Conjugate and Modulus of complex numbers Questions in English

(C) By the triangle inequality for absolute values,we have the property: $||x| - |y|| \leq |x - y| \leq |x| + |y|$.
Specifically,the identity $|x - y| = ||x| - |y||$ is not always true for all real $x$ and $y$.
However,checking the standard properties of absolute values:
$1$. $|x + y| \leq |x| + |y|$
$2$. $|x - y| \geq ||x| - |y||$
Looking at the options provided,none of the equalities are universally true for all real $x$ and $y$. However,in many textbook contexts,the question likely intends to test the reverse triangle inequality property: $||x| - |y|| \leq |x - y|$.
Given the standard form of such questions,if we assume the question implies which inequality holds or is a standard identity,there might be a typo in the options. But strictly mathematically,none of the equalities $A, B, C, D$ are identities. If we must choose the most 'standard' related property,$C$ is often discussed as the lower bound.

(D) Given the equation $(\sqrt{8} + i)^{50} = 3^{49}(a + ib)$.
Taking the modulus on both sides,we get $|(\sqrt{8} + i)^{50}| = |3^{49}(a + ib)|$.
Using the property $|z^n| = |z|^n$,we have $|\sqrt{8} + i|^{50} = 3^{49} |a + ib|$.
The modulus of $\sqrt{8} + i$ is $\sqrt{(\sqrt{8})^2 + 1^2} = \sqrt{8 + 1} = \sqrt{9} = 3$.
Substituting this into the equation,we get $3^{50} = 3^{49} \sqrt{a^2 + b^2}$.
Dividing both sides by $3^{49}$,we get $3 = \sqrt{a^2 + b^2}$.
Squaring both sides,we obtain $a^2 + b^2 = 3^2 = 9$.

(C) Let the complex number be $z = x + iy$. The conjugate is denoted by $\bar{z} = x - iy$.
Given that $\bar{z} = \frac{1}{i - 1} = \frac{1}{-1 + i}$.
To find $z$,we take the conjugate of $\bar{z}$:
$z = \overline{\left(\frac{1}{-1 + i}\right)} = \frac{1}{-1 - i}$.
Multiplying the numerator and denominator by $-1$,we get:
$z = \frac{-1}{1 + i} = -\frac{1}{i + 1}$.

(C) Let $z = \frac{\beta - \alpha}{1 - \overline{\alpha}\beta}$.
We calculate $|z|^2 = z \cdot \overline{z} = \left( \frac{\beta - \alpha}{1 - \overline{\alpha}\beta} \right) \left( \frac{\overline{\beta} - \overline{\alpha}}{1 - \alpha\overline{\beta}} \right)$.
Expanding the numerator: $(\beta - \alpha)(\overline{\beta} - \overline{\alpha}) = \beta\overline{\beta} - \beta\overline{\alpha} - \alpha\overline{\beta} + \alpha\overline{\alpha} = |\beta|^2 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2$.
Expanding the denominator: $(1 - \overline{\alpha}\beta)(1 - \alpha\overline{\beta}) = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + \alpha\overline{\alpha}\beta\overline{\beta} = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2|\beta|^2$.
Since $|\beta| = 1$,we have $|\beta|^2 = 1$.
Substituting $|\beta|^2 = 1$ into the expressions:
Numerator: $1 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2$.
Denominator: $1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2(1) = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2$.
Since the numerator and denominator are equal,$|z|^2 = 1$,which implies $|z| = 1$.

(D) We have $|\lambda_1 a_1 + \lambda_2 a_2 + ... + \lambda_n a_n| \le |\lambda_1 a_1| + |\lambda_2 a_2| + ... + |\lambda_n a_n|$ by the triangle inequality.
Since $\lambda_k \ge 0$,this equals $\lambda_1 |a_1| + \lambda_2 |a_2| + ... + \lambda_n |a_n|$.
Given $|a_k| < 1$ for all $k$,we have $\lambda_k |a_k| < \lambda_k$.
Summing these inequalities,we get $\sum_{k=1}^n \lambda_k |a_k| < \sum_{k=1}^n \lambda_k$.
Since $\sum_{k=1}^n \lambda_k = 1$,it follows that $|\lambda_1 a_1 + \lambda_2 a_2 + ... + \lambda_n a_n| < 1$.

