Mix Examples-Complex numbers Questions in English

(D) Given $x = 3 + i$,we have $x - 3 = i$.
Squaring both sides,we get $(x - 3)^2 = i^2$.
$x^2 - 6x + 9 = -1$,which simplifies to $x^2 - 6x + 10 = 0$.
Now,we divide the polynomial $x^3 - 3x^2 - 8x + 15$ by $x^2 - 6x + 10$ or express it in terms of $(x^2 - 6x + 10)$.
$x^3 - 3x^2 - 8x + 15 = x(x^2 - 6x + 10) + 3(x^2 - 6x + 10) - 15$.
Substituting $x^2 - 6x + 10 = 0$,we get $x(0) + 3(0) - 15 = -15$.

(D) Let ${z_1} = a + ib$ and ${z_2} = c + id$,where $a, b, c, d \in \mathbb{R}$.
Then ${z_1}{z_2} = (a + ib)(c + id) = (ac - bd) + i(ad + bc)$.
The real part of the product is $Re({z_1}{z_2}) = ac - bd$.
Since $a = Re(z_1)$,$c = Re(z_2)$,$b = Im(z_1)$,and $d = Im(z_2)$,we have:
$Re({z_1}{z_2}) = Re(z_1) \cdot Re(z_2) - Im(z_1) \cdot Im(z_2)$.
Comparing this with the given options,none of the options $A, B,$ or $C$ match the correct expression.
Therefore,the correct option is $D$.

(A) Let $z = \frac{\sqrt{5 + 12i} + \sqrt{5 - 12i}}{\sqrt{5 + 12i} - \sqrt{5 - 12i}}$.
Rationalizing the denominator by multiplying the numerator and denominator by $(\sqrt{5 + 12i} + \sqrt{5 - 12i})$:
$z = \frac{(\sqrt{5 + 12i} + \sqrt{5 - 12i})^2}{(\sqrt{5 + 12i})^2 - (\sqrt{5 - 12i})^2}$
$z = \frac{(5 + 12i) + (5 - 12i) + 2\sqrt{(5 + 12i)(5 - 12i)}}{(5 + 12i) - (5 - 12i)}$
$z = \frac{10 + 2\sqrt{25 - 144i^2}}{24i}$
Since $i^2 = -1$,$25 - 144i^2 = 25 + 144 = 169$.
$z = \frac{10 + 2(13)}{24i} = \frac{10 + 26}{24i} = \frac{36}{24i} = \frac{3}{2i} = -\frac{3}{2}i$.

(C) Given that ${a^2} + {b^2} = 1$.
Consider the expression $Z = \frac{{1 + b + ia}}{{1 + b - ia}}$.
Multiply the numerator and denominator by the conjugate of the denominator,$(1 + b + ia)$:
$Z = \frac{{(1 + b + ia)(1 + b + ia)}}{{(1 + b - ia)(1 + b + ia)}}$
$Z = \frac{{(1 + b)^2 + (ia)^2 + 2ia(1 + b)}}{{(1 + b)^2 + a^2}}$
Since ${a^2} + {b^2} = 1$,we have ${a^2} = 1 - {b^2}$.
$Z = \frac{{1 + 2b + {b^2} - {a^2} + 2ia(1 + b)}}{{1 + 2b + {b^2} + {a^2}}}$
Substitute ${a^2} + {b^2} = 1$ in the denominator:
$Z = \frac{{1 + 2b + {b^2} - (1 - {b^2}) + 2ia(1 + b)}}{{1 + 2b + 1}}$
$Z = \frac{{2{b^2} + 2b + 2ia(1 + b)}}{{2 + 2b}}$
$Z = \frac{{2b(b + 1) + 2ia(1 + b)}}{{2(1 + b)}}$
$Z = \frac{{2(1 + b)(b + ia)}}{{2(1 + b)}} = b + ia$
Thus,the correct option is $C$.

(C) complex number is purely imaginary if its real part is $0$.
Multiply the numerator and denominator by the conjugate of the denominator:
$\frac{3 + 2i\sin \theta}{1 - 2i\sin \theta} \times \frac{1 + 2i\sin \theta}{1 + 2i\sin \theta} = \frac{3 + 6i\sin \theta + 2i\sin \theta + 4i^2\sin^2 \theta}{1^2 + (2\sin \theta)^2} = \frac{(3 - 4\sin^2 \theta) + i(8\sin \theta)}{1 + 4\sin^2 \theta}$.
For the expression to be purely imaginary,the real part must be $0$:
$\frac{3 - 4\sin^2 \theta}{1 + 4\sin^2 \theta} = 0$
$3 - 4\sin^2 \theta = 0$
$\sin^2 \theta = \frac{3}{4}$
$\sin \theta = \pm \frac{\sqrt{3}}{2} = \sin \left( \pm \frac{\pi}{3} \right)$
Therefore,$\theta = n\pi \pm \frac{\pi}{3}$.

(B) Given $(x + iy)^{1/3} = a + ib$.
Cubing both sides,we get:
$x + iy = (a + ib)^3$
$x + iy = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3$
$x + iy = a^3 + 3a^2bi - 3ab^2 - ib^3$
$x + iy = (a^3 - 3ab^2) + i(3a^2b - b^3)$
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 $\frac{y}{b} = 3a^2 - b^2$
Adding these two:
$\frac{x}{a} + \frac{y}{b} = (a^2 - 3b^2) + (3a^2 - b^2) = 4a^2 - 4b^2 = 4(a^2 - b^2)$

(A) Let $z = x + iy$,where $x, y \in \mathbb{R}$.
Then $z^2 = (x + iy)^2 = x^2 - y^2 + i(2xy)$.
Given $\text{Re}(z) = 0$,which implies $x = 0$.
Substituting $x = 0$ into the expression for $z^2$,we get $z^2 = -y^2 + i(0) = -y^2$.
Thus,$\text{Im}(z^2) = 0$.
Therefore,$\text{Re}(z) = 0 \Rightarrow \text{Im}(z^2) = 0$ is correct.

(A) Given $z = 3 - 4i$,we have $z - 3 = -4i$.
Squaring both sides,we get $(z - 3)^2 = (-4i)^2$.
$z^2 - 6z + 9 = 16i^2$.
Since $i^2 = -1$,we have $z^2 - 6z + 9 = -16$,which simplifies to $z^2 - 6z + 25 = 0$.
Now,we perform polynomial division of $P(z) = z^4 - 3z^3 + 3z^2 + 99z - 95$ by $z^2 - 6z + 25$.
$z^4 - 3z^3 + 3z^2 + 99z - 95 = (z^2 + 3z - 4)(z^2 - 6z + 25) + 5$.
Substituting $z^2 - 6z + 25 = 0$,we get:
$P(z) = (z^2 + 3z - 4)(0) + 5 = 5$.

