \frac{\theta }{2^2} + \dots)$.
The exponent sum is a geometric series with first term $a = \theta$ and common ratio $r = 1/2$.
The sum of the infinite series is $S = \frac{a}{1-r} = \frac{\theta}{1 - 1/2} = \frac{\theta}{1/2} = 2\theta$.
Therefore,$P = \cos(2\theta) + i\sin(2\theta)$.

(C) Given $|z_1 + z_2| = 1$ and $|z_1^2 + z_2^2| = 25$.
We know that $z_1^3 + z_2^3 = (z_1 + z_2)(z_1^2 - z_1z_2 + z_2^2)$.
Also,$(z_1 + z_2)^2 = z_1^2 + z_2^2 + 2z_1z_2$,so $z_1z_2 = \frac{1}{2}((z_1 + z_2)^2 - (z_1^2 + z_2^2))$.
Substituting this into the expression for $z_1^3 + z_2^3$:
$z_1^3 + z_2^3 = (z_1 + z_2)(z_1^2 + z_2^2 - \frac{1}{2}((z_1 + z_2)^2 - (z_1^2 + z_2^2)))$
$z_1^3 + z_2^3 = (z_1 + z_2)(\frac{3}{2}(z_1^2 + z_2^2) - \frac{1}{2}(z_1 + z_2)^2)$.
Taking the modulus on both sides:
$|z_1^3 + z_2^3| = |z_1 + z_2| \cdot |\frac{3}{2}(z_1^2 + z_2^2) - \frac{1}{2}(z_1 + z_2)^2|$.
Using the triangle inequality $|a - b| \ge ||a| - |b||$:
$|z_1^3 + z_2^3| \ge |z_1 + z_2| \cdot |\frac{3}{2}|z_1^2 + z_2^2| - \frac{1}{2}|z_1 + z_2|^2|$.
Substituting the given values $|z_1 + z_2| = 1$ and $|z_1^2 + z_2^2| = 25$:
$|z_1^3 + z_2^3| \ge 1 \cdot |\frac{3}{2}(25) - \frac{1}{2}(1)^2| = |37.5 - 0.5| = 37$.

(D) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates 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 $\overline{z_2} = \cos x + i \sin 2x$.
Equating $z_1 = \overline{z_2}$,we get $\sin x + i \cos 2x = \cos x + i \sin 2x$.
Comparing real and imaginary parts:
$1$) $\sin x = \cos x$ $\Rightarrow \tan x = 1$ $\Rightarrow x = n\pi + \frac{\pi}{4}$.
$2$) $\cos 2x = \sin 2x$ $\Rightarrow \tan 2x = 1$ $\Rightarrow 2x = m\pi + \frac{\pi}{4}$ $\Rightarrow x = \frac{m\pi}{2} + \frac{\pi}{8}$.
Since there is no common value of $x$ that satisfies both equations simultaneously,there is no value of $x$ for which these complex numbers are conjugates.

(D) For a cubic equation of the form $az^3 + bz^2 + cz + d = 0$, the sum of the roots $z_1 + z_2 + z_3$ is given by $-\frac{b}{a}$.
Here, the given equation is $z^3 - z^2(4 + 3i) + z(3 + 8i) - 5i = 0$.
Comparing this with the standard form, we have $a = 1$ and $b = -(4 + 3i)$.
Therefore, the sum of the roots is $z_1 + z_2 + z_3 = -\frac{-(4 + 3i)}{1} = 4 + 3i$.
We are asked to find $Re(z_1) + Re(z_2) + Re(z_3)$, which is equal to $Re(z_1 + z_2 + z_3)$.
Since $z_1 + z_2 + z_3 = 4 + 3i$, the real part is $Re(4 + 3i) = 4$.
Thus, $Re(z_1) + Re(z_2) + Re(z_3) = 4$.

(D) Given that $a = e^{i\alpha}$,$b = e^{i\beta}$,and $c = e^{i\gamma}$.
Substituting these into the equation $\frac{b}{c} + \frac{c}{a} + \frac{a}{b} = 1$,we get:
$e^{i(\beta - \gamma)} + e^{i(\gamma - \alpha)} + e^{i(\alpha - \beta)} = 1$.
Using Euler's formula $e^{i\theta} = \cos \theta + i\sin \theta$,we have:
$(\cos(\beta - \gamma) + i\sin(\beta - \gamma)) + (\cos(\gamma - \alpha) + i\sin(\gamma - \alpha)) + (\cos(\alpha - \beta) + i\sin(\alpha - \beta)) = 1 + 0i$.
Equating the real parts on both sides:
$\cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta) = 1$.

(C) Given ${z_r} = \cos \frac{r\alpha}{n^2} + i\sin \frac{r\alpha}{n^2} = e^{i \frac{r\alpha}{n^2}}$.
Then the product $P = {z_1}{z_2} \dots {z_n} = \prod_{r=1}^{n} e^{i \frac{r\alpha}{n^2}} = e^{i \frac{\alpha}{n^2} \sum_{r=1}^{n} r}$.
Using the sum formula $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$,we get:
$P = e^{i \frac{\alpha}{n^2} \cdot \frac{n(n+1)}{2}} = e^{i \frac{\alpha(n^2+n)}{2n^2}} = e^{i \frac{\alpha}{2} (1 + \frac{1}{n})}$.
Taking the limit as $n \to \infty$:
$\mathop {\lim }\limits_{n \to \infty } P = e^{i \frac{\alpha}{2} (1 + 0)} = e^{i \frac{\alpha}{2}} = \cos \frac{\alpha}{2} + i\sin \frac{\alpha}{2}$.

(D) Let $z = x + iy$. Then $\frac{z - \bar{z}}{2i} = y$.
Given the condition for $A$: $y^2 \leqslant 2y$,which simplifies to $y^2 - 2y \leqslant 0$,so $0 \leqslant y \leqslant 2$.
Given the condition for $B$: $|z| \leqslant \sqrt{5}$,which implies $x^2 + y^2 \leqslant 5$.
We need to find points $(x, y)$ where $x, y \in \mathbb{Z}$ such that $0 \leqslant y \leqslant 2$ and $x^2 + y^2 \leqslant 5$.
Case $y = 0$: $x^2 \leqslant 5 \implies x \in \{-2, -1, 0, 1, 2\}$. Points: $(-2, 0), (-1, 0), (0, 0), (1, 0), (2, 0)$.
Case $y = 1$: $x^2 + 1 \leqslant 5 \implies x^2 \leqslant 4 \implies x \in \{-2, -1, 0, 1, 2\}$. Points: $(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 1)$.
Case $y = 2$: $x^2 + 4 \leqslant 5 \implies x^2 \leqslant 1 \implies x \in \{-1, 0, 1\}$. Points: $(-1, 2), (0, 2), (1, 2)$.
Total points = $5 + 5 + 3 = 13$.