(A) Since the coefficients of the polynomial ${x^3} - 5{x^2} + 9x - 5 = 0$ are real,complex roots must occur in conjugate pairs.
Given one root is $z_1 = 2 + i$,the second root must be $z_2 = 2 - i$.
Let the third root be $\alpha$.
According to Vieta's formulas,the sum of the roots is equal to the negative of the coefficient of ${x^2}$ divided by the coefficient of ${x^3}$.
Sum of roots $= z_1 + z_2 + \alpha = -(-5)/1 = 5$.
Substituting the known roots: $(2 + i) + (2 - i) + \alpha = 5$.
$4 + \alpha = 5 \Rightarrow \alpha = 1$.
Thus,the other roots are $1$ and $2 - i$.

(A) Given $|z| \geqslant 2$.
We want to find the minimum value of $|z + \frac{1}{2}|$.
By the triangle inequality,we know that $|z + w| \geqslant ||z| - |w||$.
Here,$|z + \frac{1}{2}| \geqslant ||z| - |-\frac{1}{2}|| = ||z| - \frac{1}{2}|$.
Since $|z| \geqslant 2$,the expression $|z| - \frac{1}{2}$ is at least $2 - \frac{1}{2} = \frac{3}{2}$.
Thus,$|z + \frac{1}{2}| \geqslant \frac{3}{2}$.
The minimum value is $\frac{3}{2}$ when $z$ is a real number such that $z = -2$.

(D) Given $|z - \frac{6}{z}| = 5$.
Using the triangle inequality property $||z_1| - |z_2|| \leq |z_1 - z_2|$,we have $|z| - |\frac{6}{z}| \leq |z - \frac{6}{z}|$.
Substituting the given value: $|z| - \frac{6}{|z|} \leq 5$.
Let $|z| = r$,where $r > 0$. Then $r - \frac{6}{r} \leq 5$.
Multiplying by $r$ (since $r > 0$): $r^2 - 6 \leq 5r \Rightarrow r^2 - 5r - 6 \leq 0$.
Factoring the quadratic: $(r - 6)(r + 1) \leq 0$.
Since $r > 0$,we must have $r \leq 6$.
Thus,the maximum value of $|z|$ is $6$.

(A) Let $|z| = r$. Given $\left| z - \frac{25}{z} \right| = 24$.
Using the triangle inequality property $||z_1| - |z_2|| \leq |z_1 - z_2| \leq |z_1| + |z_2|$,we have:
$|z| - \frac{25}{|z|} \leq \left| z - \frac{25}{z} \right| \leq |z| + \frac{25}{|z|}$.
From the right side of the inequality: $24 \leq r + \frac{25}{r} \implies r^2 - 24r + 25 \geq 0$.
From the left side of the inequality: $r - \frac{25}{r} \leq 24 \implies r^2 - 24r - 25 \leq 0$.
Solving $r^2 - 24r - 25 = 0$,we get $(r - 25)(r + 1) = 0$. Since $r > 0$,$r \leq 25$.
Thus,the maximum value of $|z|$ is $25$.

(C) Given the inequality: $\log_{\tan 30^{\circ}} \left( \frac{2|z|^2 + 2|z| - 3}{|z| + 1} \right) < -2$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,the base is $0 < \frac{1}{\sqrt{3}} < 1$.
When the base of a logarithm is between $0$ and $1$,the inequality sign reverses when removing the logarithm:
$\frac{2|z|^2 + 2|z| - 3}{|z| + 1} > (\frac{1}{\sqrt{3}})^{-2}$.
$\frac{2|z|^2 + 2|z| - 3}{|z| + 1} > 3$.
$2|z|^2 + 2|z| - 3 > 3(|z| + 1)$.
$2|z|^2 + 2|z| - 3 > 3|z| + 3$.
$2|z|^2 - |z| - 6 > 0$.
Factoring the quadratic expression: $(2|z| + 3)(|z| - 2) > 0$.
Since $|z| \geq 0$,$2|z| + 3$ is always positive.
Therefore,$|z| - 2 > 0$,which implies $|z| > 2$.

(B) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates if $z_1 = \overline{z_2}$.
Given $(x - 2y) + i(3x - y) = \overline{(2x - y) + i(x - y + 6)}$.
This implies $(x - 2y) + i(3x - y) = (2x - y) - i(x - y + 6)$.
Comparing the real and imaginary parts:
Real part: $x - 2y = 2x - y \Rightarrow x + y = 0$.
Imaginary part: $3x - y = -(x - y + 6)$ $\Rightarrow 3x - y = -x + y - 6$ $\Rightarrow 4x - 2y = -6$ $\Rightarrow 2x - y = -3$.
Adding the two equations: $(x + y) + (2x - y) = 0 - 3$ $\Rightarrow 3x = -3$ $\Rightarrow x = -1$.
Substituting $x = -1$ into $x + y = 0$,we get $y = 1$.
Thus,$|x + iy| = |-1 + i| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$.