(D) Given that $\frac{3x + 2iy}{-2 + 5i} = \frac{15}{8x + 3iy}$.
Cross-multiplying,we get $(3x + 2iy)(8x + 3iy) = 15(-2 + 5i)$.
$24x^2 + 9ixy + 16ixy + 6i^2y^2 = -30 + 75i$.
Since $i^2 = -1$,we have $24x^2 - 6y^2 + 25ixy = -30 + 75i$.
Equating real and imaginary parts:
$24x^2 - 6y^2 = -30 \implies 4x^2 - y^2 = -5$ (Equation $1$).
$25xy = 75 \implies xy = 3$ (Equation $2$).
From Equation $2$,$y = \frac{3}{x}$. Substituting into Equation $1$:
$4x^2 - (\frac{3}{x})^2 = -5 \implies 4x^2 - \frac{9}{x^2} = -5$.
Let $t = x^2$,then $4t - \frac{9}{t} = -5 \implies 4t^2 + 5t - 9 = 0$.
$(4t + 9)(t - 1) = 0$. Since $x$ is real,$x^2 = t = 1$,so $x = \pm 1$.
If $x = 1$,$y = 3$. If $x = -1$,$y = -3$.
Thus,$(x, y) = (1, 3)$ or $(-1, -3)$.

(A) Given $z = \frac{b + ic}{1 + a}$.
Substituting $z$ into the expression $\frac{1 + iz}{1 - iz}$:
$\frac{1 + i(\frac{b + ic}{1 + a})}{1 - i(\frac{b + ic}{1 + a})} = \frac{1 + a + ib - c}{1 + a - ib + c} = \frac{(1 + a - c) + ib}{(1 + a + c) - ib}$.
Multiplying numerator and denominator by the conjugate of the denominator,$(1 + a + c) + ib$:
$= \frac{((1 + a - c) + ib)((1 + a + c) + ib)}{(1 + a + c)^2 + b^2}$.
Expanding the numerator:
$= \frac{(1 + a)^2 - c^2 + ib(1 + a - c) + ib(1 + a + c) - b^2}{(1 + a + c)^2 + b^2} = \frac{1 + 2a + a^2 - c^2 - b^2 + 2ib(1 + a)}{(1 + a + c)^2 + b^2}$.
Since $a^2 + b^2 + c^2 = 1$,we have $a^2 - b^2 - c^2 = 2a^2 - 1$.
Numerator $= 1 + 2a + 2a^2 - 1 + 2ib(1 + a) = 2a(1 + a) + 2ib(1 + a) = 2(1 + a)(a + ib)$.
Denominator $= (1 + a)^2 + c^2 + 2c(1 + a) + b^2 = (1 + a)^2 + (b^2 + c^2) + 2c(1 + a) = (1 + a)^2 + (1 - a^2) + 2c(1 + a) = (1 + a)^2 + (1 - a)(1 + a) + 2c(1 + a) = (1 + a)(1 + a + 1 - a + 2c) = (1 + a)(2 + 2c) = 2(1 + a)(1 + c)$.
Thus,the expression simplifies to $\frac{2(1 + a)(a + ib)}{2(1 + a)(1 + c)} = \frac{a + ib}{1 + c}$.

(A) Given expression: $Z = \frac{(\cos x + i\sin x)(\cos y + i\sin y)}{(\cot u + i)(1 + i\tan v)}$
Using Euler's formula,the numerator is $e^{i(x+y)}$.
For the denominator:
$(\cot u + i) = \frac{\cos u + i\sin u}{\sin u} = \frac{e^{iu}}{\sin u}$
$(1 + i\tan v) = \frac{\cos v + i\sin v}{\cos v} = \frac{e^{iv}}{\cos v}$
Substituting these into the expression:
$Z = \frac{e^{i(x+y)}}{\frac{e^{iu}}{\sin u} \cdot \frac{e^{iv}}{\cos v}} = \sin u \cos v \cdot \frac{e^{i(x+y)}}{e^{i(u+v)}}$
$Z = \sin u \cos v \cdot e^{i(x+y-u-v)}$
$Z = \sin u \cos v [\cos(x+y-u-v) + i\sin(x+y-u-v)]$

(B) Given that $(x + iy)^{1/3} = a - ib$.
By cubing both sides,we get $x + iy = (a - ib)^3$.
Expanding the right side using $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$,we have:
$x + iy = a^3 - 3a^2(ib) + 3a(ib)^2 - (ib)^3$
$x + iy = a^3 - 3ia^2b - 3ab^2 + ib^3$
$x + iy = (a^3 - 3ab^2) + i(b^3 - 3a^2b)$.
Equating real and imaginary parts:
$x = a^3 - 3ab^2$ and $y = b^3 - 3a^2b$.
Now,$\frac{x}{a} = a^2 - 3b^2$ and $\frac{y}{b} = b^2 - 3a^2$.
Calculating $\frac{x}{a} - \frac{y}{b}$:
$\frac{x}{a} - \frac{y}{b} = (a^2 - 3b^2) - (b^2 - 3a^2)$
$\frac{x}{a} - \frac{y}{b} = a^2 - 3b^2 - b^2 + 3a^2 = 4a^2 - 4b^2 = 4(a^2 - b^2)$.
Comparing this with $k(a^2 - b^2)$,we get $k = 4$.

(D) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates if $a = c$ and $b = -d$.
Given $z_1 = \sin x + i\cos 2x$ and $z_2 = \cos x - i\sin 2x$,they are conjugates if:
$1) \sin x = \cos x \implies \tan x = 1 \implies x = n\pi + \frac{\pi}{4}$
$2) \cos 2x = -(-\sin 2x) = \sin 2x \implies \tan 2x = 1 \implies 2x = m\pi + \frac{\pi}{4} \implies x = \frac{m\pi}{2} + \frac{\pi}{8}$
Comparing the two conditions:
For $n=0, x = \frac{\pi}{4} = 0.25\pi$.
For $m=0, x = \frac{\pi}{8} = 0.125\pi$.
For $m=1, x = \frac{\pi}{2} + \frac{\pi}{8} = 0.625\pi$.
There is no integer $n$ and $m$ such that $n\pi + \frac{\pi}{4} = \frac{m\pi}{2} + \frac{\pi}{8}$.
Thus,there is no value of $x$ for which the complex numbers are conjugate.