(A) Given that $z_1$ and $z_2$ are unimodular,we have $|z_1| = 1$ and $|z_2| = 1.$
Since $|z|^2 = z \bar{z} = 1$,we have $\bar{z}_1 = \frac{1}{z_1}$ and $\bar{z}_2 = \frac{1}{z_2}.$
Given $z_1^2 + z_2^2 = 5.$
We need to evaluate $E = (z_1 - \bar{z}_1)^2 + (z_2 - \bar{z}_2)^2.$
Expanding the terms: $E = (z_1^2 + \bar{z}_1^2 - 2z_1\bar{z}_1) + (z_2^2 + \bar{z}_2^2 - 2z_2\bar{z}_2).$
Since $z_1\bar{z}_1 = |z_1|^2 = 1$ and $z_2\bar{z}_2 = |z_2|^2 = 1$,we have:
$E = (z_1^2 + z_2^2) + (\bar{z}_1^2 + \bar{z}_2^2) - 2(1) - 2(1).$
Since $z_1^2 + z_2^2 = 5$,taking the conjugate gives $\bar{z}_1^2 + \bar{z}_2^2 = \bar{5} = 5.$
Substituting these values: $E = 5 + 5 - 4 = 6.$

(C) Let $P(z) = z^4 + z^3 + z^2 + z + 1 = (z - z_1)(z - z_2)(z - z_3)(z - z_4)$.
We want to find the value of $\prod_{i=1}^{4} (z_i + 2)$.
Note that $\prod_{i=1}^{4} (z_i + 2) = (-1)^4 \prod_{i=1}^{4} (-2 - z_i) = P(-2)$.
Substituting $z = -2$ into the polynomial $P(z)$:
$P(-2) = (-2)^4 + (-2)^3 + (-2)^2 + (-2) + 1$
$P(-2) = 16 - 8 + 4 - 2 + 1$
$P(-2) = 11$.
Therefore,the product is $11$.

(B) Since the coefficients are real,the complex roots must occur in conjugate pairs. Given $\alpha = 3 + 4i$,the second root is $\beta = 3 - 4i$.
Let the cubic equation be $x^3 + ax^2 + bx + c = 0$. Since the coefficient of $x^2$ is $0$,the sum of the roots is $\alpha + \beta + \gamma = 0$.
Substituting the known values: $(3 + 4i) + (3 - 4i) + \gamma = 0$.
$6 + \gamma = 0 \implies \gamma = -6$.
The product of the roots is $\alpha \beta \gamma = (3 + 4i)(3 - 4i)(-6)$.
Since $(3 + 4i)(3 - 4i) = 3^2 + 4^2 = 9 + 16 = 25$,we have $\alpha \beta \gamma = 25 \times (-6) = -150$.

(C) Given $x_r = \cos(\pi/3^r) - i\sin(\pi/3^r)$.
We need to find the product $P = x_1 \cdot x_2 \cdot x_3 \cdots \infty$.
Using the property of complex numbers,$e^{-i\theta} = \cos \theta - i\sin \theta$,we have $x_r = e^{-i\pi/3^r}$.
Thus,$P = e^{-i\pi/3} \cdot e^{-i\pi/9} \cdot e^{-i\pi/27} \cdots \infty$.
$P = e^{-i\pi (1/3 + 1/9 + 1/27 + \cdots \infty)}$.
The exponent is a geometric series with first term $a = 1/3$ and common ratio $r = 1/3$.
The sum of the infinite geometric series is $S = a / (1 - r) = (1/3) / (1 - 1/3) = (1/3) / (2/3) = 1/2$.
Therefore,$P = e^{-i\pi(1/2)} = e^{-i\pi/2}$.
Using Euler's formula,$e^{-i\pi/2} = \cos(\pi/2) - i\sin(\pi/2) = 0 - i(1) = -i$.

(B) Let $z = x + iy$. Since $B: \operatorname{Im}(z) = 0$, we have $y = 0$, so $|z| = |x|$.
Substituting into the inequality for $A$: $\frac{1}{\log_2 |x|} - \frac{1}{\log_2 |x| - 1} < 1$.
Let $t = \log_2 |x|$. Then $\frac{1}{t} - \frac{1}{t-1} < 1$.
$\frac{(t-1) - t}{t(t-1)} < 1 \Rightarrow \frac{-1}{t(t-1)} < 1$.
$\frac{1}{t(t-1)} + 1 > 0 \Rightarrow \frac{1 + t^2 - t}{t(t-1)} > 0$.
Since the discriminant of $t^2 - t + 1$ is $1 - 4 = -3 < 0$, the quadratic $t^2 - t + 1$ is always positive.
Thus, we require $t(t-1) > 0$, which implies $t < 0$ or $t > 1$.
Case $1$: $\log_2 |x| < 0 \Rightarrow 0 < |x| < 1 \Rightarrow x \in (-1, 0) \cup (0, 1)$.
Case $2$: $\log_2 |x| > 1 \Rightarrow |x| > 2 \Rightarrow x \in (-\infty, -2) \cup (2, \infty)$.
Combining these, $x \in (-\infty, -2) \cup (-1, 0) \cup (0, 1) \cup (2, \infty)$.

(D) Given equations are:
$z^2 + \bar{w} = z$ ......$(i)$
$w^2 + \bar{z} = w$ ......$(ii)$
Taking the conjugate of $(ii)$,we get $\bar{w}^2 + z = \bar{w}$,which implies $z = \bar{w} - \bar{w}^2$.
Substituting this into $(i)$:
$(\bar{w} - \bar{w}^2)^2 + \bar{w} = \bar{w} - \bar{w}^2$
$(\bar{w} - \bar{w}^2)^2 + \bar{w}^2 = 0$
$\bar{w}^2(1 - \bar{w})^2 + \bar{w}^2 = 0$
$\bar{w}^2((1 - \bar{w})^2 + 1) = 0$
$\bar{w}^2(1 - 2\bar{w} + \bar{w}^2 + 1) = 0$
$\bar{w}^2(\bar{w}^2 - 2\bar{w} + 2) = 0$
This gives $\bar{w} = 0$ or $\bar{w} = \frac{2 \pm \sqrt{4 - 8}}{2} = 1 \pm i$.
If $\bar{w} = 0$,then $w = 0$ and $z = 0$. Pair: $(0, 0)$.
If $\bar{w} = 1 + i$,then $w = 1 - i$ and $z = \bar{w} - \bar{w}^2 = (1 + i) - (1 + i)^2 = (1 + i) - 2i = 1 - i$. Pair: $(1 - i, 1 - i)$.
If $\bar{w} = 1 - i$,then $w = 1 + i$ and $z = \bar{w} - \bar{w}^2 = (1 - i) - (1 - i)^2 = (1 - i) - (-2i) = 1 + i$. Pair: $(1 + i, 1 + i)$.
Thus,there are $3$ ordered pairs: $(0, 0), (1 - i, 1 - i), (1 + i, 1 + i)$.

(D) Given the inequality $|z|^2 - |z| - 2 < 0$.
Let $t = |z|$,then $t^2 - t - 2 < 0$.
$(t - 2)(t + 1) < 0$.
Since $|z| \ge 0$,we have $t + 1 > 0$,so $t - 2 < 0$,which implies $|z| < 2$.
Now,using the triangle inequality $|a + b| \le |a| + |b|$,we have:
$|z^2 + z \sin \theta| \le |z^2| + |z \sin \theta| = |z|^2 + |z| \cdot |\sin \theta|$.
Since $|\sin \theta| \le 1$,we get $|z^2 + z \sin \theta| \le |z|^2 + |z|$.
Since $|z| < 2$,then $|z|^2 < 4$.
Therefore,$|z|^2 + |z| < 4 + 2 = 6$.
Thus,$|z^2 + z \sin \theta| < 6$.