(B) Let $z = x + iy$.
Given equation: $|z| + z - 3\bar{z} = 0$.
Substituting $z = x + iy$ and $\bar{z} = x - iy$:
$\sqrt{x^2 + y^2} + (x + iy) - 3(x - iy) = 0$.
$\sqrt{x^2 + y^2} + x + iy - 3x + 3iy = 0$.
$\sqrt{x^2 + y^2} - 2x + 4iy = 0$.
Equating the real and imaginary parts to zero:
Imaginary part: $4y = 0 \implies y = 0$.
Real part: $\sqrt{x^2 + 0^2} - 2x = 0 \implies |x| = 2x$.
If $x \ge 0$,then $x = 2x \implies x = 0$.
If $x < 0$,then $-x = 2x \implies 3x = 0$,which contradicts $x < 0$.
Thus,the only solution is $x = 0$ and $y = 0$,which means $z = 0$.
Therefore,there is only $1$ such complex number.

(B) Let $z = \frac{1 + i \cos \theta}{1 - 2i \cos \theta}$.
To make $z$ a real number,we multiply the numerator and denominator by the conjugate of the denominator,which is $1 + 2i \cos \theta$:
$z = \frac{(1 + i \cos \theta)(1 + 2i \cos \theta)}{(1 - 2i \cos \theta)(1 + 2i \cos \theta)}$
$z = \frac{1 + 2i \cos \theta + i \cos \theta + 2i^2 \cos^2 \theta}{1 + 4 \cos^2 \theta}$
Since $i^2 = -1$,we have:
$z = \frac{1 - 2 \cos^2 \theta + 3i \cos \theta}{1 + 4 \cos^2 \theta}$
For $z$ to be a real number,the imaginary part must be zero:
$\frac{3 \cos \theta}{1 + 4 \cos^2 \theta} = 0$
This implies $\cos \theta = 0$.
The general solution for $\cos \theta = 0$ is $\theta = (2n + 1)\frac{\pi}{2}$ for $n \in I$.
Thus,the correct option is $(B)$.

(D) Given,$z_1 = 1+2i$,$z_2 = 3+5i$,and $\bar{z}_2 = 3-5i$.
First,calculate the numerator: $\bar{z}_2 z_1 = (3-5i)(1+2i) = 3 + 6i - 5i - 10i^2 = 3 + i + 10 = 13+i$.
Now,calculate the expression: $\frac{\bar{z}_2 z_1}{z_2} = \frac{13+i}{3+5i}$.
Multiply the numerator and denominator by the conjugate of the denominator $(3-5i)$:
$\frac{13+i}{3+5i} \times \frac{3-5i}{3-5i} = \frac{39 - 65i + 3i - 5i^2}{3^2 + 5^2} = \frac{39 - 62i + 5}{9 + 25} = \frac{44 - 62i}{34}$.
Simplifying the fraction: $\frac{44}{34} - \frac{62}{34}i = \frac{22}{17} - \frac{31}{17}i$.
Therefore,$\text{Re} \left( \frac{\bar{z}_2 z_1}{z_2} \right) = \frac{22}{17}$.

(B) Statement $1$ is the triangle inequality for subtraction,which states $|Z_1 - Z_2| \ge ||Z_1| - |Z_2||$,and since $||Z_1| - |Z_2|| \ge |Z_1| - |Z_2|$,Statement $1$ is true.
Statement $2$ is the standard triangle inequality for addition,which is also a fundamental property of complex numbers and is true.
However,Statement $2$ does not explain Statement $1$ as they are independent properties of the modulus of complex numbers.

