Conjugate and Modulus of complex numbers Questions in English

(D) Let $z = (1 - \cos \theta ) + i(2\sin \theta )$. We want the real part of $z^{-1} = \frac{1}{z}$.
Using trigonometric identities,$1 - \cos \theta = 2\sin^2(\frac{\theta}{2})$ and $2\sin \theta = 4\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$.
So,$z = 2\sin(\frac{\theta}{2}) [\sin(\frac{\theta}{2}) + 2i\cos(\frac{\theta}{2})]$.
Then,$z^{-1} = \frac{1}{2\sin(\frac{\theta}{2})} \cdot \frac{1}{\sin(\frac{\theta}{2}) + 2i\cos(\frac{\theta}{2})}$.
Multiply numerator and denominator by the conjugate $\sin(\frac{\theta}{2}) - 2i\cos(\frac{\theta}{2})$:
$z^{-1} = \frac{1}{2\sin(\frac{\theta}{2})} \cdot \frac{\sin(\frac{\theta}{2}) - 2i\cos(\frac{\theta}{2})}{\sin^2(\frac{\theta}{2}) + 4\cos^2(\frac{\theta}{2})}$.
The real part is $\frac{\sin(\frac{\theta}{2})}{2\sin(\frac{\theta}{2}) [\sin^2(\frac{\theta}{2}) + 4\cos^2(\frac{\theta}{2})]} = \frac{1}{2[\sin^2(\frac{\theta}{2}) + 4\cos^2(\frac{\theta}{2})]}$.
Using $\sin^2(\frac{\theta}{2}) = \frac{1 - \cos \theta}{2}$ and $\cos^2(\frac{\theta}{2}) = \frac{1 + \cos \theta}{2}$:
Real part $= \frac{1}{2[\frac{1 - \cos \theta}{2} + 4(\frac{1 + \cos \theta}{2})]} = \frac{1}{1 - \cos \theta + 4 + 4\cos \theta} = \frac{1}{5 + 3\cos \theta }$.

(B) Given $x + iy = \frac{3}{2 + \cos \theta + i\sin \theta}$.
Taking the modulus on both sides,we have $|x + iy| = \left| \frac{3}{2 + \cos \theta + i\sin \theta} \right|$.
$|x + iy| = \frac{|3|}{|(2 + \cos \theta) + i\sin \theta|} = \frac{3}{\sqrt{(2 + \cos \theta)^2 + \sin^2 \theta}}$.
$|x + iy| = \frac{3}{\sqrt{4 + \cos^2 \theta + 4\cos \theta + \sin^2 \theta}} = \frac{3}{\sqrt{5 + 4\cos \theta}}$.
Squaring both sides,${x^2} + {y^2} = \frac{9}{5 + 4\cos \theta}$.
From the real part,$x = \frac{3(2 + \cos \theta)}{5 + 4\cos \theta} = \frac{6 + 3\cos \theta}{5 + 4\cos \theta}$.
Thus,$4x - 3 = 4\left( \frac{6 + 3\cos \theta}{5 + 4\cos \theta} \right) - 3 = \frac{24 + 12\cos \theta - 15 - 12\cos \theta}{5 + 4\cos \theta} = \frac{9}{5 + 4\cos \theta}$.
Therefore,${x^2} + {y^2} = 4x - 3$.

(D) Given $\mu + i\lambda = \frac{(p + i)^2}{2p - i}$.
Taking the modulus on both sides,we get $|\mu + i\lambda| = \left|\frac{(p + i)^2}{2p - i}\right|$.
Since $|z_1/z_2| = |z_1|/|z_2|$ and $|z^n| = |z|^n$,we have $\sqrt{\mu^2 + \lambda^2} = \frac{|p + i|^2}{|2p - i|}$.
Calculating the moduli: $|p + i| = \sqrt{p^2 + 1}$ and $|2p - i| = \sqrt{(2p)^2 + (-1)^2} = \sqrt{4p^2 + 1}$.
Thus,$\sqrt{\mu^2 + \lambda^2} = \frac{(\sqrt{p^2 + 1})^2}{\sqrt{4p^2 + 1}} = \frac{p^2 + 1}{\sqrt{4p^2 + 1}}$.
Squaring both sides,we get $\mu^2 + \lambda^2 = \frac{(p^2 + 1)^2}{4p^2 + 1}$.