(D) Let $z = x + iy$,so that $\bar{z} = x - iy$. Therefore,the equation becomes:
$z^2 + \bar{z} = 0 \iff (x + iy)^2 + (x - iy) = 0$
$(x^2 - y^2 + x) + i(2xy - y) = 0$
Equating real and imaginary parts to zero,we get:
$x^2 - y^2 + x = 0$ .....$(i)$
$2xy - y = 0$ .....$(ii)$
From $(ii)$,$y(2x - 1) = 0$,which implies $y = 0$ or $x = \frac{1}{2}$.
Case $1$: If $y = 0$,then $(i)$ gives $x^2 + x = 0$,so $x(x + 1) = 0$,which gives $x = 0$ or $x = -1$. This provides two solutions: $z = 0$ and $z = -1$.
Case $2$: If $x = \frac{1}{2}$,then $(i)$ gives $(\frac{1}{2})^2 - y^2 + \frac{1}{2} = 0$,so $\frac{1}{4} - y^2 + \frac{1}{2} = 0$,which means $y^2 = \frac{3}{4}$,so $y = \pm \frac{\sqrt{3}}{2}$. This provides two solutions: $z = \frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $z = \frac{1}{2} - i\frac{\sqrt{3}}{2}$.
Thus,there are $4$ solutions in total.

(B) Let $\frac{2z_1}{3z_2} = ki$,where $k \in \mathbb{R}$ and $k \neq 0$.
Then $\frac{z_1}{z_2} = \frac{3}{2}ki$.
Let $\frac{z_1}{z_2} = iy$,where $y = \frac{3}{2}k$.
Now,consider the expression $\left| \frac{z_1 - z_2}{z_1 + z_2} \right|$.
Dividing the numerator and denominator by $z_2$,we get $\left| \frac{\frac{z_1}{z_2} - 1}{\frac{z_1}{z_2} + 1} \right|$.
Substituting $\frac{z_1}{z_2} = iy$,we have $\left| \frac{iy - 1}{iy + 1} \right| = \frac{|iy - 1|}{|iy + 1|} = \frac{|-(1 - iy)|}{|1 + iy|} = \frac{|1 - iy|}{|1 + iy|}$.
Since the modulus of a complex number $z$ is equal to the modulus of its conjugate $\overline{z}$,and $\overline{1 + iy} = 1 - iy$,we have $|1 - iy| = |1 + iy|$.
Therefore,$\frac{|1 - iy|}{|1 + iy|} = 1$.

(D) Let the two complex numbers be $z_1$ and $z_2$ such that $|z_1| < 1$ and $|z_2| < 1$.
By the triangle inequality,we have $|z_1 + z_2| \leq |z_1| + |z_2|$.
Since $|z_1| < 1$ and $|z_2| < 1$,the sum $|z_1| + |z_2| < 2$.
This implies $|z_1 + z_2| < 2$.
However,the value of $|z_1 + z_2|$ can be less than,greater than,or equal to $1$.
For example,if $z_1 = 0.1$ and $z_2 = 0.1$,then $|z_1 + z_2| = 0.2 < 1$.
If $z_1 = 0.8$ and $z_2 = 0.8$,then $|z_1 + z_2| = 1.6 > 1$.
If $z_1 = 0.5$ and $z_2 = 0.5$,then $|z_1 + z_2| = 1$.
Thus,the modulus of the sum can be any value in the interval $[0, 2)$.

(B) Let $z = r(\cos \theta + i \sin \theta)$.
Given $\left| z + \frac{1}{z} \right| = 1$,we have $\left| z + \frac{1}{z} \right|^2 = 1$.
Substituting $z = r e^{i\theta}$,we get $\left| r e^{i\theta} + \frac{1}{r} e^{-i\theta} \right|^2 = 1$.
This expands to $\left( r + \frac{1}{r} \right)^2 \cos^2 \theta + \left( r - \frac{1}{r} \right)^2 \sin^2 \theta = 1$.
Simplifying,we get $r^2 + \frac{1}{r^2} + 2 \cos(2\theta) = 1$.
To maximize $r$,we differentiate with respect to $\theta$ and set $\frac{dr}{d\theta} = 0$.
This yields $-4 \sin(2\theta) = 0$,so $\sin(2\theta) = 0$,which means $\theta = 0, \frac{\pi}{2}, \pi, \dots$.
Since $z$ does not lie on the $X$-axis,$\theta \neq 0, \pi$. Thus,$\theta = \frac{\pi}{2}$ or $\frac{3\pi}{2}$.
In these cases,$z$ is purely imaginary,so $\text{Re}(z) = 0$.

(A) Let $z_1 + \sqrt{z_1^2 - z_2^2} = u$ and $z_1 - \sqrt{z_1^2 - z_2^2} = v$.
Then $u + v = 2z_1$ and $uv = z_1^2 - (z_1^2 - z_2^2) = z_2^2$.
We want to find $|u| + |v|$.
Note that $(|u| + |v|)^2 = |u|^2 + |v|^2 + 2|uv| = u\bar{u} + v\bar{v} + 2|z_2^2|$.
Alternatively,using the property $|u+v| + |u-v| = |u+v| + |u-v|$,we know that for any complex numbers $u, v$,$|u+v| + |u-v| = |z_1 + \sqrt{z_1^2 - z_2^2} + z_1 - \sqrt{z_1^2 - z_2^2}| + |z_1 + \sqrt{z_1^2 - z_2^2} - (z_1 - \sqrt{z_1^2 - z_2^2})| = |2z_1| + |2\sqrt{z_1^2 - z_2^2}|$.
Testing with $z_1 = 5, z_2 = 3$: $|5 + \sqrt{25-9}| + |5 - \sqrt{25-9}| = |5+4| + |5-4| = 9 + 1 = 10$.
Checking options: $|z_1+z_2| + |z_1-z_2| = |5+3| + |5-3| = 8 + 2 = 10$.
Thus,the correct expression is $|z_1 + z_2| + |z_1 - z_2|$.

(C) Let $z_1 = a^2$ and $z_2 = b^2$,where $a = \sqrt{z_1}$ and $b = \sqrt{z_2}$.
Then the expression becomes $\left| \frac{1}{2}(a^2 + b^2) + ab \right| + \left| \frac{1}{2}(a^2 + b^2) - ab \right|$.
This simplifies to $\frac{1}{2} |a^2 + b^2 + 2ab| + \frac{1}{2} |a^2 + b^2 - 2ab|$.
$= \frac{1}{2} |(a+b)^2| + \frac{1}{2} |(a-b)^2|$.
Since $|z^2| = |z|^2$,this is $\frac{1}{2} |a+b|^2 + \frac{1}{2} |a-b|^2$.
Using the parallelogram law $|u+v|^2 + |u-v|^2 = 2(|u|^2 + |v|^2)$,we get:
$= \frac{1}{2} \cdot 2(|a|^2 + |b|^2) = |a|^2 + |b|^2$.
Substituting back $a^2 = z_1$ and $b^2 = z_2$,we get $|z_1| + |z_2|$.