(B) We use the property of the modulus of complex numbers: $\left| z \right|^2 = z \bar{z}$.
Expanding the terms:
${\left| {{z_1} + {z_2}} \right|^2} = (z_1 + z_2)(\bar{z_1} + \bar{z_2}) = z_1\bar{z_1} + z_1\bar{z_2} + z_2\bar{z_1} + z_2\bar{z_2} = \left| {{z_1}} \right|^2 + \left| {{z_2}} \right|^2 + z_1\bar{z_2} + z_2\bar{z_1}$.
Similarly:
${\left| {{z_1} - {z_2}} \right|^2} = (z_1 - z_2)(\bar{z_1} - \bar{z_2}) = z_1\bar{z_1} - z_1\bar{z_2} - z_2\bar{z_1} + z_2\bar{z_2} = \left| {{z_1}} \right|^2 + \left| {{z_2}} \right|^2 - z_1\bar{z_2} - z_2\bar{z_1}$.
Adding these two expressions:
${\left| {{z_1} + {z_2}} \right|^2} + {\left| {{z_1} - {z_2}} \right|^2} = (\left| {{z_1}} \right|^2 + \left| {{z_2}} \right|^2 + z_1\bar{z_2} + z_2\bar{z_1}) + (\left| {{z_1}} \right|^2 + \left| {{z_2}} \right|^2 - z_1\bar{z_2} - z_2\bar{z_1})$
$= 2\left| {{z_1}} \right|^2 + 2\left| {{z_2}} \right|^2 = 2(\left| {{z_1}} \right|^2 + \left| {{z_2}} \right|^2)$.

(B) Let $z = x + iy$.
Given $|z| + z = 3 + i$.
Substituting $z = x + iy$,we get $\sqrt{x^2 + y^2} + x + iy = 3 + i$.
Comparing the imaginary parts,we get $y = 1$.
Comparing the real parts,we get $\sqrt{x^2 + 1} + x = 3$.
$\sqrt{x^2 + 1} = 3 - x$.
Squaring both sides,$x^2 + 1 = (3 - x)^2 = 9 - 6x + x^2$.
$1 = 9 - 6x$,which implies $6x = 8$,so $x = \frac{4}{3}$.
Now,$|z| = \sqrt{x^2 + y^2} = \sqrt{(\frac{4}{3})^2 + 1^2} = \sqrt{\frac{16}{9} + 1} = \sqrt{\frac{25}{9}} = \frac{5}{3}$.

(B) Given $z = \frac{(1 + i)^2}{a - i} = \frac{2i}{a - i}$.
Multiplying numerator and denominator by the conjugate $(a + i)$,we get $z = \frac{2i(a + i)}{a^2 + 1} = \frac{-2 + 2ai}{a^2 + 1}$.
The magnitude is $|z| = \frac{|2i|}{|a - i|} = \frac{2}{\sqrt{a^2 + 1}}$.
Given $|z| = \sqrt{\frac{2}{5}}$,so $\frac{2}{\sqrt{a^2 + 1}} = \sqrt{\frac{2}{5}} \implies \frac{4}{a^2 + 1} = \frac{2}{5} \implies a^2 + 1 = 10 \implies a^2 = 9$.
Since $a > 0$,we have $a = 3$.
Substituting $a = 3$ into $z$,we get $z = \frac{2i(3 + i)}{3^2 + 1} = \frac{6i - 2}{10} = -\frac{1}{5} + \frac{3}{5}i$.
The conjugate $\bar{z}$ is $-\frac{1}{5} - \frac{3}{5}i$.

(A) Let $z = 2-3i$.
The multiplicative inverse of $z$ is given by $z^{-1} = \frac{1}{z} = \frac{1}{2-3i}$.
To simplify,multiply the numerator and denominator by the conjugate of $z$,which is $2+3i$:
$z^{-1} = \frac{1(2+3i)}{(2-3i)(2+3i)}$
Using the identity $(a-bi)(a+bi) = a^2+b^2$:
$z^{-1} = \frac{2+3i}{2^2 + (-3)^2} = \frac{2+3i}{4+9} = \frac{2+3i}{13}$
Thus,$z^{-1} = \frac{2}{13} + \frac{3}{13}i$.

(A) Let $z = 4 - 3i$.
The multiplicative inverse of $z$ is given by $z^{-1} = \frac{1}{z} = \frac{1}{4 - 3i}$.
To simplify,multiply the numerator and denominator by the conjugate $(4 + 3i)$:
$z^{-1} = \frac{1(4 + 3i)}{(4 - 3i)(4 + 3i)}$.
Using the identity $(a - b)(a + b) = a^2 - b^2$:
$z^{-1} = \frac{4 + 3i}{4^2 - (3i)^2} = \frac{4 + 3i}{16 - 9i^2}$.
Since $i^2 = -1$:
$z^{-1} = \frac{4 + 3i}{16 + 9} = \frac{4 + 3i}{25}$.
Thus,$z^{-1} = \frac{4}{25} + \frac{3}{25}i$.