(B) Let $z_1 = a + ib$ and $z_2 = c + id$,where $a, b, c, d \in \mathbb{R}$.
Since $z_1 + z_2 = (a + c) + i(b + d)$ is real,the imaginary part must be zero: $b + d = 0$,which implies $d = -b$.
Since $z_1 z_2 = (a + ib)(c + id) = (ac - bd) + i(ad + bc)$ is real,the imaginary part must be zero: $ad + bc = 0$.
Substituting $d = -b$ into the equation $ad + bc = 0$,we get $a(-b) + bc = 0$,which simplifies to $b(c - a) = 0$.
Case $1$: If $b = 0$,then $d = 0$. Thus $z_1 = a$ and $z_2 = c$,which are real numbers. In this case,$z_1 = \bar{z}_1$ and $z_2 = \bar{z}_2$. This does not uniquely determine the relationship between $z_1$ and $z_2$ unless they are equal or conjugate.
Case $2$: If $b \neq 0$,then $c = a$. Since $d = -b$,we have $z_2 = a - ib = \bar{z}_1$,which implies $z_1 = \bar{z}_2$.
Thus,the condition $z_1 = \bar{z}_2$ satisfies the given requirements.

(C) Given $a = \cos \theta + i \sin \theta$.
We need to evaluate $\frac{1 + a}{1 - a}$.
$\frac{1 + a}{1 - a} = \frac{1 + \cos \theta + i \sin \theta}{1 - (\cos \theta + i \sin \theta)} = \frac{(1 + \cos \theta) + i \sin \theta}{(1 - \cos \theta) - i \sin \theta}$.
Using half-angle formulas: $1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}$,$1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$,and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$.
$\frac{1 + a}{1 - a} = \frac{2 \cos^2 \frac{\theta}{2} + i (2 \sin \frac{\theta}{2} \cos \frac{\theta}{2})}{2 \sin^2 \frac{\theta}{2} - i (2 \sin \frac{\theta}{2} \cos \frac{\theta}{2})}$.
Factor out $2 \cos \frac{\theta}{2}$ from the numerator and $2 \sin \frac{\theta}{2}$ from the denominator:
$\frac{1 + a}{1 - a} = \frac{2 \cos \frac{\theta}{2} (\cos \frac{\theta}{2} + i \sin \frac{\theta}{2})}{2 \sin \frac{\theta}{2} (\sin \frac{\theta}{2} - i \cos \frac{\theta}{2})}$.
Note that $\sin \frac{\theta}{2} - i \cos \frac{\theta}{2} = -i (\cos \frac{\theta}{2} + i \sin \frac{\theta}{2})$.
$\frac{1 + a}{1 - a} = \frac{\cos \frac{\theta}{2} (\cos \frac{\theta}{2} + i \sin \frac{\theta}{2})}{\sin \frac{\theta}{2} (-i) (\cos \frac{\theta}{2} + i \sin \frac{\theta}{2})} = \frac{\cot \frac{\theta}{2}}{-i} = i \cot \frac{\theta}{2}$.

(B) Given expression: $z = \frac{1}{(1 - \cos \theta) + i \sin \theta}$.
Multiply the numerator and denominator by the conjugate $(1 - \cos \theta) - i \sin \theta$:
$z = \frac{(1 - \cos \theta) - i \sin \theta}{(1 - \cos \theta)^2 + \sin^2 \theta}$.
Simplify the denominator:
$(1 - \cos \theta)^2 + \sin^2 \theta = 1 - 2 \cos \theta + \cos^2 \theta + \sin^2 \theta = 1 - 2 \cos \theta + 1 = 2 - 2 \cos \theta = 2(1 - \cos \theta)$.
Thus,$z = \frac{1 - \cos \theta}{2(1 - \cos \theta)} - i \frac{\sin \theta}{2(1 - \cos \theta)}$.
The real part is $\frac{1 - \cos \theta}{2(1 - \cos \theta)} = \frac{1}{2}$.

(A) We know that for any complex number $z$,the conjugate of the inverse is the inverse of the conjugate,i.e.,$\overline{z^{-1}} = (\overline{z})^{-1}$.
Therefore,$(\overline{z^{-1}})(\overline{z}) = (\overline{z})^{-1}(\overline{z})$.
Since $(\overline{z})^{-1} = \frac{1}{\overline{z}}$,we have $\frac{1}{\overline{z}} \times \overline{z} = 1$,provided $z \neq 0$.

(A) Let $z = x + iy$,where $x, y \in \mathbb{R}$. Then $\overline{z} = x - iy$.
Given $z \cdot \overline{z} = 0$.
Substituting the values,we get $(x + iy)(x - iy) = 0$.
This simplifies to $x^2 - (iy)^2 = 0$,which is $x^2 + y^2 = 0$.
Since $x^2 \ge 0$ and $y^2 \ge 0$ for all real $x$ and $y$,the sum $x^2 + y^2 = 0$ is possible if and only if $x = 0$ and $y = 0$.
Therefore,$z = 0 + 0i = 0$.