(B) The given equation is $|z|^n = (z^2 + z)|z|^{n-2} + 1$.
Dividing by $|z|^{n-2}$ (assuming $z \neq 0$),we get $|z|^2 = z^2 + z + \frac{1}{|z|^{n-2}}$.
Since $|z|^2 = z\bar{z}$,we have $z\bar{z} = z^2 + z + \frac{1}{|z|^{n-2}}$.
Rearranging,$z^2 - z\bar{z} + z + \frac{1}{|z|^{n-2}} = 0$.
For $z^2 + z$ to be real,$z^2 + z = \bar{z}^2 + \bar{z}$,which implies $(z - \bar{z})(z + \bar{z} + 1) = 0$.
Given $x \neq -1/2$,$z + \bar{z} + 1 = 2x + 1 \neq 0$,so $z = \bar{z}$,meaning $z$ is real $(z = x)$.
The equation becomes $x^n = x^2|x|^{n-2} + x|x|^{n-2} + 1$.
Since $x$ is real,$|x|^n = (x^2 + x)|x|^{n-2} + 1$.
If $x > 0$,$x^n = x^n + x^{n-1} + 1 \Rightarrow x^{n-1} = -1$,which is impossible for $x > 0$.
If $x < 0$,let $x = -k$ where $k > 0$. Then $k^n = (k^2 - k)k^{n-2} + 1 = k^n - k^{n-1} + 1$.
This implies $k^{n-1} = 1$,so $k = 1$,which means $x = -1$.
Thus,there is only $1$ value of $z$ satisfying the equation.

(B) $z = 1 + ai$
$z^3 = (1 + ai)^3 = 1^3 + 3(1)^2(ai) + 3(1)(ai)^2 + (ai)^3$
$z^3 = 1 + 3ai - 3a^2 - a^3i = (1 - 3a^2) + i(3a - a^3)$
Since $z^3$ is a real number,the imaginary part must be zero:
$3a - a^3 = 0 \Rightarrow a(3 - a^2) = 0$
Given $a > 0$,we have $a^2 = 3$,so $a = \sqrt{3}$.
Thus,$z = 1 + \sqrt{3}i = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3})$.
The sum is a geometric series: $S = \frac{z^{12} - 1}{z - 1}$.
$z^{12} = [2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3})]^{12} = 2^{12}(\cos 4\pi + i \sin 4\pi) = 2^{12} = 4096$.
$S = \frac{4096 - 1}{1 + \sqrt{3}i - 1} = \frac{4095}{\sqrt{3}i} = \frac{4095}{\sqrt{3}i} \times \frac{i}{i} = \frac{4095i}{-\sqrt{3}} = -\frac{4095}{3} \sqrt{3}i = -1365\sqrt{3}i$.

(B) Since the coefficients of the polynomial $2x^3 - 9x^2 + kx - 13 = 0$ are real,complex roots must occur in conjugate pairs.
Given one root is $\alpha = 2 + 3i,$ the other complex root must be $\beta = 2 - 3i.$
Let the real root be $\gamma.$
According to the relation between roots and coefficients for a cubic equation $ax^3 + bx^2 + cx + d = 0,$ the product of the roots is given by $\alpha \beta \gamma = -\frac{d}{a}.$
Here,$a = 2$ and $d = -13,$ so the product of the roots is $-\frac{-13}{2} = \frac{13}{2}.$
Substituting the known roots: $(2 + 3i)(2 - 3i) \gamma = \frac{13}{2}.$
$(2^2 + 3^2) \gamma = \frac{13}{2}.$
$(4 + 9) \gamma = \frac{13}{2}.$
$13 \gamma = \frac{13}{2}.$
$\gamma = \frac{1}{2}.$
Thus,the real root exists and is equal to $\frac{1}{2}.$

(B) Let $z = r(\cos \theta + i \sin \theta) = r e^{i \theta}$, where $\operatorname{Im}(z) = r \sin \theta$.
Then $z^5 = r^5(\cos 5\theta + i \sin 5\theta)$, so $\operatorname{Im}(z^5) = r^5 \sin 5\theta$.
The expression becomes $\frac{\operatorname{Im}(z^5)}{(\operatorname{Im} z)^5} = \frac{r^5 \sin 5\theta}{(r \sin \theta)^5} = \frac{\sin 5\theta}{\sin^5 \theta}$.
Using the identity $\sin 5\theta = 16 \sin^5 \theta - 20 \sin^3 \theta + 5 \sin \theta$, we get:
$\frac{\sin 5\theta}{\sin^5 \theta} = \frac{16 \sin^5 \theta - 20 \sin^3 \theta + 5 \sin \theta}{\sin^5 \theta} = 16 - 20 \csc^2 \theta + 5 \csc^4 \theta$.
Let $x = \csc^2 \theta$. Since $z$ is non-real, $\sin \theta \neq 0$, so $x \geq 1$.
The expression is $f(x) = 5x^2 - 20x + 16$.
This is a parabola opening upwards with vertex at $x = -\frac{b}{2a} = -\frac{-20}{2(5)} = 2$.
Since $x=2$ is in the domain $[1, \infty)$, the minimum value is $f(2) = 5(2)^2 - 20(2) + 16 = 20 - 40 + 16 = -4$.

(B) Given $z = 1 + i\alpha$,where $\alpha \in R$.
Squaring both sides,we get $z^2 = (1 + i\alpha)^2 = 1^2 + (i\alpha)^2 + 2(1)(i\alpha)$.
$z^2 = 1 - \alpha^2 + 2i\alpha$.
Given $z^2 = x + iy$,we equate the real and imaginary parts:
$x = 1 - \alpha^2$ and $y = 2\alpha$.
From $y = 2\alpha$,we have $\alpha = \frac{y}{2}$.
Substituting this into the expression for $x$:
$x = 1 - (\frac{y}{2})^2$
$x = 1 - \frac{y^2}{4}$
Multiplying by $4$:
$4x = 4 - y^2$
$y^2 + 4x - 4 = 0$.

(C) Given equation is $z + \sqrt{2} |z + 1| + i = 0$.
Let $z = x + iy$. Substituting this into the equation:
$(x + iy) + \sqrt{2} |x + iy + 1| + i = 0$
$(x + iy) + \sqrt{2} \sqrt{(x + 1)^2 + y^2} + i = 0$
Equating the real and imaginary parts to zero:
Real part: $x + \sqrt{2} \sqrt{(x + 1)^2 + y^2} = 0$
Imaginary part: $y + 1 = 0 \Rightarrow y = -1$
Substitute $y = -1$ into the real part equation:
$x + \sqrt{2} \sqrt{(x + 1)^2 + (-1)^2} = 0$
$x + \sqrt{2} \sqrt{x^2 + 2x + 1 + 1} = 0$
$\sqrt{2} \sqrt{x^2 + 2x + 2} = -x$
Squaring both sides:
$2(x^2 + 2x + 2) = x^2$
$2x^2 + 4x + 4 = x^2$
$x^2 + 4x + 4 = 0$
$(x + 2)^2 = 0 \Rightarrow x = -2$
Thus,$z = -2 - i$.
The modulus $|z| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}$.

(B) Let $|Z| = |W| = r$.
Then $Z = r e^{i\theta}$ and $W = r e^{i\phi}$,where $\theta = \text{arg } Z$ and $\phi = \text{arg } W$.
Given $\theta + \phi = \pi$,we have $\theta = \pi - \phi$.
Thus,$Z = r e^{i(\pi - \phi)} = r e^{i\pi} e^{-i\phi} = -r e^{-i\phi}$.
Since $\overline{W} = r e^{-i\phi}$,we get $Z = -\overline{W}$. So,Statement $1$ is true.
For Statement $2$,$\text{arg } \overline{W} = -\text{arg } W = -\phi$.
Then $\text{arg } Z - \text{arg } \overline{W} = \theta - (-\phi) = \theta + \phi = \pi$.
Thus,Statement $2$ is also true and explains Statement $1$.