(A) Let $z = \sqrt{5} + 3i$.
The multiplicative inverse of $z$ is given by $z^{-1} = \frac{1}{z} = \frac{1}{\sqrt{5} + 3i}$.
To simplify,multiply the numerator and denominator by the conjugate $\sqrt{5} - 3i$:
$z^{-1} = \frac{\sqrt{5} - 3i}{(\sqrt{5} + 3i)(\sqrt{5} - 3i)}$.
Using the identity $(a+bi)(a-bi) = a^2 + b^2$:
$z^{-1} = \frac{\sqrt{5} - 3i}{(\sqrt{5})^2 + 3^2} = \frac{\sqrt{5} - 3i}{5 + 9} = \frac{\sqrt{5} - 3i}{14}$.
Thus,$z^{-1} = \frac{\sqrt{5}}{14} - \frac{3i}{14}$.

(A) Let $z = \frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$.
First,simplify the numerator: $(3-2 i)(2+3 i) = 6 + 9i - 4i - 6i^2 = 6 + 5i + 6 = 12 + 5i$.
Next,simplify the denominator: $(1+2 i)(2-i) = 2 - i + 4i - 2i^2 = 2 + 3i + 2 = 4 + 3i$.
So,$z = \frac{12+5i}{4+3i}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator $(4-3i)$:
$z = \frac{(12+5i)(4-3i)}{(4+3i)(4-3i)} = \frac{48 - 36i + 20i - 15i^2}{16 - 9i^2} = \frac{48 - 16i + 15}{16 + 9} = \frac{63 - 16i}{25} = \frac{63}{25} - \frac{16}{25}i$.
The conjugate of $z = a + bi$ is $\bar{z} = a - bi$.
Therefore,the conjugate of $\frac{63}{25} - \frac{16}{25}i$ is $\frac{63}{25} + \frac{16}{25}i$.

(A) Let $z = \frac{1}{1+i}$.
Multiplying the numerator and denominator by the conjugate $(1-i)$,we get:
$z = \frac{1(1-i)}{(1+i)(1-i)} = \frac{1-i}{1^2 - i^2} = \frac{1-i}{1+1} = \frac{1}{2} - \frac{1}{2}i$.
Here,the real part $x = \frac{1}{2}$ and the imaginary part $y = -\frac{1}{2}$.
The modulus $r = |z| = \sqrt{x^2 + y^2} = \sqrt{(\frac{1}{2})^2 + (-\frac{1}{2})^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
For the argument $\theta$,since $x > 0$ and $y < 0$,the complex number lies in the fourth quadrant.
$\tan \theta = |\frac{y}{x}| = |\frac{-1/2}{1/2}| = |-1| = 1$.
Since $\theta$ is in the fourth quadrant,$\theta = -\tan^{-1}(1) = -\frac{\pi}{4}$.
Thus,the modulus is $\frac{1}{\sqrt{2}}$ and the argument is $-\frac{\pi}{4}$.

(N/A) Given that $x+iy = \frac{a+ib}{a-ib}$.
Taking the modulus on both sides,we have $|x+iy| = \left|\frac{a+ib}{a-ib}\right|$.
Since $|z_1/z_2| = |z_1|/|z_2|$,we get $\sqrt{x^2+y^2} = \frac{|a+ib|}{|a-ib|}$.
We know that $|a+ib| = \sqrt{a^2+b^2}$ and $|a-ib| = \sqrt{a^2+(-b)^2} = \sqrt{a^2+b^2}$.
Thus,$\sqrt{x^2+y^2} = \frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}} = 1$.
Squaring both sides,we get $x^2+y^2 = 1^2 = 1$.

(N/A) Given $x-iy = \sqrt{\frac{a-ib}{c-id}}$.
Taking the modulus on both sides,we have $|x-iy| = \left|\sqrt{\frac{a-ib}{c-id}}\right|$.
Since $|z_1/z_2| = |z_1|/|z_2|$ and $|\sqrt{z}| = \sqrt{|z|}$,we get $|x-iy| = \sqrt{\frac{|a-ib|}{|c-id|}}$.
We know that $|x-iy| = \sqrt{x^2+y^2}$,$|a-ib| = \sqrt{a^2+b^2}$,and $|c-id| = \sqrt{c^2+d^2}$.
Substituting these values,we get $\sqrt{x^2+y^2} = \sqrt{\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}}$.
Squaring both sides,we get $x^2+y^2 = \sqrt{\frac{a^2+b^2}{c^2+d^2}}$.
Squaring again,we obtain $(x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2}$.
Hence,proved.