(A) Given that $(a + ib)(c + id)(e + if)(g + ih) = A + iB$ ..... $(i)$
Taking the conjugate on both sides,we get $(a - ib)(c - id)(e - if)(g - ih) = A - iB$ ..... $(ii)$
Multiplying equation $(i)$ and equation $(ii)$,we get:
$[(a + ib)(a - ib)] \times [(c + id)(c - id)] \times [(e + if)(e - if)] \times [(g + ih)(g - ih)] = (A + iB)(A - iB)$
Using the property $(x + iy)(x - iy) = x^2 + y^2$,we have:
$({a^2} + {b^2})({c^2} + {d^2})({e^2} + {f^2})({g^2} + {h^2}) = {A^2} + {B^2}$

(C) Let $z = x + iy$,where $x, y \in \mathbb{R}$.
Then the conjugate is $\bar z = x - iy$.
First,$z + \bar z = (x + iy) + (x - iy) = 2x$,which is a real number.
Second,$z\,\bar z = (x + iy)(x - iy) = x^2 - (iy)^2 = x^2 + y^2$,which is also a real number.
Therefore,both $z + \bar z$ and $z\,\bar z$ are real numbers.

(A) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates if $z_1 = \overline{z_2}$.
Given $z_1 = 3 + i(x^2y)$ and $z_2 = (x^2 + y) + 4i$.
The conjugate of $z_2$ is $\overline{z_2} = (x^2 + y) - 4i$.
Equating $z_1 = \overline{z_2}$,we get $3 + i(x^2y) = (x^2 + y) - 4i$.
Comparing real and imaginary parts:
$x^2 + y = 3$ (Equation $1$)
$x^2y = - 4$ (Equation $2$)
From Equation $1$,$x^2 = 3 - y$. Substituting this into Equation $2$:
$(3 - y)y = - 4 \implies 3y - y^2 = - 4 \implies y^2 - 3y - 4 = 0$.
Factoring the quadratic: $(y - 4)(y + 1) = 0$,so $y = 4$ or $y = - 1$.
If $y = 4$,$x^2 = 3 - 4 = - 1$,which gives no real values for $x$.
If $y = - 1$,$x^2 = 3 - (- 1) = 4$,so $x = \pm 2$.
Thus,the possible values are $(x, y) = (2, - 1)$ or $( - 2, - 1)$.

(B) Let $z = \frac{2 + 5i}{4 - 3i}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator,$4 + 3i$:
$z = \frac{(2 + 5i)(4 + 3i)}{(4 - 3i)(4 + 3i)}$
$z = \frac{8 + 6i + 20i + 15i^2}{16 + 9}$
Since $i^2 = -1$,we have:
$z = \frac{8 + 26i - 15}{25} = \frac{-7 + 26i}{25}$
The conjugate of $z = a + bi$ is $\bar{z} = a - bi$.
Therefore,the conjugate of $\frac{-7 + 26i}{25}$ is $\frac{-7 - 26i}{25}$.

(C) Given that $a$ is a real number,we have $a = \bar{a}$.
Now,consider the expression $(z + a)(\bar{z} + a)$.
Since $a = \bar{a}$,we can write this as $(z + a)(\bar{z} + \bar{a})$.
By the property of complex conjugates,$\bar{z} + \bar{a} = \overline{z + a}$.
Therefore,$(z + a)(\bar{z} + a) = (z + a)(\overline{z + a})$.
Using the property $|w|^2 = w \cdot \bar{w}$,we get $(z + a)(\overline{z + a}) = |z + a|^2$.

(B) Let $z = x + iy$. Then $\frac{z - i}{z + i} = \frac{x + i(y - 1)}{x + i(y + 1)}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator: $\frac{x + i(y - 1)}{x + i(y + 1)} \times \frac{x - i(y + 1)}{x - i(y + 1)}$.
$= \frac{x^2 - ix(y + 1) + ix(y - 1) + (y - 1)(y + 1)}{x^2 + (y + 1)^2} = \frac{(x^2 + y^2 - 1) - 2ix}{x^2 + (y + 1)^2}$.
Since the expression is purely imaginary,the real part must be zero:
$x^2 + y^2 - 1 = 0 \implies x^2 + y^2 = 1$.
Since $z \cdot \bar{z} = |z|^2 = x^2 + y^2$,we have $z \cdot \bar{z} = 1$.