(D) For any two complex numbers $z_1$ and $z_2$:
$1$. The modulus of the product is the product of the moduli: $|z_1 z_2| = |z_1| |z_2|$. Thus,$(a)$ is correct.
$2$. The argument of the product is the sum of the arguments: $arg(z_1 z_2) = arg(z_1) + arg(z_2)$. Thus,$(b)$ is incorrect.
$3$. The triangle inequality for moduli states that $|z_1 - z_2| \geqslant ||z_1| - |z_2||$. Thus,$(c)$ is correct.
Therefore,both $(a)$ and $(c)$ are correct.

(D) Given ${z_1} = 1 + 2i$ and ${z_2} = 3 + 5i$.
First, find the conjugate of ${z_2}$: ${\bar{z}_2} = 3 - 5i$.
Now, calculate the numerator: ${\bar{z}_2}{z_1} = (3 - 5i)(1 + 2i) = 3 + 6i - 5i - 10i^2 = 3 + i + 10 = 13 + i$.
Next, divide by ${z_2}$: $\frac{13 + i}{3 + 5i}$.
To simplify, multiply the numerator and denominator by the conjugate of the denominator $(3 - 5i)$:
$\frac{13 + i}{3 + 5i} \times \frac{3 - 5i}{3 - 5i} = \frac{39 - 65i + 3i - 5i^2}{3^2 + 5^2} = \frac{39 - 62i + 5}{9 + 25} = \frac{44 - 62i}{34}$.
Simplify the fraction: $\frac{44}{34} - \frac{62}{34}i = \frac{22}{17} - \frac{31}{17}i$.
The real part $\operatorname{Re} \left( \frac{{\bar{z}_2}{z_1}}{{z_2}} \right)$ is $\frac{22}{17}$.

(A) Given equation: $(3 + i)z = (3 - i)\bar{z}$.
Let $z = x(3 - i)$ where $x \in R$.
Then $\bar{z} = x(3 + i)$.
Substituting these into the left-hand side $(LHS)$: $(3 + i)z = (3 + i)x(3 - i) = x(3^2 + 1^2) = 10x$.
Substituting these into the right-hand side $(RHS)$: $(3 - i)\bar{z} = (3 - i)x(3 + i) = x(3^2 + 1^2) = 10x$.
Since $LHS = RHS$,the complex number $z$ is of the form $x(3 - i)$ for $x \in R$.

(A) Given expression: $z = \frac{1 + 7i}{(2 - i)^2}$
First,expand the denominator: $(2 - i)^2 = 4 - 4i + i^2 = 4 - 4i - 1 = 3 - 4i$
Now,simplify the fraction: $z = \frac{1 + 7i}{3 - 4i}$
Multiply the numerator and denominator by the conjugate of the denominator $(3 + 4i)$:
$z = \frac{(1 + 7i)(3 + 4i)}{(3 - 4i)(3 + 4i)} = \frac{3 + 4i + 21i + 28i^2}{3^2 + 4^2} = \frac{3 + 25i - 28}{9 + 16} = \frac{-25 + 25i}{25} = -1 + i$
To convert to polar form $r(\cos \theta + i\sin \theta)$,find the modulus $r$ and argument $\theta$:
$r = |z| = \sqrt{(-1)^2 + (1)^2} = \sqrt{2}$
Since $z$ is in the second quadrant (real part negative,imaginary part positive),$\theta = \pi - \tan^{-1}\left| \frac{1}{-1} \right| = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$
Therefore,$z = \sqrt{2} \left( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right)$

(C) Given that ${e^{iA}} \cdot {e^{iB}} \cdot {e^{iC}} = {e^{i(A + B + C)}}$.
Since the sum of angles in a triangle is $A + B + C = \pi$,we have:
${e^{i(A + B + C)}} = {e^{i\pi }}$.
Using Euler's formula ${e^{i\theta }} = \cos \theta + i\sin \theta$:
${e^{i\pi }} = \cos \pi + i\sin \pi$.
Since $\cos \pi = -1$ and $\sin \pi = 0$,we get:
$-1 + i(0) = -1$.

(D) Given $z = \frac{7 - i}{3 - 4i}$.
Multiply the numerator and denominator by the conjugate of the denominator $(3 + 4i)$:
$z = \frac{(7 - i)(3 + 4i)}{(3 - 4i)(3 + 4i)} = \frac{21 + 28i - 3i - 4i^2}{3^2 + 4^2} = \frac{21 + 25i + 4}{9 + 16} = \frac{25 + 25i}{25} = 1 + i$.
Now,calculate $z^{14}$:
$z^{14} = (1 + i)^{14} = ((1 + i)^2)^7$.
Since $(1 + i)^2 = 1^2 + i^2 + 2i = 1 - 1 + 2i = 2i$,
$z^{14} = (2i)^7 = 2^7 \times i^7$.
Since $i^7 = i^4 \times i^3 = 1 \times (-i) = -i$,
$z^{14} = 2^7 \times (-i) = -2^7i$.

(C) Given the product of complex numbers in polar form: $(\cos \theta + i\sin \theta )(\cos 2\theta + i\sin 2\theta ) \dots (\cos n\theta + i\sin n\theta ) = 1$.
Using the property $(\cos \alpha + i\sin \alpha)(\cos \beta + i\sin \beta) = \cos(\alpha + \beta) + i\sin(\alpha + \beta)$,we get:
$\cos(\theta + 2\theta + 3\theta + \dots + n\theta) + i\sin(\theta + 2\theta + 3\theta + \dots + n\theta) = 1$.
The sum of the first $n$ natural numbers is $\frac{n(n + 1)}{2}$,so the expression becomes:
$\cos\left(\frac{n(n + 1)}{2}\theta\right) + i\sin\left(\frac{n(n + 1)}{2}\theta\right) = 1$.
For this to equal $1$ (which is $\cos(2m\pi) + i\sin(2m\pi)$),the argument must be an integer multiple of $2\pi$:
$\frac{n(n + 1)}{2}\theta = 2m\pi$,where $m \in \mathbb{Z}$.
Solving for $\theta$:
$\theta = \frac{4m\pi}{n(n + 1)}$.

(A) Using the property of complex numbers in polar form,$\left( \cos \theta_1 + i\sin \theta_1 \right) \left( \cos \theta_2 + i\sin \theta_2 \right) = \cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)$.
Given expression $= \cos \left( \frac{\pi }{2} + \frac{\pi }{{{2^2}}} + \frac{\pi }{{{2^3}}} + \dots \right) + i\sin \left( \frac{\pi }{2} + \frac{\pi }{{{2^2}}} + \frac{\pi }{{{2^3}}} + \dots \right)$.
The sum of the infinite geometric series in the argument is $S = \frac{\pi}{2} + \frac{\pi}{4} + \frac{\pi}{8} + \dots = \frac{\pi/2}{1 - 1/2} = \frac{\pi/2}{1/2} = \pi$.
Thus,the expression becomes $\cos(\pi) + i\sin(\pi)$.
Since $\cos(\pi) = -1$ and $\sin(\pi) = 0$,the value is $-1 + i(0) = -1$.