(B) Given $z_{1} = 2 - i$ and $z_{2} = 1 + i$.
First,calculate the numerator: $z_{1} + z_{2} + 1 = (2 - i) + (1 + i) + 1 = 4$.
Next,calculate the denominator: $z_{1} - z_{2} + 1 = (2 - i) - (1 + i) + 1 = 2 - i - 1 - i + 1 = 2 - 2i$.
Now,substitute these into the expression: $\left| \frac{4}{2 - 2i} \right| = \left| \frac{4}{2(1 - i)} \right| = \left| \frac{2}{1 - i} \right|$.
To simplify,multiply the numerator and denominator by the conjugate $(1 + i)$: $\left| \frac{2(1 + i)}{(1 - i)(1 + i)} \right| = \left| \frac{2(1 + i)}{1^2 - i^2} \right|$.
Since $i^2 = -1$,we have $\left| \frac{2(1 + i)}{1 + 1} \right| = \left| \frac{2(1 + i)}{2} \right| = |1 + i|$.
The modulus is $|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}$.

(A) Given $z_{1} = 2 - i$ and $z_{2} = -2 + i$.
First, calculate the product $z_{1} z_{2} = (2 - i)(-2 + i) = -4 + 2i + 2i - i^{2} = -4 + 4i - (-1) = -3 + 4i$.
The conjugate $\bar{z}_{1} = 2 + i$.
Now, consider the expression $\frac{z_{1} z_{2}}{\bar{z}_{1}} = \frac{-3 + 4i}{2 + i}$.
To simplify, multiply the numerator and denominator by the conjugate of the denominator, $(2 - i)$:
$\frac{(-3 + 4i)(2 - i)}{(2 + i)(2 - i)} = \frac{-6 + 3i + 8i - 4i^{2}}{2^{2} + 1^{2}} = \frac{-6 + 11i - 4(-1)}{4 + 1} = \frac{-6 + 11i + 4}{5} = \frac{-2 + 11i}{5} = \frac{-2}{5} + \frac{11}{5}i$.
Comparing the real part, we get $\operatorname{Re}\left(\frac{z_{1} z_{2}}{\bar{z}_{1}}\right) = \frac{-2}{5}$.

(A) Given $z_{1} = 2 - i$.
Then the conjugate $\bar{z}_{1} = 2 + i$.
We know that $z_{1} \bar{z}_{1} = |z_{1}|^2 = (2)^2 + (-1)^2 = 4 + 1 = 5$.
Therefore,$\frac{1}{z_{1} \bar{z}_{1}} = \frac{1}{5}$.
Since $\frac{1}{5}$ is a purely real number,its imaginary part is $0$.
Thus,$\operatorname{Im}\left(\frac{1}{z_{1} \bar{z}_{1}}\right) = 0$.

(N/A) Let $z = \frac{1+2i}{1-3i}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator,$1+3i$:
$z = \frac{1+2i}{1-3i} \times \frac{1+3i}{1+3i} = \frac{1 + 3i + 2i + 6i^2}{1^2 + 3^2} = \frac{1 + 5i - 6}{1 + 9} = \frac{-5 + 5i}{10} = -\frac{1}{2} + \frac{1}{2}i$.
Let $z = r(\cos \theta + i \sin \theta)$.
Then $r = |z| = \sqrt{(-\frac{1}{2})^2 + (\frac{1}{2})^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
For the argument $\theta$,we have $\cos \theta = \frac{-1/2}{1/\sqrt{2}} = -\frac{1}{\sqrt{2}}$ and $\sin \theta = \frac{1/2}{1/\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Since $\cos \theta < 0$ and $\sin \theta > 0$,$\theta$ lies in the $II$ quadrant.
Thus,$\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
The modulus is $\frac{1}{\sqrt{2}}$ and the argument is $\frac{3\pi}{4}$.