(A) Given $\frac{c + i}{c - i} = a + ib$ $(i)$
Taking the conjugate on both sides,we get $\frac{c - i}{c + i} = a - ib$ $(ii)$
Multiplying equation $(i)$ and equation $(ii)$,we get:
$\left(\frac{c + i}{c - i}\right) \times \left(\frac{c - i}{c + i}\right) = (a + ib)(a - ib)$
$1 = a^2 - (ib)^2$
$1 = a^2 - i^2b^2$
Since $i^2 = -1$,we have $1 = a^2 + b^2$
Therefore,$a^2 + b^2 = 1$.

(C) Given that $\overline{(x + iy)(1 - 2i)} = 1 + i$.
Using the property $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$,we have $\overline{(x + iy)} \cdot \overline{(1 - 2i)} = 1 + i$.
This implies $(x - iy)(1 + 2i) = 1 + i$.
Therefore,$x - iy = \frac{1 + i}{1 + 2i}$.
Taking the conjugate on both sides,we get $\overline{x - iy} = \overline{\left(\frac{1 + i}{1 + 2i}\right)}$.
Thus,$x + iy = \frac{1 - i}{1 - 2i}$.

(C) Let $z = \frac{(2 + i)^2}{3 + i}$.
First,expand the numerator: $(2 + i)^2 = 4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i$.
So,$z = \frac{3 + 4i}{3 + i}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator $(3 - i)$:
$z = \frac{(3 + 4i)(3 - i)}{(3 + i)(3 - i)} = \frac{9 - 3i + 12i - 4i^2}{3^2 + 1^2} = \frac{9 + 9i + 4}{9 + 1} = \frac{13 + 9i}{10} = \frac{13}{10} + i\frac{9}{10}$.
The conjugate of $z = a + ib$ is $\bar{z} = a - ib$.
Therefore,the conjugate is $\frac{13}{10} - i\frac{9}{10}$.

(C) Given $z = 3 + 5i$,the conjugate is $\bar{z} = 3 - 5i$.
First,calculate $z^3$:
$z^3 = (3 + 5i)^3 = 3^3 + (5i)^3 + 3(3)(5i)(3 + 5i)$
$z^3 = 27 + 125i^3 + 45i(3 + 5i)$
$z^3 = 27 - 125i + 135i + 225i^2$
Since $i^2 = -1$,$z^3 = 27 + 10i - 225 = -198 + 10i$.
Now,substitute into the expression:
$z^3 + \bar{z} + 198 = (-198 + 10i) + (3 - 5i) + 198$
$= (-198 + 198) + (10i - 5i) + 3$
$= 3 + 5i$.

(B) Let $z = \frac{2 - 3i}{4 - i}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator,which is $4 + i$:
$z = \frac{(2 - 3i)(4 + i)}{(4 - i)(4 + i)}$
$z = \frac{8 + 2i - 12i - 3i^2}{16 - i^2}$
Since $i^2 = -1$,we have:
$z = \frac{8 - 10i + 3}{16 + 1} = \frac{11 - 10i}{17}$
The conjugate of $z = a + bi$ is $\bar{z} = a - bi$.
Therefore,the conjugate of $\frac{11 - 10i}{17}$ is $\frac{11 + 10i}{17}$.

(C) Let $z = 1 + i$.
The conjugate of a complex number $z = a + ib$ is defined as $\bar{z} = a - ib$.
Therefore,for $z = 1 + i$,the conjugate is $\bar{z} = 1 - i$.

(B) We use the property of complex numbers: $|z|^2 = z \bar{z}$.
$|{z_1} + {z_2}|^2 + |{z_1} - {z_2}|^2 = ({z_1} + {z_2})(\overline{{z_1} + {z_2}}) + ({z_1} - {z_2})(\overline{{z_1} - {z_2}})$
$= ({z_1} + {z_2})(\bar{{z_1}} + \bar{{z_2}}) + ({z_1} - {z_2})(\bar{{z_1}} - \bar{{z_2}})$
$= ({z_1}\bar{{z_1}} + {z_1}\bar{{z_2}} + {z_2}\bar{{z_1}} + {z_2}\bar{{z_2}}) + ({z_1}\bar{{z_1}} - {z_1}\bar{{z_2}} - {z_2}\bar{{z_1}} + {z_2}\bar{{z_2}})$
$= 2{z_1}\bar{{z_1}} + 2{z_2}\bar{{z_2}}$
$= 2|{z_1}|^2 + 2|{z_2}|^2$