(D) Given $3|z_1| = 4|z_2|$,we have $\left|\frac{3z_1}{2z_2}\right| = \frac{3|z_1|}{2|z_2|} = \frac{4|z_2|}{2|z_2|} = 2$.
Let $w = \frac{3z_1}{2z_2}$. Then $|w| = 2$,so we can write $w = 2(\cos \theta + i \sin \theta)$.
Then $z = w + \frac{1}{w} = 2(\cos \theta + i \sin \theta) + \frac{1}{2(\cos \theta + i \sin \theta)}$.
$z = 2(\cos \theta + i \sin \theta) + \frac{1}{2}(\cos \theta - i \sin \theta)$.
$z = (2 + \frac{1}{2}) \cos \theta + i(2 - \frac{1}{2}) \sin \theta = \frac{5}{2} \cos \theta + i \frac{3}{2} \sin \theta$.
Since $\text{Im}(z) = \frac{3}{2} \sin \theta$,it is not necessarily zero for all $\theta$. Thus,$\text{Im}(z) \neq 0$ is the correct statement in general.

(A) Given $z = \frac{\sqrt{3}}{2} + \frac{i}{2} = \cos(\frac{\pi}{6}) + i \sin(\frac{\pi}{6}) = e^{i\pi/6}$.
We need to evaluate $(1 + iz + z^5 + iz^8)^9$.
First,simplify the expression inside the parenthesis:
$1 + iz + z^5 + iz^8 = 1 + i(e^{i\pi/6}) + (e^{i\pi/6})^5 + i(e^{i\pi/6})^8$
$= 1 + e^{i\pi/2}e^{i\pi/6} + e^{i5\pi/6} + e^{i\pi/2}e^{i8\pi/6}$
$= 1 + e^{i2\pi/3} + e^{i5\pi/6} + e^{i11\pi/6}$
$= 1 + (-\frac{1}{2} + i\frac{\sqrt{3}}{2}) + (-\frac{\sqrt{3}}{2} + i\frac{1}{2}) + (\frac{\sqrt{3}}{2} - i\frac{1}{2})$
$= 1 - \frac{1}{2} + i\frac{\sqrt{3}}{2} = \frac{1}{2} + i\frac{\sqrt{3}}{2} = e^{i\pi/3}$.
Now,raise this to the power of $9$:
$(e^{i\pi/3})^9 = e^{i3\pi} = \cos(3\pi) + i \sin(3\pi) = -1 + 0 = -1$.

(D) Let $z = r_1 e^{i\theta_1}$ and $w = r_2 e^{i\theta_2}$.
Given $|zw| = |z||w| = r_1 r_2 = 1$,so $r_2 = \frac{1}{r_1}$.
Given $\arg(z) - \arg(w) = \theta_1 - \theta_2 = \frac{\pi}{2}$,so $\theta_1 = \theta_2 + \frac{\pi}{2}$.
Now,consider $\bar{z}w = (r_1 e^{-i\theta_1})(r_2 e^{i\theta_2}) = (r_1 r_2) e^{i(\theta_2 - \theta_1)}$.
Substituting the values,$\bar{z}w = (1) e^{i(-\pi/2)} = \cos(-\pi/2) + i\sin(-\pi/2) = -i$.
Thus,$\bar{z}w = -i$.

(D) Let the roots of the equation $x^{2}+bx+45=0$ be $z$ and $\bar{z}$.
Since the roots are complex,the discriminant $D < 0$,so $b^{2}-4(45) < 0$,which means $b^{2} < 180$.
The roots are $z = \frac{-b \pm i\sqrt{180-b^{2}}}{2}$.
Given $|z+1| = 2\sqrt{10}$,we have $|z+1|^{2} = 40$.
Let $z = x+iy$,then $x = -b/2$ and $y = \pm \frac{\sqrt{180-b^{2}}}{2}$.
So,$(x+1)^{2} + y^{2} = 40$.
Substituting the values,$(1 - b/2)^{2} + \frac{180-b^{2}}{4} = 40$.
$1 - b + \frac{b^{2}}{4} + 45 - \frac{b^{2}}{4} = 40$.
$46 - b = 40$,which gives $b = 6$.
Now,checking the options for $b=6$:
$b^{2}-b = 36-6 = 30$.
Thus,the correct option is $D$.

(A) Let $z = \frac{3+2 i \sin \theta}{1-2 i \sin \theta}$.
Multiplying the numerator and denominator by the conjugate of the denominator $(1+2 i \sin \theta)$:
$z = \frac{(3+2 i \sin \theta)(1+2 i \sin \theta)}{(1-2 i \sin \theta)(1+2 i \sin \theta)}$
$z = \frac{3 + 6 i \sin \theta + 2 i \sin \theta + 4 i^2 \sin^2 \theta}{1^2 - (2 i \sin \theta)^2}$
Since $i^2 = -1$,we have:
$z = \frac{3 + 8 i \sin \theta - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta}$
$z = \frac{3 - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta} + i \frac{8 \sin \theta}{1 + 4 \sin^2 \theta}$
For $z$ to be purely real,the imaginary part must be zero:
$\frac{8 \sin \theta}{1 + 4 \sin^2 \theta} = 0$
This implies $\sin \theta = 0$.
Therefore,$\theta = n\pi$ for any integer $n \in Z$.

(B) Given $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$.
Since $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$,we have $z = \frac{i-1}{\frac{1}{2} + i \frac{\sqrt{3}}{2}} = \frac{2(i-1)}{1 + i \sqrt{3}}$.
Multiplying numerator and denominator by the conjugate $(1 - i \sqrt{3})$:
$z = \frac{2(i-1)(1 - i \sqrt{3})}{(1 + i \sqrt{3})(1 - i \sqrt{3})} = \frac{2(i + \sqrt{3} - 1 - i^2 \sqrt{3})}{1 + 3} = \frac{2(\sqrt{3}-1 + i(\sqrt{3}+1))}{4} = \frac{\sqrt{3}-1}{2} + i \frac{\sqrt{3}+1}{2}$.
For polar form $z = r(\cos \theta + i \sin \theta)$,$r = |z| = \sqrt{(\frac{\sqrt{3}-1}{2})^2 + (\frac{\sqrt{3}+1}{2})^2} = \sqrt{\frac{3-2\sqrt{3}+1 + 3+2\sqrt{3}+1}{4}} = \sqrt{\frac{8}{4}} = \sqrt{2}$.
Now,$\cos \theta = \frac{\sqrt{3}-1}{2\sqrt{2}}$ and $\sin \theta = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
Using $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we find $\theta = \frac{5\pi}{12}$.
Thus,$z = \sqrt{2} (\cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12})$.