(C) Given $\cos (u + iv) = \alpha + i\beta$.
Using the identity $\cos (A + B) = \cos A \cos B - \sin A \sin B$,we have:
$\cos u \cos (iv) - \sin u \sin (iv) = \alpha + i\beta$.
Since $\cos (iv) = \cosh v$ and $\sin (iv) = i \sinh v$,we get:
$\cos u \cosh v - i \sin u \sinh v = \alpha + i\beta$.
Comparing real and imaginary parts:
$\alpha = \cos u \cosh v$ and $\beta = - \sin u \sinh v$.
Now,calculate ${\alpha ^2} + {\beta ^2} + 1$:
${\alpha ^2} + {\beta ^2} + 1 = (\cos u \cosh v)^2 + (- \sin u \sinh v)^2 + 1$
$= \cos ^2 u \cosh ^2 v + \sin ^2 u \sinh ^2 v + 1$.
Using the identity $\cosh ^2 v = 1 + \sinh ^2 v$,we substitute $1 = \cosh ^2 v - \sinh ^2 v$:
$= \cos ^2 u \cosh ^2 v + \sin ^2 u \sinh ^2 v + (\cosh ^2 v - \sinh ^2 v)$
$= \cosh ^2 v (\cos ^2 u + 1) + \sinh ^2 v (\sin ^2 u - 1)$
Alternatively,using $\sin ^2 u = 1 - \cos ^2 u$:
$= \cos ^2 u \cosh ^2 v + (1 - \cos ^2 u) \sinh ^2 v + 1$
$= \cos ^2 u \cosh ^2 v + \sinh ^2 v - \cos ^2 u \sinh ^2 v + 1$
$= \cos ^2 u (\cosh ^2 v - \sinh ^2 v) + (\sinh ^2 v + 1)$
$= \cos ^2 u (1) + \cosh ^2 v = \cos ^2 u + \cosh ^2 v$.

(C) We use the identity $\cosh(x \pm y) = \cosh x \cosh y \pm \sinh x \sinh y$.
Also,$\cosh(i\beta) = \cos \beta$ and $\sinh(i\beta) = i \sin \beta$.
$\cosh (\alpha + i\beta ) - \cosh (\alpha - i\beta ) = (\cosh \alpha \cosh (i\beta ) + \sinh \alpha \sinh (i\beta )) - (\cosh \alpha \cosh (i\beta ) - \sinh \alpha \sinh (i\beta ))$
$= \cosh \alpha \cosh (i\beta ) + \sinh \alpha \sinh (i\beta ) - \cosh \alpha \cosh (i\beta ) + \sinh \alpha \sinh (i\beta )$
$= 2 \sinh \alpha \sinh (i\beta )$
Since $\sinh (i\beta ) = i \sin \beta$,the expression becomes $2i \sinh \alpha \sin \beta $.

(B) Using the addition formula for hyperbolic functions: $\cosh(\alpha + i\beta) = \cosh \alpha \cosh(i\beta) + \sinh \alpha \sinh(i\beta)$.
Since $\cosh(i\beta) = \cos \beta$ and $\sinh(i\beta) = i \sin \beta$,we have:
$\cosh(\alpha + i\beta) = \cosh \alpha \cos \beta + i \sinh \alpha \sin \beta$.
The imaginary part is $\sinh \alpha \sin \beta$.

(C) We use the trigonometric identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$.
Substituting $A = x$ and $B = iy$,we get:
$\cos(x + iy) = \cos x \cos(iy) - \sin x \sin(iy)$.
Using the relations $\cos(iy) = \cosh y$ and $\sin(iy) = i \sinh y$,we have:
$\cos(x + iy) = \cos x \cosh y - \sin x (i \sinh y)$.
Therefore,$\cos(x + iy) = \cos x \cosh y - i \sin x \sinh y$.

(B) Given $\tan (u + iv) = i$.
Using the identity $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have $\frac{\tan u + \tan (iv)}{1 - \tan u \tan (iv)} = i$.
Since $\tan (iv) = i \tanh v$,the equation becomes $\frac{\tan u + i \tanh v}{1 - i \tan u \tanh v} = i$.
Multiplying both sides by the denominator: $\tan u + i \tanh v = i(1 - i \tan u \tanh v) = i + \tan u \tanh v$.
Rearranging terms: $\tan u - i = \tanh v (\tan u - i)$.
This implies $(\tan u - i)(1 - \tanh v) = 0$.
For this to hold,$1 - \tanh v = 0$,which means $\tanh v = 1$.
Since $\tanh v = \frac{e^v - e^{-v}}{e^v + e^{-v}}$,setting this equal to $1$ leads to $e^v - e^{-v} = e^v + e^{-v}$,which simplifies to $2e^{-v} = 0$.
This is only possible as $v \to \infty$.

(D) Given the expression: $(1 + i)^{n_1} + (1 + i^3)^{n_1} + (1 + i^5)^{n_2} + (1 + i^7)^{n_2}$.
Since $i^3 = -i$, $i^5 = i$, and $i^7 = -i$, the expression becomes:
$(1 + i)^{n_1} + (1 - i)^{n_1} + (1 + i)^{n_2} + (1 - i)^{n_2}$.
Let $z = (1 + i)^{n} + (1 - i)^{n}$.
Using the polar form $1 + i = \sqrt{2} e^{i\pi/4}$ and $1 - i = \sqrt{2} e^{-i\pi/4}$, we get:
$z = (\sqrt{2})^n (e^{in\pi/4} + e^{-in\pi/4}) = 2^{n/2} \cdot 2 \cos(n\pi/4) = 2^{n/2+1} \cos(n\pi/4)$.
Since $2^{n/2+1} \cos(n\pi/4)$ is always a real number for any positive integer $n$, the sum of two such terms is also always real.
Therefore, the expression is a real number for all positive integers $n_1$ and $n_2$.

(D) Given the equation $z^2 + (p + iq)z + r + is = 0$ ... $(i)$
Let $z = \alpha$ (where $\alpha$ is real) be a root of $(i)$.
Then,$\alpha^2 + (p + iq)\alpha + r + is = 0$.
This can be written as $(\alpha^2 + p\alpha + r) + i(q\alpha + s) = 0$.
Equating the real and imaginary parts to zero,we get:
$1) \alpha^2 + p\alpha + r = 0$
$2) q\alpha + s = 0$
From $(2)$,we have $\alpha = -\frac{s}{q}$.
Substituting this into $(1)$,we get $\left(-\frac{s}{q}\right)^2 + p\left(-\frac{s}{q}\right) + r = 0$.
$\frac{s^2}{q^2} - \frac{ps}{q} + r = 0$.
Multiplying by $q^2$,we get $s^2 - pqs + q^2r = 0$.
Therefore,$pqs = s^2 + q^2r$.