(C) For any complex number $z = x + iy$,we have $|z| = \sqrt{x^2 + y^2}$.
$(a)$ $|z^2| = |(x + iy)^2| = |x^2 - y^2 + 2ixy| = \sqrt{(x^2 - y^2)^2 + (2xy)^2} = \sqrt{x^4 + y^4 - 2x^2y^2 + 4x^2y^2} = \sqrt{(x^2 + y^2)^2} = x^2 + y^2 = |z|^2$. This is true.
$(b)$ Since $|z| = |\bar{z}|$,it follows that $|z|^2 = |\bar{z}|^2$. Thus,$|z^2| = |\bar{z}|^2$ is true.
$(c)$ $z = \bar{z}$ implies $x + iy = x - iy$,which means $2iy = 0$,or $y = 0$. This is only true if $z$ is purely real. It is not true for all complex numbers $z$.
$(d)$ $\bar{z^2} = (\bar{z})^2$ is a standard property of complex conjugates. This is true.
Therefore,the statement that is not true is $z = \bar{z}$.

(A) Let the two complex numbers be $z_1$ and $z_2$ such that $|z_1| = 1$ and $|z_2| = 1$.
Using the property of the modulus of the product of complex numbers,we have $|z_1 z_2| = |z_1| |z_2|$.
Substituting the given values,we get $|z_1 z_2| = 1 \times 1 = 1$.
Thus,the product $z_1 z_2$ is also a complex number with unit modulus.

(A) Suppose there exists a complex number $z$ such that $z^4 + z + 2 = 0$ and $|z| < 1$.
Then $z^4 + z = -2$.
Taking the modulus on both sides,we get $|z^4 + z| = |-2| = 2$.
By the triangle inequality,$|z^4 + z| \le |z^4| + |z| = |z|^4 + |z|$.
Since $|z| < 1$,it follows that $|z|^4 < 1$ and $|z| < 1$.
Therefore,$|z^4 + z| < 1 + 1 = 2$.
This implies $2 < 2$,which is a contradiction.
Thus,the equation cannot have a root such that $|z| < 1$.

(C) Given $|z_k| = 1$ for $k = 1, 2, \dots, n$.
Since $|z_k|^2 = z_k \overline{z_k} = 1$,we have $\overline{z_k} = \frac{1}{z_k}$.
We know that $|z| = |\overline{z}|$.
Therefore,$|z_1 + z_2 + \dots + z_n| = |\overline{z_1 + z_2 + \dots + z_n}|$.
Using the property of conjugates,$|\overline{z_1} + \overline{z_2} + \dots + \overline{z_n}| = \left| \frac{1}{z_1} + \frac{1}{z_2} + \dots + \frac{1}{z_n} \right|$.
Thus,$|z_1 + z_2 + \dots + z_n| = \left| \frac{1}{z_1} + \frac{1}{z_2} + \dots + \frac{1}{z_n} \right|$.

(B) Given that $\bar{z} = \frac{1}{z}$.
Multiplying both sides by $z$,we get $z \cdot \bar{z} = 1$.
We know that for any complex number $z$,$z \cdot \bar{z} = |z|^2$.
Therefore,$|z|^2 = 1$.
Since the modulus $|z|$ must be non-negative,we have $|z| = 1$.

(A) According to the triangle inequality for complex numbers,for any two complex numbers ${z_1}$ and ${z_2}$,we have the property $|{z_1} - {z_2}| \ge ||{z_1}| - |{z_2}||$.
This implies that $|{z_1} - {z_2}| \ge |{z_1}| - |{z_2}|$ and $|{z_1} - {z_2}| \ge |{z_2}| - |{z_1}|$.
Therefore,the correct relation is $|{z_1} - {z_2}| \ge |{z_1}| - |{z_2}|$.

(A) Given $z = x + iy$.
Then,$z - 5 = (x + iy) - 5 = (x - 5) + iy$.
The modulus of a complex number $a + ib$ is defined as $\sqrt{a^2 + b^2}$.
Therefore,$|z - 5| = |(x - 5) + iy| = \sqrt{(x - 5)^2 + y^2}$.

(C) We use the property of the modulus of complex numbers: $|z_1 z_2| = |z_1| |z_2|$ and $|\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|}$.
$\left| {(1 + i)\frac{{(2 + i)}}{{(3 + i)}}} \right| = |1 + i| \times \frac{|2 + i|}{|3 + i|}$
$= \sqrt{1^2 + 1^2} \times \frac{\sqrt{2^2 + 1^2}}{\sqrt{3^2 + 1^2}}$
$= \sqrt{2} \times \frac{\sqrt{5}}{\sqrt{10}}$
$= \frac{\sqrt{10}}{\sqrt{10}} = 1$.