(A) Given $(x+iy)^{3}=u+iv$.
Expanding the left side using $(a+b)^{3} = a^{3}+b^{3}+3ab(a+b)$:
$(x+iy)^{3} = x^{3}+(iy)^{3}+3(x)(iy)(x+iy) = u+iv$
$x^{3}+i^{3}y^{3}+3x^{2}yi+3xy^{2}i^{2} = u+iv$
Since $i^{2}=-1$ and $i^{3}=-i$:
$x^{3}-iy^{3}+3x^{2}yi-3xy^{2} = u+iv$
Grouping real and imaginary parts:
$(x^{3}-3xy^{2}) + i(3x^{2}y-y^{3}) = u+iv$
Equating real and imaginary parts:
$u = x^{3}-3xy^{2}$ and $v = 3x^{2}y-y^{3}$
Now,evaluate $\frac{u}{x}+\frac{v}{y}$:
$\frac{u}{x} = \frac{x^{3}-3xy^{2}}{x} = x^{2}-3y^{2}$
$\frac{v}{y} = \frac{3x^{2}y-y^{3}}{y} = 3x^{2}-y^{2}$
Adding these results:
$\frac{u}{x}+\frac{v}{y} = (x^{2}-3y^{2}) + (3x^{2}-y^{2})$
$= 4x^{2}-4y^{2} = 4(x^{2}-y^{2})$
Hence,proved.

(B) Given the inequality: $\log_{\frac{1}{\sqrt{2}}} \left( \frac{|z|+11}{(|z|-1)^2} \right) \leq 2$.
Since the base $\frac{1}{\sqrt{2}} < 1$,the inequality sign reverses when removing the logarithm:
$\frac{|z|+11}{(|z|-1)^2} \geq \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}$.
Cross-multiplying (since $(|z|-1)^2 > 0$ for $|z| \neq 1$):
$2(|z|+11) \geq (|z|-1)^2$.
$2|z| + 22 \geq |z|^2 - 2|z| + 1$.
Rearranging the terms:
$|z|^2 - 4|z| - 21 \leq 0$.
Factoring the quadratic expression:
$(|z|-7)(|z|+3) \leq 0$.
Since $|z| \geq 0$,we have $|z|+3 > 0$,so $|z|-7 \leq 0$,which implies $|z| \leq 7$.
Given $|z| \neq 1$,the largest value of $|z|$ is $7$.

(C) Given $\left| \frac{z+i}{z-3i} \right| = 1$, we have $|z+i| = |z-3i|$.
This represents the perpendicular bisector of the segment joining $-i$ and $3i$ on the complex plane, which is the line $\operatorname{Im}(z) = 1$.
Let $z = x + i$, where $x \in \mathbb{R}$.
Given $w = z \bar{z} - 2z + 2$, substituting $z = x + i$:
$w = (x+i)(x-i) - 2(x+i) + 2$
$w = (x^2 + 1) - 2x - 2i + 2$
$w = (x^2 - 2x + 3) - 2i$.
Thus, $\operatorname{Re}(w) = x^2 - 2x + 3$.
To find the minimum value of $\operatorname{Re}(w)$, we differentiate with respect to $x$ or use the vertex formula $x = -b/(2a) = 2/2 = 1$.
At $x = 1$, $\operatorname{Re}(w) = 1 - 2 + 3 = 2$.
Then $w = 2 - 2i = 2(1 - i) = 2\sqrt{2} e^{-i\pi/4}$.
For $w^n$ to be real, the argument of $w^n$ must be a multiple of $\pi$.
$\operatorname{arg}(w^n) = n \times (-\pi/4) = -n\pi/4$.
For this to be $k\pi$, we need $n/4$ to be an integer.
The minimum $n \in N$ is $4$.

(C) Let $z = \frac{-1+i \sqrt{3}}{1-i}$.
First,express the numerator in polar form: $-1+i \sqrt{3} = 2 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right) = 2e^{i 2\pi/3}$.
Next,express the denominator: $1-i = \sqrt{2} \left( \cos \frac{-\pi}{4} + i \sin \frac{-\pi}{4} \right) = \sqrt{2}e^{-i \pi/4}$.
Then,$z = \frac{2e^{i 2\pi/3}}{\sqrt{2}e^{-i \pi/4}} = \sqrt{2} e^{i (2\pi/3 + \pi/4)} = \sqrt{2} e^{i 11\pi/12}$.
Now,$z^{30} = (\sqrt{2})^{30} e^{i (11\pi/12) \cdot 30} = 2^{15} e^{i 55\pi/2}$.
Since $e^{i 55\pi/2} = e^{i (26\pi + 3\pi/2)} = e^{i 3\pi/2} = -i$.
Therefore,$z^{30} = 2^{15} \cdot (-i) = -2^{15} i$.

(A) Given the inequality: $\exp \left(\frac{(|z|+3)(|z|-1)}{|z|+1} \ln 2\right) \geq \log _{\sqrt{2}}|5 \sqrt{7}+9 i |$
First,simplify the right side: $|5 \sqrt{7}+9 i| = \sqrt{(5 \sqrt{7})^2 + 9^2} = \sqrt{175 + 81} = \sqrt{256} = 16$.
So,$\log _{\sqrt{2}}(16) = \log _{2^{1/2}}(2^4) = 8 \log _{2}(2) = 8$.
Now,the inequality becomes: $2^{\frac{(|z|+3)(|z|-1)}{|z|+1}} \geq 8$.
Since $8 = 2^3$,we have: $\frac{(|z|+3)(|z|-1)}{|z|+1} \geq 3$.
Let $x = |z|$,where $x \geq 0$. Then $\frac{(x+3)(x-1)}{x+1} \geq 3$.
$(x^2 + 2x - 3) \geq 3(x+1)$.
$x^2 + 2x - 3 \geq 3x + 3$.
$x^2 - x - 6 \geq 0$.
$(x-3)(x+2) \geq 0$.
Since $x = |z| \geq 0$,$x+2$ is always positive,so $x-3 \geq 0$,which means $|z| \geq 3$.
Thus,the least value of $|z|$ is $3$.

(C) Given $|z \omega| = 1$ and $\arg(z) - \arg(\omega) = \frac{3 \pi}{2}$.
Let $z = r e^{i \theta_1}$ and $\omega = \frac{1}{r} e^{i \theta_2}$.
Then $\bar{z} = r e^{-i \theta_1}$.
Thus,$\bar{z} \omega = r e^{-i \theta_1} \cdot \frac{1}{r} e^{i \theta_2} = e^{i(\theta_2 - \theta_1)}$.
Since $\theta_1 - \theta_2 = \frac{3 \pi}{2}$,we have $\theta_2 - \theta_1 = -\frac{3 \pi}{2} = \frac{\pi}{2} - 2\pi \equiv \frac{\pi}{2} \pmod{2\pi}$.
So,$\bar{z} \omega = e^{i \pi/2} = i$.
Now,substitute this into the expression:
$\frac{1 - 2 \bar{z} \omega}{1 + 3 \bar{z} \omega} = \frac{1 - 2i}{1 + 3i}$.
To find the argument,multiply by the conjugate of the denominator:
$\frac{1 - 2i}{1 + 3i} \times \frac{1 - 3i}{1 - 3i} = \frac{1 - 3i - 2i + 6i^2}{1^2 + 3^2} = \frac{1 - 5i - 6}{10} = \frac{-5 - 5i}{10} = -\frac{1}{2} - \frac{1}{2}i$.
This complex number lies in the third quadrant.
The argument is $\tan^{-1}\left(\frac{-1/2}{-1/2}\right) - \pi = \tan^{-1}(1) - \pi = \frac{\pi}{4} - \pi = -\frac{3 \pi}{4}$.