(B) Given $x = -5 + 2\sqrt{-4} = -5 + 4i$.
Then $x + 5 = 4i$.
Squaring both sides,we get $(x + 5)^2 = (4i)^2$.
$x^2 + 10x + 25 = -16$,which implies $x^2 + 10x + 41 = 0$.
Now,we perform polynomial division of $x^4 + 9x^3 + 35x^2 - x + 4$ by $x^2 + 10x + 41$.
$x^4 + 9x^3 + 35x^2 - x + 4 = (x^2 + 10x + 41)(x^2 - x + 4) - 160$.
Since $x^2 + 10x + 41 = 0$,the expression becomes $0 \times (x^2 - x + 4) - 160 = -160$.

(B) Given that $(1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni) = a + ib$ ..... $(i)$
Taking the conjugate on both sides,we get $(1 - i)(1 - 2i)(1 - 3i) \dots (1 - ni) = a - ib$ ..... $(ii)$
Multiplying equation $(i)$ and equation $(ii)$,we get:
$[(1 + i)(1 - i)] \times [(1 + 2i)(1 - 2i)] \times \dots \times [(1 + ni)(1 - ni)] = (a + ib)(a - ib)$
Since $(1 + ki)(1 - ki) = 1^2 - (ki)^2 = 1 + k^2$,the expression becomes:
$(1 + 1^2)(1 + 2^2)(1 + 3^2) \dots (1 + n^2) = a^2 + b^2$
$2 \times 5 \times 10 \times \dots \times (1 + n^2) = a^2 + b^2$

(D) Given $|{z_1}| = |{z_2}| = 1$,we have ${z_1} = \cos {\theta _1} + i\sin {\theta _1}$ and ${z_2} = \cos {\theta _2} + i\sin {\theta _2}$,where ${\theta _1} = \arg({z_1})$ and ${\theta _2} = \arg({z_2})$.
Since ${z_1} = a + ib$ and ${z_2} = c + id$,we have $a = \cos {\theta _1}, b = \sin {\theta _1}, c = \cos {\theta _2}, d = \sin {\theta _2}$.
Given $R({z_1}\overline {{z_2}} ) = 0$,we have $R[(\cos {\theta _1} + i\sin {\theta _1})(\cos {\theta _2} - i\sin {\theta _2})] = \cos({\theta _1} - {\theta _2}) = 0$.
Thus,${\theta _1} - {\theta _2} = \pm \frac{\pi }{2}$,which implies ${\theta _1} = {\theta _2} \pm \frac{\pi }{2}$.
Now,${w_1} = a + ic = \cos {\theta _1} + i\cos {\theta _2}$. Since $\cos {\theta _2} = \sin {\theta _1}$ (or $-\sin {\theta _1}$),we have $|{w_1}|^2 = \cos^2 {\theta _1} + \cos^2 {\theta _2} = \cos^2 {\theta _1} + \sin^2 {\theta _1} = 1$,so $|{w_1}| = 1$.
Similarly,$|{w_2}|^2 = b^2 + d^2 = \sin^2 {\theta _1} + \sin^2 {\theta _2} = \sin^2 {\theta _1} + \cos^2 {\theta _1} = 1$,so $|{w_2}| = 1$.
Finally,$R({w_1}\overline {{w_2}} ) = R((a + ic)(b - id)) = R(ab - iad + ibc + cd) = ab + cd = \cos {\theta _1}\sin {\theta _1} + \cos {\theta _2}\sin {\theta _2} = \frac{1}{2}(\sin 2{\theta _1} + \sin 2{\theta _2})$.
Since ${\theta _1} = {\theta _2} + \frac{\pi }{2}$,$\sin 2{\theta _1} = \sin(2{\theta _2} + \pi) = -\sin 2{\theta _2}$,so $R({w_1}\overline {{w_2}} ) = 0$.
Thus,all the above are satisfied.

(C) Given $|z| \le 1$ and $|w| \le 1$.
We are given $|z + iw| = 2$ and $|z - i\overline{w}| = 2$.
Let $z = a + ib$ and $w = c + id$. Then $|z|^2 = a^2 + b^2 \le 1$ and $|w|^2 = c^2 + d^2 \le 1$.
$|z + iw| = |(a + ib) + i(c + id)| = |(a - d) + i(b + c)| = 2$.
Squaring both sides: $(a - d)^2 + (b + c)^2 = 4$ $(i)$.
$|z - i\overline{w}| = |(a + ib) - i(c - id)| = |(a - d) + i(b - c)| = 2$.
Squaring both sides: $(a - d)^2 + (b - c)^2 = 4$ $(ii)$.
Subtracting $(ii)$ from $(i)$,we get $(b + c)^2 - (b - c)^2 = 0$,which simplifies to $4bc = 0$,so $bc = 0$.
If $b = 0$,then $(a - d)^2 + c^2 = 4$. Since $a^2 \le 1$ and $c^2 + d^2 \le 1$,the maximum value of $(a - d)^2 + c^2$ is $(1 - (-1))^2 + 1 = 5$ (not helpful).
However,for $(a - d)^2 + c^2 = 4$ with $a^2 \le 1$ and $c^2 + d^2 \le 1$,the only way to reach $4$ is if $a = 1, d = -1, c = 0$ or $a = -1, d = 1, c = 0$.
In both cases,$z = a + i(0) = \pm 1$.

(C) Given equations are $\left| \frac{z - 12}{z - 8i} \right| = \frac{5}{3}$ and $\left| \frac{z - 4}{z - 8} \right| = 1$.
Let $z = x + iy$.
From the second equation,$|z - 4| = |z - 8|$,which implies $|(x - 4) + iy| = |(x - 8) + iy|$.
Squaring both sides,$(x - 4)^2 + y^2 = (x - 8)^2 + y^2$.
$x^2 - 8x + 16 = x^2 - 16x + 64$,which gives $8x = 48$,so $x = 6$.
Now substitute $x = 6$ into the first equation: $3|z - 12| = 5|z - 8i|$.
$3|(6 - 12) + iy| = 5|6 + (y - 8)i|$.
$3|-6 + iy| = 5|6 + (y - 8)i|$.
Squaring both sides,$9(36 + y^2) = 25(36 + (y - 8)^2)$.
$324 + 9y^2 = 25(36 + y^2 - 16y + 64)$.
$324 + 9y^2 = 25(y^2 - 16y + 100)$.
$324 + 9y^2 = 25y^2 - 400y + 2500$.
$16y^2 - 400y + 2176 = 0$.
Dividing by $16$,$y^2 - 25y + 136 = 0$.
$(y - 8)(y - 17) = 0$.
Thus,$y = 8$ or $y = 17$.
Therefore,$z = 6 + 8i$ or $z = 6 + 17i$.