(A) Let $z = \frac{3 + 2i}{3 - 2i}$.
Using the property of modulus,$|\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|}$.
$|z| = \left| \frac{3 + 2i}{3 - 2i} \right| = \frac{|3 + 2i|}{|3 - 2i|}$.
Since $|a + bi| = \sqrt{a^2 + b^2}$,we have $|3 + 2i| = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$.
Similarly,$|3 - 2i| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$.
Therefore,$|z| = \frac{\sqrt{13}}{\sqrt{13}} = 1$.

(A) Given $|z| = 1$,let $z = x + iy$,so $x^2 + y^2 = 1$.
$\omega = \frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}$.
To find the real part,multiply the numerator and denominator by the conjugate of the denominator $(x + 1) - iy$:
$\omega = \frac{((x - 1) + iy)((x + 1) - iy)}{(x + 1)^2 + y^2} = \frac{(x^2 - 1) + i(y(x + 1) - y(x - 1)) + y^2}{(x + 1)^2 + y^2}$.
$\omega = \frac{(x^2 + y^2 - 1) + i(xy + y - xy + y)}{(x + 1)^2 + y^2} = \frac{(1 - 1) + 2iy}{(x + 1)^2 + y^2} = \frac{2iy}{(x + 1)^2 + y^2}$.
Since the real part is $0$,$\text{Re}(\omega) = 0$.

(B) Given $|{z_1}| = 1$ and ${z_2}$ is any complex number.
Consider the expression $\left| \frac{{z_1 - z_2}}{{1 - z_1 \bar{z}_2}} \right|$.
Since $|{z_1}| = 1$,we have ${z_1} \bar{z}_1 = |{z_1}|^2 = 1$,which implies $\bar{z}_1 = \frac{1}{z_1}$.
Substituting this into the denominator: $\left| 1 - z_1 \bar{z}_2 \right| = \left| z_1 \bar{z}_1 - z_1 \bar{z}_2 \right| = |z_1| |\bar{z}_1 - \bar{z}_2|$.
Since $|z_1| = 1$,this simplifies to $|\bar{z}_1 - \bar{z}_2| = |\overline{z_1 - z_2}|$.
Using the property $|\bar{z}| = |z|$,we get $|\overline{z_1 - z_2}| = |z_1 - z_2|$.
Therefore,$\left| \frac{{z_1 - z_2}}{{1 - z_1 \bar{z}_2}} \right| = \frac{|z_1 - z_2|}{|z_1| |\bar{z}_1 - \bar{z}_2|} = \frac{|z_1 - z_2|}{|z_1 - z_2|} = 1$.

(D) Let $z = \frac{1 + i}{1 - i}$.
Multiplying the numerator and denominator by the conjugate of the denominator $(1 + i)$:
$z = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{1^2 - i^2} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i$.
We can write $z = 0 + 1i$.
The modulus is $|z| = \sqrt{0^2 + 1^2} = 1$.
The argument $\theta$ satisfies $\cos \theta = 0$ and $\sin \theta = 1$,which gives $\theta = \frac{\pi}{2}$.
Thus,the argument is $\frac{\pi}{2}$ and the modulus is $1$.

(D) Let $z = x + iy$,then $\bar{z} = x - iy$.
Since $\arg(z) = \theta = \tan^{-1}(\frac{y}{x})$,
$\arg(\bar{z}) = \tan^{-1}(\frac{-y}{x}) = -\theta$.
Thus,$\arg(z) \neq \arg(\bar{z})$.
Therefore,the relation $\arg(z) = \arg(\bar{z})$ is false.

(B) Let $z_1 = r(\cos \theta_1 + i \sin \theta_1)$.
Given $|z_1| = |z_2|$,so $|z_2| = r$.
Given $\text{amp}(z_1) + \text{amp}(z_2) = 0$,so $\text{amp}(z_2) = -\theta_1$.
Thus,$z_2 = r(\cos(-\theta_1) + i \sin(-\theta_1)) = r(\cos \theta_1 - i \sin \theta_1)$.
Since $\bar{z}_1 = r(\cos \theta_1 - i \sin \theta_1)$,it follows that $\bar{z}_1 = z_2$.

(B) Given $z = \cos \frac{\pi}{6} + i\sin \frac{\pi}{6}$.
This is in the polar form $z = r(\cos \theta + i\sin \theta)$,where $r = |z|$ and $\theta = \arg(z)$.
Comparing the given expression with the polar form,we get $r = |z| = 1$ and $\theta = \arg(z) = \frac{\pi}{6}$.
Thus,$|z| = 1$ and $\arg(z) = \frac{\pi}{6}$.