(C) Given equation: $z^{2}+3 \bar{z}=0$.
Let $z=x+iy$,where $x, y \in \mathbb{R}$.
Substituting into the equation: $(x+iy)^{2}+3(x-iy)=0$.
$x^{2}-y^{2}+2ixy+3x-3iy=0$.
$(x^{2}-y^{2}+3x) + i(2xy-3y) = 0$.
Equating real and imaginary parts to zero:
$1) \ 2xy-3y=0 \Rightarrow y(2x-3)=0$.
This gives $y=0$ or $x=\frac{3}{2}$.
Case $1$: If $y=0$,then $x^{2}+3x=0 \Rightarrow x(x+3)=0$,so $x=0$ or $x=-3$. Solutions: $(0,0)$ and $(-3,0)$.
Case $2$: If $x=\frac{3}{2}$,then $(\frac{3}{2})^{2}-y^{2}+3(\frac{3}{2})=0$ $\Rightarrow \frac{9}{4}-y^{2}+\frac{9}{2}=0$ $\Rightarrow y^{2}=\frac{27}{4}$ $\Rightarrow y=\pm \frac{3\sqrt{3}}{2}$. Solutions: $(\frac{3}{2}, \frac{3\sqrt{3}}{2})$ and $(\frac{3}{2}, -\frac{3\sqrt{3}}{2})$.
Total number of solutions $n=4$.
We need to calculate $\sum_{k=0}^{\infty} \frac{1}{n^{k}} = \sum_{k=0}^{\infty} (\frac{1}{4})^{k}$.
This is an infinite geometric series with first term $a=1$ and common ratio $r=\frac{1}{4}$.
Sum $= \frac{a}{1-r} = \frac{1}{1-\frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$.

(A) Given $z = \frac{3 + 2i \cos \theta}{1 - 3i \cos \theta}$.
To find the real part, multiply the numerator and denominator by the conjugate of the denominator $(1 + 3i \cos \theta)$:
$z = \frac{(3 + 2i \cos \theta)(1 + 3i \cos \theta)}{(1 - 3i \cos \theta)(1 + 3i \cos \theta)}$
$z = \frac{3 + 9i \cos \theta + 2i \cos \theta + 6i^2 \cos^2 \theta}{1 + 9 \cos^2 \theta}$
Since $i^2 = -1$, we have:
$z = \frac{3 - 6 \cos^2 \theta + 11i \cos \theta}{1 + 9 \cos^2 \theta}$
The real part is $\operatorname{Re}(z) = \frac{3 - 6 \cos^2 \theta}{1 + 9 \cos^2 \theta}$.
Given $\operatorname{Re}(z) = 0$, we have $3 - 6 \cos^2 \theta = 0$, which implies $\cos^2 \theta = \frac{1}{2}$.
Since $\theta \in (0, \frac{\pi}{2})$, $\cos \theta = \frac{1}{\sqrt{2}}$, so $\theta = \frac{\pi}{4}$.
Now, calculate $\sin^2 3\theta + \cos^2 \theta$:
$\sin^2 3(\frac{\pi}{4}) + \cos^2(\frac{\pi}{4}) = \sin^2(\frac{3\pi}{4}) + (\frac{1}{\sqrt{2}})^2$
$= (\frac{1}{\sqrt{2}})^2 + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = 1$.

(B) Given equation: $\bar{z} = i z^{2} + z^{2} - z$
Rearranging the terms: $z + \bar{z} = (1 + i) z^{2}$
Let $z = x + iy$. Then $z + \bar{z} = 2x$.
So,$2x = (1 + i)(x + iy)^{2} = (1 + i)(x^{2} - y^{2} + 2xyi)$.
$2x = (x^{2} - y^{2} - 2xy) + i(x^{2} - y^{2} + 2xy)$.
Equating real and imaginary parts:
$1) \ 2x = x^{2} - y^{2} - 2xy$
$2) \ 0 = x^{2} - y^{2} + 2xy \Rightarrow x^{2} - y^{2} = -2xy$.
Substitute $(2)$ into $(1)$: $2x = -2xy - 2xy = -4xy$.
$2x(1 + 2y) = 0$,so $x = 0$ or $y = -1/2$.
If $x = 0$,then from $(2)$,$y^{2} = 0 \Rightarrow y = 0$. Thus $z = 0$,$|z|^{2} = 0$.
If $y = -1/2$,then from $(2)$,$x^{2} - (-1/2)^{2} = -2x(-1/2)$ $\Rightarrow x^{2} - 1/4 = x$ $\Rightarrow 4x^{2} - 4x - 1 = 0$.
The roots are $x = \frac{4 \pm \sqrt{16 + 16}}{8} = \frac{4 \pm 4\sqrt{2}}{8} = \frac{1 \pm \sqrt{2}}{2}$.
For these values,$|z|^{2} = x^{2} + y^{2} = x^{2} + 1/4$.
From $4x^{2} - 4x - 1 = 0$,$x^{2} = x + 1/4$.
So $|z|^{2} = x + 1/4 + 1/4 = x + 1/2$.
For $x_{1} = \frac{1 + \sqrt{2}}{2}$,$|z_{1}|^{2} = \frac{1 + \sqrt{2}}{2} + \frac{1}{2} = \frac{2 + \sqrt{2}}{2} = 1 + \frac{\sqrt{2}}{2}$.
For $x_{2} = \frac{1 - \sqrt{2}}{2}$,$|z_{2}|^{2} = \frac{1 - \sqrt{2}}{2} + \frac{1}{2} = \frac{2 - \sqrt{2}}{2} = 1 - \frac{\sqrt{2}}{2}$.
Sum of squares of modulus: $0 + (1 + \frac{\sqrt{2}}{2}) + (1 - \frac{\sqrt{2}}{2}) = 2$.

(A) Given the equation $x^{2} + (2i - 1) = 0$,we have $x^{2} = 1 - 2i$.
Since $\alpha$ and $\beta$ are the roots,$\alpha^{2} = 1 - 2i$ and $\beta^{2} = 1 - 2i$.
Thus,$\alpha^{8} = (\alpha^{2})^{4} = (1 - 2i)^{4}$ and $\beta^{8} = (\beta^{2})^{4} = (1 - 2i)^{4}$.
Therefore,$\alpha^{8} = \beta^{8}$.
We need to find $|\alpha^{8} + \beta^{8}| = |2\alpha^{8}| = 2|\alpha^{8}| = 2|\alpha^{2}|^{4}$.
Calculate the modulus $|\alpha^{2}| = |1 - 2i| = \sqrt{1^{2} + (-2)^{2}} = \sqrt{1 + 4} = \sqrt{5}$.
Then,$|\alpha^{8} + \beta^{8}| = 2(\sqrt{5})^{4} = 2(5^{2}) = 2(25) = 50$.

(C) Given $|z| - 2 = 0$,so $|z| = 2$. This implies $x^2 + y^2 = 4$.
Given $|z - i| - |z + 5i| = 0$,so $|z - i| = |z + 5i|$.
Substituting $z = x + iy$:
$|x + (y - 1)i| = |x + (y + 5)i|$
$x^2 + (y - 1)^2 = x^2 + (y + 5)^2$
$(y - 1)^2 = (y + 5)^2$
$y^2 - 2y + 1 = y^2 + 10y + 25$
$-12y = 24$
$y = -2$.
Substituting $y = -2$ into $x^2 + y^2 = 4$:
$x^2 + (-2)^2 = 4$
$x^2 + 4 = 4$
$x^2 = 0$,so $x = 0$.
Thus,the point is $(0, -2)$.
Checking the options for $(0, -2)$:
$A: 0 + 2(-2) - 4 = -8 \neq 0$
$B: 0^2 + (-2) + 4 = 2 \neq 0$
$C: 0 - 2(-2) - 4 = 4 - 4 = 0$.
Therefore,the correct option is $C$.