(B) The given inequality $|z - 25i| \le 15$ represents a disk centered at $25i$ with radius $15$. Let $O$ be the origin $(0,0)$ and $C$ be the center $(0, 25)$. The tangents from the origin to the circle touch at points $z_1$ and $z_2$.
In the right-angled triangle formed by the origin,the center of the circle,and a point of tangency,the hypotenuse is $25$ and the opposite side to the angle $\alpha$ at the origin is $15$.
Thus,$\sin \alpha = \frac{15}{25} = \frac{3}{5}$,which implies $\alpha = \sin^{-1}\left(\frac{3}{5}\right)$.
The argument of the center is $\frac{\pi}{2}$. The range of arguments is $\left[\frac{\pi}{2} - \alpha, \frac{\pi}{2} + \alpha\right]$.
Therefore,$\max \text{amp}(z) = \frac{\pi}{2} + \alpha$ and $\min \text{amp}(z) = \frac{\pi}{2} - \alpha$.
The difference is $(\frac{\pi}{2} + \alpha) - (\frac{\pi}{2} - \alpha) = 2\alpha = 2\sin^{-1}\left(\frac{3}{5}\right)$.
Using the identity $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$,we have $2\sin^{-1}\left(\frac{3}{5}\right) = 2\left(\frac{\pi}{2} - \cos^{-1}\left(\frac{3}{5}\right)\right) = \pi - 2\cos^{-1}\left(\frac{3}{5}\right)$.

(A) Let the vertices of the triangle be $O(0), A(z_1),$ and $B(z_2)$. Since $\Delta OAB$ is an equilateral triangle,the rotation of vector $OA$ by $60^\circ$ ($\pi/3$ radians) about the origin $O$ must coincide with the vector $OB$ (or vice versa).
Thus,we have $z_2 = z_1 e^{\pm i\pi/3}$.
This implies $\frac{z_2}{z_1} = \cos(\pi/3) \pm i \sin(\pi/3) = \frac{1}{2} \pm i \frac{\sqrt{3}}{2}$.
Rearranging gives $2z_2 = z_1(1 \pm i\sqrt{3})$,or $2z_2 - z_1 = \pm i\sqrt{3} z_1$.
Squaring both sides: $(2z_2 - z_1)^2 = -3z_1^2$.
$4z_2^2 - 4z_1 z_2 z_1^2 = -3z_1^2$.
$4z_1^2 4z_2^2 = 4z_1 z_2$.
Dividing by $4$,we get $z_1^2 z_2^2 = z_1 z_2$.

(D) Given $a = \cos (2\pi /7) + i\sin (2\pi /7)$.
Since $a^7 = \cos (2\pi) + i\sin (2\pi) = 1$,we have $a^7 = 1$.
Sum of roots $S = \alpha + \beta = (a + a^2 + a^4) + (a^3 + a^5 + a^6) = a + a^2 + a^3 + a^4 + a^5 + a^6$.
Using the sum of a geometric series,$S = \frac{a(1 - a^6)}{1 - a} = \frac{a - a^7}{1 - a} = \frac{a - 1}{1 - a} = -1$.
Product of roots $P = \alpha \beta = (a + a^2 + a^4)(a^3 + a^5 + a^6) = a^4 + a^6 + a^7 + a^5 + a^7 + a^8 + a^7 + a^9 + a^{10}$.
Using $a^7 = 1$,we get $P = a^4 + a^6 + 1 + a^5 + 1 + a + 1 + a^2 + a^3 = 3 + (a + a^2 + a^3 + a^4 + a^5 + a^6) = 3 + S$.
Substituting $S = -1$,we get $P = 3 - 1 = 2$.
The quadratic equation is $x^2 - Sx + P = 0$,which becomes $x^2 - (-1)x + 2 = 0$,or $x^2 + x + 2 = 0$.

(A) The given series is $S = i - 2 - 3i + 4 + 5i - 6 - 7i + 8 + \dots$ up to $100$ terms.
We can group the terms in pairs: $(i - 2) + (-3i + 4) + (5i - 6) + (-7i + 8) + \dots$
Each pair consists of $2$ terms,so there are $100 / 2 = 50$ pairs.
Let's look at the $k$-th pair where $k = 1, 2, \dots, 50$.
The $k$-th pair is $((4k-3)i^{4k-3} + (4k-2)i^{4k-2} + (4k-1)i^{4k-1} + 4ki^{4k})$ is not the pattern here.
Actually,the pattern is: $(i - 2) + (-3i + 4) + (5i - 6) + (-7i + 8) + \dots$
Sum of each pair: $(i - 2) = -2 + i$,$(-3i + 4) = 4 - 3i$,$(5i - 6) = -6 + 5i$,$(-7i + 8) = 8 - 7i$.
Sum of $k$-th pair: $P_k = (-1)^k [2k - (2k-1)i]$.
Sum $S = \sum_{k=1}^{50} P_k = (-2+i) + (4-3i) + (-6+5i) + (8-7i) + \dots + (100-99i)$.
$S = (-2+4-6+8+\dots+100) + i(1-3+5-7+\dots-99)$.
Sum of real part: $-2+4-6+8+\dots+100 = (-2+4) + (-6+8) + \dots + (-98+100) = 2 \times 25 = 50$.
Sum of imaginary part: $i(1-3+5-7+\dots-99) = i((1-3) + (5-7) + \dots + (97-99)) = i(-2 \times 25) = -50i$.
Thus,$S = 50 - 50i = 50(1 - i)$.

(C) Since the coefficients of the polynomial are real,complex roots must occur in conjugate pairs. Thus,$3 - i\sqrt{6}$ is also a root.
Let the quadratic factor corresponding to these roots be $(x - (3 + i\sqrt{6}))(x - (3 - i\sqrt{6})) = (x - 3)^2 + 6 = x^2 - 6x + 15 = 0$.
Dividing the given equation $4x^4 - 24x^3 + 57x^2 + 18x - 45 = 0$ by $(x^2 - 6x + 15)$,we get:
$4x^4 - 24x^3 + 57x^2 + 18x - 45 = (x^2 - 6x + 15)(4x^2 - 3) = 0$.
Setting the second factor to zero,$4x^2 - 3 = 0$,we get $x^2 = \frac{3}{4}$,which implies $x = \pm \frac{\sqrt{3}}{2}$.
Therefore,the roots are $3 \pm i\sqrt{6}$ and $\pm \frac{\sqrt{3}}{2}$.