(C) Given: $(\sqrt{8} + i)^{50} = 3^{49}(a + ib)$.
Taking the modulus on both sides:
$|(\sqrt{8} + i)^{50}| = |3^{49}(a + ib)|$
$|\sqrt{8} + i|^{50} = 3^{49} |a + ib|$
Since $|\sqrt{8} + i| = \sqrt{(\sqrt{8})^2 + 1^2} = \sqrt{8 + 1} = \sqrt{9} = 3$,we have:
$3^{50} = 3^{49} \sqrt{a^2 + b^2}$
Squaring both sides:
$(3^{50})^2 = (3^{49})^2 (a^2 + b^2)$
$3^{100} = 3^{98} (a^2 + b^2)$
$a^2 + b^2 = \frac{3^{100}}{3^{98}} = 3^2 = 9$.

(A) Given $x + iy = \sqrt{\frac{a + ib}{c + id}}$.
Taking the conjugate on both sides,we get $x - iy = \sqrt{\frac{a - ib}{c - id}}$.
Now,multiplying these two equations:
$(x + iy)(x - iy) = \sqrt{\frac{a + ib}{c + id}} \times \sqrt{\frac{a - ib}{c - id}}$
$x^2 + y^2 = \sqrt{\frac{(a + ib)(a - ib)}{(c + id)(c - id)}}$
$x^2 + y^2 = \sqrt{\frac{a^2 + b^2}{c^2 + d^2}}$
Squaring both sides,we get $(x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2}$.

(D) Given the equation $|1 - i|^x = 2^x$.
First,calculate the modulus of the complex number $1 - i$:
$|1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.
Substitute this into the equation:
$(\sqrt{2})^x = 2^x$.
This can be written as:
$(2^{1/2})^x = 2^x$
$2^{x/2} = 2^x$.
Equating the exponents:
$\frac{x}{2} = x$
$x = 0$.
The only solution is $x = 0$,which is not a non-zero integer.
Therefore,the number of non-zero integral solutions is $0$.

(B) Given $z = r{e^{i\theta }} = r(\cos \theta + i\sin \theta )$.
Then $iz = ir(\cos \theta + i\sin \theta ) = -r\sin \theta + ir\cos \theta$.
Therefore,${e^{iz}} = {e^{-r\sin \theta + ir\cos \theta}} = {e^{-r\sin \theta}} \cdot {e^{i(r\cos \theta)}}$.
Taking the modulus on both sides:
$|{e^{iz}}| = |{e^{-r\sin \theta}}| \cdot |{e^{i(r\cos \theta)}}|$.
Since ${e^{-r\sin \theta}}$ is a real number,$|{e^{-r\sin \theta}}| = {e^{-r\sin \theta}}$.
Also,$|{e^{i(r\cos \theta)}}| = |\cos(r\cos \theta) + i\sin(r\cos \theta)| = \sqrt{\cos^2(r\cos \theta) + \sin^2(r\cos \theta)} = 1$.
Thus,$|{e^{iz}}| = {e^{-r\sin \theta}} \cdot 1 = {e^{-r\sin \theta}}$.

(B) To simplify $\frac{1 - i}{1 + i}$,we multiply the numerator and denominator by the conjugate of the denominator,which is $(1 - i)$:
$\frac{1 - i}{1 + i} = \frac{(1 - i)(1 - i)}{(1 + i)(1 - i)}$
$= \frac{1 - 2i + i^2}{1^2 - i^2}$
$= \frac{1 - 2i - 1}{1 + 1}$
$= \frac{-2i}{2} = -i$
Now,expressing $-i$ in polar form $\cos \theta + i \sin \theta$:
Since $\cos \frac{\pi}{2} = 0$ and $\sin \frac{\pi}{2} = 1$,we have:
$-i = 0 - i(1) = \cos \frac{\pi}{2} - i \sin \frac{\pi}{2}$

(D) Given $x + \frac{1}{x} = \sqrt{3}$.
Multiplying by $x$,we get $x^2 - \sqrt{3}x + 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we have:
$x = \frac{\sqrt{3} \pm \sqrt{(\sqrt{3})^2 - 4(1)(1)}}{2(1)}$
$x = \frac{\sqrt{3} \pm \sqrt{3 - 4}}{2} = \frac{\sqrt{3} \pm \sqrt{-1}}{2} = \frac{\sqrt{3} \pm i}{2}$.
Thus,$x = \frac{\sqrt{3}}{2} \pm \frac{i}{2}$.
Taking the positive sign,$x = \frac{\sqrt{3}}{2} + i \frac{1}{2} = \cos \frac{\pi}{6} + i \sin \frac{\pi}{6}$.