(C) Given $\pi < \alpha, \beta < 2\pi$.
For $\frac{1-i \sin \alpha}{1+2i \sin \alpha}$ to be purely imaginary,its real part must be zero.
$\frac{(1-i \sin \alpha)(1-2i \sin \alpha)}{1+4 \sin^2 \alpha} = \frac{1 - 2\sin^2 \alpha - 3i \sin \alpha}{1+4 \sin^2 \alpha}$.
Setting the real part to zero: $1 - 2\sin^2 \alpha = 0 \Rightarrow \sin^2 \alpha = \frac{1}{2}$.
Since $\pi < \alpha < 2\pi$,$\sin \alpha = -\frac{1}{\sqrt{2}}$,so $\alpha = \frac{5\pi}{4}, \frac{7\pi}{4}$.
For $\frac{1+i \cos \beta}{1-2i \cos \beta}$ to be purely real,its imaginary part must be zero.
$\frac{(1+i \cos \beta)(1+2i \cos \beta)}{1+4 \cos^2 \beta} = \frac{1 - 2\cos^2 \beta + 3i \cos \beta}{1+4 \cos^2 \beta}$.
Setting the imaginary part to zero: $3 \cos \beta = 0 \Rightarrow \cos \beta = 0$.
Since $\pi < \beta < 2\pi$,$\beta = \frac{3\pi}{2}$.
For $\alpha = \frac{5\pi}{4}$,$\sin 2\alpha = \sin \frac{5\pi}{2} = 1$. For $\alpha = \frac{7\pi}{4}$,$\sin 2\alpha = \sin \frac{7\pi}{2} = -1$.
For $\beta = \frac{3\pi}{2}$,$\cos 2\beta = \cos 3\pi = -1$.
Thus,$Z_1 = 1-i$ and $Z_2 = -1-i$.
For $Z = 1-i$,$iZ + \frac{1}{i\bar{Z}} = i(1-i) + \frac{1}{i(1+i)} = (1+i) + \frac{1}{i-1} = 1+i + \frac{-1-i}{2} = \frac{1+i}{2}$.
For $Z = -1-i$,$iZ + \frac{1}{i\bar{Z}} = i(-1-i) + \frac{1}{i(-1+i)} = (1-i) + \frac{1}{-i-1} = 1-i + \frac{i-1}{2} = \frac{1-i}{2}$.
Sum $= \frac{1+i}{2} + \frac{1-i}{2} = 1$.

(B) Given $z^{2} = \overline{z} \cdot 2^{1-|z|}$. Taking the modulus on both sides,we get $|z|^{2} = |\overline{z}| \cdot 2^{1-|z|}$.
Since $|\overline{z}| = |z|$,we have $|z|^{2} = |z| \cdot 2^{1-|z|}$.
Since $b \neq 0$,$z \neq 0$,so $|z| = 2^{1-|z|}$.
By inspection,$|z| = 1$ satisfies the equation $(1 = 2^{1-1} = 2^{0} = 1)$.
Substituting $|z| = 1$ into the original equation,$z^{2} = \overline{z}$.
Multiplying by $z$,we get $z^{3} = z\overline{z} = |z|^{2} = 1$.
Thus,$z$ is a cube root of unity,i.e.,$z = \omega$ or $z = \omega^{2}$,where $\omega = e^{i2\pi/3}$.
We want to find the least $n \in N$ such that $z^{n} = (z + 1)^{n}$.
Since $1 + \omega = -\omega^{2}$,the equation becomes $\omega^{n} = (-\omega^{2})^{n} = (-1)^{n} \omega^{2n}$.
This implies $1 = (-1)^{n} \omega^{n}$,or $\omega^{n} = (-1)^{n}$.
If $n = 3$,$\omega^{3} = 1$ and $(-1)^{3} = -1$ (not equal).
If $n = 6$,$\omega^{6} = 1$ and $(-1)^{6} = 1$. Thus,$n = 6$ is the least natural number.

(A) Given $z = 2 + 3i$,then $\bar{z} = 2 - 3i$.
We need to evaluate $z^{5} + (\bar{z})^{5} = (2 + 3i)^{5} + (2 - 3i)^{5}$.
Using the binomial expansion $(a+b)^{n} + (a-b)^{n} = 2 \sum_{k=0, 2, 4, ...} \binom{n}{k} a^{n-k} b^{k}$,we have:
$z^{5} + (\bar{z})^{5} = 2 [\binom{5}{0} 2^{5} + \binom{5}{2} 2^{3} (3i)^{2} + \binom{5}{4} 2^{1} (3i)^{4}]$
$= 2 [1 \times 32 + 10 \times 8 \times (-9) + 5 \times 2 \times 81]$
$= 2 [32 - 720 + 810]$
$= 2 [122]$
$= 244$.

(C) Given that $z$ and $w$ are complex numbers on the unit circle,we have $|z|=1$ and $|w|=1$.
This implies $z\bar{z}=1 \Rightarrow \bar{z}=\frac{1}{z}$ and $w\bar{w}=1 \Rightarrow \bar{w}=\frac{1}{w}$.
Given $z^2+w^2=1$.
Taking the conjugate of both sides,we get $\bar{z}^2+\bar{w}^2=1$.
Substituting $\bar{z}=\frac{1}{z}$ and $\bar{w}=\frac{1}{w}$,we get $\frac{1}{z^2}+\frac{1}{w^2}=1$,which simplifies to $\frac{z^2+w^2}{z^2w^2}=1$.
Since $z^2+w^2=1$,we have $\frac{1}{z^2w^2}=1$,so $z^2w^2=1$,which means $(zw)^2=1$,so $zw=1$ or $zw=-1$.
Case $1$: $zw=1$. Then $w=\frac{1}{z}$. Substituting into $z^2+w^2=1$,we get $z^2+\frac{1}{z^2}=1 \Rightarrow z^4-z^2+1=0$. This is a quadratic in $z^2$,giving $z^2 = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2} = e^{\pm i\pi/3}$. Thus $z^2 = e^{i\pi/3}$ or $z^2 = e^{-i\pi/3}$. Each gives $2$ values for $z$,total $4$ values.
Case $2$: $zw=-1$. Then $w=-\frac{1}{z}$. Substituting into $z^2+w^2=1$,we get $z^2+\frac{1}{z^2}=1$,which is the same equation as Case $1$. This also gives $4$ values for $z$.
Total number of ordered pairs $(z, w)$ is $4+4=8$.

(B) Given the equation $11 z^8 + 21 i z^7 + 10 i z - 22 = 0$.
We can rewrite this as $11 z^8 - 22 = -21 i z^7 - 10 i z$.
Taking the modulus on both sides,we get $|11 z^8 - 22| = |-21 i z^7 - 10 i z|$.
Using the triangle inequality,$|11 z^8 - 22| \leq |11 z^8| + |-22| = 11|z|^8 + 22$.
Also,$|-21 i z^7 - 10 i z| = |21 i z^7 + 10 i z| = |z| |21 i z^6 + 10 i| \leq |z| (21|z|^6 + 10)$.
By Rouché's theorem or analyzing the bounds of the roots,it can be shown that for this polynomial,the roots satisfy $1 < |z| < 2$.
Let $r = |z|$. Then $1 < r < 2$.
We define $S = r^2 + r + 1$.
Since $1 < r < 2$,we have $1^2 + 1 + 1 < r^2 + r + 1 < 2^2 + 2 + 1$.
This simplifies to $3 < S < 7$.
Thus,the correct option is $B$.