$(A)$ We know that the expansion of $e^z$ is $1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \dots \infty$.
Substituting $z = ix$, we get $e^{ix} = 1 + \frac{ix}{1!} + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \dots \infty = 1 + \frac{ix}{1!} - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \dots \infty$.
Similarly, $e^{-ix} = 1 - \frac{ix}{1!} + \frac{(-ix)^2}{2!} + \frac{(-ix)^3}{3!} + \frac{(-ix)^4}{4!} + \dots \infty = 1 - \frac{ix}{1!} - \frac{x^2}{2!} + \frac{ix^3}{3!} + \frac{x^4}{4!} + \dots \infty$.
Adding these two expressions:
$e^{ix} + e^{-ix} = 2 \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \infty \right)$.
Therefore, $\frac{e^{ix} + e^{-ix}}{2} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \infty$.

(A) Let $z_1 = -1 + i\sqrt{3} = 2\omega$ and $z_2 = -1 - i\sqrt{3} = 2\omega^2$,where $\omega$ is the cube root of unity.
Then,the expression becomes $\frac{(2\omega)^{15}}{(1-i)^{20}} + \frac{(2\omega^2)^{15}}{(1+i)^{20}}$.
Since $\omega^3 = 1$,$\omega^{15} = 1$ and $\omega^{30} = 1$.
So,the expression is $\frac{2^{15}}{(1-i)^{20}} + \frac{2^{15}}{(1+i)^{20}} = 2^{15} \left[ \frac{(1+i)^{20} + (1-i)^{20}}{(1-i^2)^{20}} \right]$.
Note that $(1-i^2) = (1 - (-1)) = 2$,so $(1-i^2)^{20} = 2^{20}$.
Thus,$\frac{2^{15}}{2^{20}} [(1+i)^{20} + (1-i)^{20}] = \frac{1}{2^5} [(1+i)^2]^{10} + [(1-i)^2]^{10}$.
$(1+i)^2 = 1 + i^2 + 2i = 2i$ and $(1-i)^2 = 1 + i^2 - 2i = -2i$.
Substituting these,we get $\frac{1}{32} [(2i)^{10} + (-2i)^{10}] = \frac{1}{32} [2^{10} i^{10} + 2^{10} i^{10}] = \frac{1}{32} [2 \times 2^{10} \times (-1)]$.
$= \frac{1}{32} [-2^{11}] = -\frac{2048}{32} = -64$.

(A) Given $z = x - iy$ and $z^{1/3} = p + iq$.
Taking the cube on both sides,we get $z = (p + iq)^3$.
$z = p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3$.
$z = p^3 + 3ip^2q - 3pq^2 - iq^3$.
$z = (p^3 - 3pq^2) + i(3p^2q - q^3)$.
Comparing the real and imaginary parts with $z = x - iy$,we have:
$x = p^3 - 3pq^2 = p(p^2 - 3q^2)$ and $-y = 3p^2q - q^3$,which implies $y = q^2 - 3p^2q = q(q^2 - 3p^2)$.
Now,$\frac{x}{p} = p^2 - 3q^2$ and $\frac{y}{q} = q^2 - 3p^2$.
Adding these two expressions:
$\frac{x}{p} + \frac{y}{q} = (p^2 - 3q^2) + (q^2 - 3p^2) = -2p^2 - 2q^2 = -2(p^2 + q^2)$.
Therefore,$\frac{\frac{x}{p} + \frac{y}{q}}{p^2 + q^2} = \frac{-2(p^2 + q^2)}{p^2 + q^2} = -2$.

(C) Given ${x_n} = \cos \left( \frac{\pi }{3^n} \right) + i\sin \left( \frac{\pi }{3^n} \right) = e^{i\pi / 3^n}$.
The product is ${x_1} \cdot {x_2} \cdot {x_3} \cdots {x_\infty } = e^{i\pi / 3^1} \cdot e^{i\pi / 3^2} \cdot e^{i\pi / 3^3} \cdots e^{i\pi / 3^\infty }$.
Using the property of exponents,this equals $e^{i\pi \left( \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots \right)}$.
The sum of the infinite geometric series in the exponent is $S = \frac{a}{1-r} = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = \frac{1}{2}$.
Therefore,the product is $e^{i\pi (1/2)} = e^{i\pi / 2}$.
Using Euler's formula,$e^{i\pi / 2} = \cos \left( \frac{\pi }{2} \right) + i\sin \left( \frac{\pi }{2} \right) = 0 + i(1) = i$.

(B) Given $\frac{1 - ix}{1 + ix} = a - ib$.
Multiplying numerator and denominator by $(1 - ix)$,we get:
$\frac{(1 - ix)^2}{1 + x^2} = a - ib$
$\frac{1 - x^2 - 2ix}{1 + x^2} = a - ib$
Equating real and imaginary parts:
$a = \frac{1 - x^2}{1 + x^2}$ and $b = \frac{2x}{1 + x^2}$.
We know that $a^2 + b^2 = 1$.
Consider the expression $\frac{b}{1 + a} = \frac{\frac{2x}{1 + x^2}}{1 + \frac{1 - x^2}{1 + x^2}} = \frac{2x}{1 + x^2 + 1 - x^2} = \frac{2x}{2} = x$.
Thus,$x = \frac{b}{1 + a}$.
Since $a^2 + b^2 = 1$,we can write $1 = a^2 + b^2$.
Substituting this into the denominator:
$x = \frac{b}{1 + a} = \frac{2b}{2(1 + a)} = \frac{2b}{1 + 1 + 2a} = \frac{2b}{1 + (a^2 + b^2) + 2a} = \frac{2b}{(1 + a)^2 + b^2}$.

(C) The condition $|z - 5i| \le 1$ represents a disk centered at $C(0, 5)$ with radius $r = 1$.
To minimize $\text{amp } z$,the line $OA$ must be tangent to the circle at point $A$.
In $\triangle OAC$,$\angle OAC = 90^\circ$ (radius is perpendicular to the tangent).
$OC = 5$ and $AC = 1$.
In $\triangle OAC$,$\sin(\angle AOC) = \frac{AC}{OC} = \frac{1}{5}$.
Let $\alpha = \angle AOC$. Then $\sin \alpha = \frac{1}{5}$,so $\cos \alpha = \sqrt{1 - (\frac{1}{5})^2} = \frac{\sqrt{24}}{5} = \frac{2\sqrt{6}}{5}$.
The angle of $z$ is $\theta = 90^\circ - \alpha$. Thus,$\cos \theta = \sin \alpha = \frac{1}{5}$ and $\sin \theta = \cos \alpha = \frac{2\sqrt{6}}{5}$.
The distance $OA = \sqrt{OC^2 - AC^2} = \sqrt{25 - 1} = \sqrt{24} = 2\sqrt{6}$.
Thus,$z = OA(\cos \theta + i \sin \theta) = 2\sqrt{6}(\frac{1}{5} + i\frac{2\sqrt{6}}{5}) = \frac{2\sqrt{6}}{5} + i\frac{4 \times 6}{5} = \frac{2\sqrt{6}}{5} + \frac{24}{5}i$.