(B) Given $\left| \omega + \frac{1}{\omega} \right| = 2$.
Using the triangle inequality $\left| z_1 + z_2 \right| \le \left| z_1 \right| + \left| z_2 \right|$,we have:
$\left| \omega \right| = \left| \left( \omega + \frac{1}{\omega} \right) - \frac{1}{\omega} \right| \le \left| \omega + \frac{1}{\omega} \right| + \left| \frac{1}{\omega} \right|$.
Substituting the given value:
$\left| \omega \right| \le 2 + \frac{1}{\left| \omega \right|}$.
Let $r = \left| \omega \right|$. Then $r \le 2 + \frac{1}{r}$,which implies $r^2 - 2r - 1 \le 0$.
Solving the quadratic equation $r^2 - 2r - 1 = 0$ using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$r = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$.
Since $r = \left| \omega \right| > 0$,we take $r \le 1 + \sqrt{2}$.
Thus,the maximum distance of $\omega$ from the origin is $1 + \sqrt{2}$.

(A) 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.
Mathematically,this is expressed as $|{z_1} + {z_2}| \le |{z_1}| + |{z_2}|$.
Thus,the correct option is $A$.

(D) The triangle inequality for complex numbers states that for any two complex numbers ${z_1}$ and ${z_2}$,the following holds:
$|{z_1} + {z_2}| \le |{z_1}| + |{z_2}|$
$|{z_1} - {z_2}| \ge ||{z_1}| - |{z_2}||$
$|{z_1} + {z_2}| \ge ||{z_1}| - |{z_2}||$
Among the given options,the inequality $|{z_1} + {z_2}| \ge ||{z_1}| - |{z_2}||$ is a standard property derived from the triangle inequality.

(C) The triangle inequality for absolute values states that for any real numbers $x$ and $y$,$|x - y| \ge |x| - |y|$.
This is a fundamental property of the modulus function.
Therefore,the correct statement is $|x - y| \ge |x| - |y|$.

(B) We use the property $|z|^2 = z \cdot \overline{z}$.
$|az_1 - bz_2|^2 + |bz_1 + az_2|^2 = (az_1 - bz_2)(\overline{az_1 - bz_2}) + (bz_1 + az_2)(\overline{bz_1 + az_2})$
$= (az_1 - bz_2)(a\overline{z_1} - b\overline{z_2}) + (bz_1 + az_2)(b\overline{z_1} + a\overline{z_2})$
$= (a^2|z_1|^2 - ab z_1\overline{z_2} - ab \overline{z_1}z_2 + b^2|z_2|^2) + (b^2|z_1|^2 + ab z_1\overline{z_2} + ab \overline{z_1}z_2 + a^2|z_2|^2)$
$= a^2|z_1|^2 + b^2|z_2|^2 + b^2|z_1|^2 + a^2|z_2|^2$
$= (a^2 + b^2)|z_1|^2 + (a^2 + b^2)|z_2|^2$
$= (a^2 + b^2)(|z_1|^2 + |z_2|^2)$

(A) Given that $|{z_1}| = |{z_2}| = |{z_3}| = 1$.
Since $|{z_i}| = 1$,we have $|{z_i}|^2 = {z_i} \overline{z_i} = 1$,which implies $\frac{1}{z_i} = \overline{z_i}$ for $i = 1, 2, 3$.
Given $\left| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right| = 1$.
Substituting $\frac{1}{z_i} = \overline{z_i}$,we get $|\overline{z_1} + \overline{z_2} + \overline{z_3}| = 1$.
Using the property of conjugates $|\overline{z}| = |z|$,we have $|\overline{z_1 + z_2 + z_3}| = |z_1 + z_2 + z_3| = 1$.
Therefore,$|{z_1} + {z_2} + {z_3}| = 1$.

(A) To find the value of the expression $\left| \frac{3x^3 + 1}{2x^2 + 2} \right|$ at $x = -3$,we substitute $x = -3$ into the expression:
Step $1$: Calculate the numerator: $3(-3)^3 + 1 = 3(-27) + 1 = -81 + 1 = -80$.
Step $2$: Calculate the denominator: $2(-3)^2 + 2 = 2(9) + 2 = 18 + 2 = 20$.
Step $3$: Divide the numerator by the denominator: $\frac{-80}{20} = -4$.
Step $4$: Apply the absolute value: $|-4| = 4$.
Therefore,the numerical value is $4$.