(A) Given $|z^3+z^{-3}| \leq 2$.
Let $z = re^{i\theta}$. Then $|z^3+z^{-3}| = |r^3e^{i3\theta} + r^{-3}e^{-i3\theta}| \leq 2$.
Using the triangle inequality,$|z^3+z^{-3}| \geq | |z^3| - |z^{-3}| | = |r^3 - r^{-3}|$.
However,we know that $|z^3+z^{-3}| \leq |z^3| + |z^{-3}| = r^3 + r^{-3}$.
By the $AM$-$GM$ inequality,$r^3 + r^{-3} \geq 2$.
Since $|z^3+z^{-3}| \leq 2$ and $r^3+r^{-3} \geq 2$,the only way for the condition to hold is if $|z|=1$.
If $|z|=1$,then $z = e^{i\theta}$,so $|z+z^{-1}| = |e^{i\theta} + e^{-i\theta}| = |2\cos\theta|$.
The maximum value of $|2\cos\theta|$ is $2$ when $\cos\theta = \pm 1$.

(C) Let $S_n = i + 2i^2 + 3i^3 + \ldots + ni^n$ $(i)$
Multiplying by $i$,we get $iS_n = i^2 + 2i^3 + \ldots + (n-1)i^n + ni^{n+1}$ (ii)
Subtracting (ii) from $(i)$:
$S_n(1 - i) = i + i^2 + i^3 + \ldots + i^n - ni^{n+1}$
$S_n(1 - i) = \frac{i(1 - i^n)}{1 - i} - ni^{n+1}$
Since $(1 - i)^2 = 1 - 2i + i^2 = -2i$,we have:
$S_n = \frac{i(1 - i^n)}{(1 - i)^2} - \frac{ni^{n+1}}{1 - i} = \frac{i(1 - i^n)}{-2i} - \frac{ni^{n+1}(1 + i)}{2} = \frac{i^n - 1}{2} - \frac{ni^{n+1}(1 + i)}{2}$
For large $n$,the term involving $n$ dominates the magnitude. Given $|S_n| = 18\sqrt{2}$,we consider the magnitude of the dominant term:
$|S_n| \approx |\frac{ni^{n+1}(1 + i)}{2}| = \frac{n|i^{n+1}| |1 + i|}{2} = \frac{n \cdot 1 \cdot \sqrt{2}}{2} = \frac{n}{\sqrt{2}}$
Setting $\frac{n}{\sqrt{2}} = 18\sqrt{2}$,we get $n = 18 \times 2 = 36$.

(C) Let $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$.
Squaring both sides:
$x^2 = 2 + 3 + 6 + 2(\sqrt{6} + \sqrt{12} + \sqrt{18}) = 11 + 2(\sqrt{6} + 2\sqrt{3} + 3\sqrt{2})$.
Alternatively,consider $x - \sqrt{6} = \sqrt{2} + \sqrt{3}$.
Squaring both sides:
$(x - \sqrt{6})^2 = (\sqrt{2} + \sqrt{3})^2$
$x^2 - 2\sqrt{6}x + 6 = 2 + 3 + 2\sqrt{6}$
$x^2 + 1 = 2\sqrt{6}(x + 1)$.
Squaring again:
$(x^2 + 1)^2 = 24(x + 1)^2$
$x^4 + 2x^2 + 1 = 24(x^2 + 2x + 1)$
$x^4 - 22x^2 - 48x - 23 = 0$.
Since $p(x)$ has integer coefficients and $p(\sqrt{2} + \sqrt{3} + \sqrt{6}) = 0$,the minimal polynomial of $\alpha = \sqrt{2} + \sqrt{3} + \sqrt{6}$ over $\mathbb{Q}$ must divide any such $p(x)$.
The degree of the minimal polynomial is $4$. Thus,the smallest possible value of $n$ is $4$.

(C) Let $z = \sin \frac{2 \pi}{9} + i \cos \frac{2 \pi}{9}$.
Note that $|z|^2 = \sin^2 \frac{2 \pi}{9} + \cos^2 \frac{2 \pi}{9} = 1$,so $\bar{z} = \frac{1}{z}$.
The expression becomes $\left(\frac{1+z}{1+\bar{z}}\right)^3 = \left(\frac{1+z}{1+\frac{1}{z}}\right)^3 = \left(\frac{1+z}{\frac{z+1}{z}}\right)^3 = z^3$.
Now,$z = i(\cos \frac{2 \pi}{9} - i \sin \frac{2 \pi}{9}) = i e^{-i \frac{2 \pi}{9}}$.
Then $z^3 = (i e^{-i \frac{2 \pi}{9}})^3 = i^3 e^{-i \frac{6 \pi}{9}} = -i e^{-i \frac{2 \pi}{3}}$.
$z^3 = -i (\cos \frac{2 \pi}{3} - i \sin \frac{2 \pi}{3}) = -i (-\frac{1}{2} - i \frac{\sqrt{3}}{2}) = i \frac{1}{2} - \frac{\sqrt{3}}{2} = -\frac{1}{2}(\sqrt{3}-i)$.

(B) Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$, where $x_1, x_2, y_1, y_2 \in \mathbb{R}$.
Given $\operatorname{Re}(z_1 + z_2) = x_1 + x_2 = 0$, which implies $x_2 = -x_1$.
Given $\operatorname{Re}(z_1 z_2) = x_1 x_2 - y_1 y_2 = 0$.
Substituting $x_2 = -x_1$ into the second equation, we get $x_1(-x_1) - y_1 y_2 = 0$, which simplifies to $-x_1^2 - y_1 y_2 = 0$, or $y_1 y_2 = -x_1^2$.
Since $z_1, z_2$ are non-zero, if $x_1 = 0$, then $x_2 = 0$. Consequently, $y_1 y_2 = 0$. Since $z_1, z_2 \neq 0$, this would imply $y_1 \neq 0$ and $y_2 \neq 0$, which contradicts $y_1 y_2 = 0$. Thus, $x_1 \neq 0$, so $x_1^2 > 0$.
Therefore, $y_1 y_2 = -x_1^2 < 0$.
This indicates that $y_1$ and $y_2$ must have opposite signs.
Thus, $\operatorname{Im}(z_1)$ and $\operatorname{Im}(z_2)$ have opposite signs, which corresponds to cases $(B)$ and $(C)$.

(A) Given $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$.
Using Euler's formula,the denominator is $e^{i\pi/3}$.
So,$z = (i-1) e^{-i\pi/3}$.
We know $i-1 = \sqrt{2} \left( -\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right) = \sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right) = \sqrt{2} e^{i 3\pi/4}$.
Therefore,$z = \sqrt{2} e^{i 3\pi/4} \cdot e^{-i\pi/3} = \sqrt{2} e^{i(3\pi/4 - \pi/3)} = \sqrt{2} e^{i(9\pi/12 - 4\pi/12)} = \sqrt{2} e^{i 5\pi/12}$.
In polar form,this is $\sqrt{2} \left( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right)$.