(A) Let $z = (x-iy)(3+5i)$.
$z = 3x + 5xi - 3yi - 5yi^2 = 3x + 5xi - 3yi + 5y = (3x+5y) + i(5x-3y)$.
Therefore,the conjugate $\bar{z} = (3x+5y) - i(5x-3y)$.
It is given that $\bar{z} = -6 - 24i$.
Equating real and imaginary parts,we get:
$3x + 5y = -6$ $(i)$
$5x - 3y = 24$ $(ii)$
Multiplying $(i)$ by $3$ and $(ii)$ by $5$,we get:
$9x + 15y = -18$
$25x - 15y = 120$
Adding these equations: $34x = 102$,so $x = 3$.
Substituting $x=3$ into $(i)$: $3(3) + 5y = -6$ $\Rightarrow 9 + 5y = -6$ $\Rightarrow 5y = -15$ $\Rightarrow y = -3$.
Thus,$x=3$ and $y=-3$.

(C) Let $z = \frac{1+i}{1-i}-\frac{1-i}{1+i}$.
Simplifying the expression:
$z = \frac{(1+i)^2 - (1-i)^2}{(1-i)(1+i)}$
Using the identities $(1+i)^2 = 1 + i^2 + 2i = 2i$ and $(1-i)^2 = 1 + i^2 - 2i = -2i$:
$z = \frac{2i - (-2i)}{1^2 - i^2} = \frac{4i}{1 - (-1)} = \frac{4i}{2} = 2i$.
The modulus is $|z| = |2i| = \sqrt{0^2 + 2^2} = 2$.

(B) Let $z = \frac{\beta-\alpha}{1-\bar{\alpha} \beta}.$ We want to find $|z|.$
Consider $|z|^2 = z \cdot \bar{z} = \left( \frac{\beta-\alpha}{1-\bar{\alpha} \beta} \right) \left( \frac{\bar{\beta}-\bar{\alpha}}{1-\alpha \bar{\beta}} \right).$
Since $|\beta|=1,$ we have $\beta \bar{\beta} = |\beta|^2 = 1,$ so $\bar{\beta} = \frac{1}{\beta}.$
Substituting this,$|z|^2 = \frac{(\beta-\alpha)(\bar{\beta}-\bar{\alpha})}{(1-\bar{\alpha} \beta)(1-\alpha \bar{\beta})} = \frac{\beta \bar{\beta} - \beta \bar{\alpha} - \alpha \bar{\beta} + \alpha \bar{\alpha}}{1 - \alpha \bar{\beta} - \bar{\alpha} \beta + \alpha \bar{\alpha} \beta \bar{\beta}}.$
Using $\beta \bar{\beta} = 1,$ the numerator becomes $1 - \beta \bar{\alpha} - \alpha \bar{\beta} + |\alpha|^2.$
The denominator becomes $1 - \alpha \bar{\beta} - \bar{\alpha} \beta + |\alpha|^2 (1) = 1 - \alpha \bar{\beta} - \bar{\alpha} \beta + |\alpha|^2.$
Since the numerator and denominator are equal,$|z|^2 = 1,$ which implies $|z| = 1.$

(A) Given equation: $|1-i|^{x} = 2^{x}$
Calculate the modulus: $|1-i| = \sqrt{1^{2} + (-1)^{2}} = \sqrt{2}$
Substitute the modulus into the equation: $(\sqrt{2})^{x} = 2^{x}$
Rewrite $\sqrt{2}$ as $2^{1/2}$: $(2^{1/2})^{x} = 2^{x}$
Simplify the exponents: $2^{x/2} = 2^{x}$
Equate the exponents: $\frac{x}{2} = x$
Multiply by $2$: $x = 2x$
Rearrange: $2x - x = 0 \Rightarrow x = 0$
The only integral solution is $x = 0$.
Since $0$ is not a non-zero integer,the number of non-zero integral solutions is $0$.

(N/A) Given: $(a+ib)(c+id)(e+if)(g+ih)=A+iB$
Taking the modulus on both sides:
$|(a+ib)(c+id)(e+if)(g+ih)| = |A+iB|$
Using the property $|z_{1}z_{2}z_{3}z_{4}| = |z_{1}||z_{2}||z_{3}||z_{4}|$,we get:
$|a+ib| \times |c+id| \times |e+if| \times |g+ih| = |A+iB|$
Since $|x+iy| = \sqrt{x^{2}+y^{2}}$,we have:
$\sqrt{a^{2}+b^{2}} \times \sqrt{c^{2}+d^{2}} \times \sqrt{e^{2}+f^{2}} \times \sqrt{g^{2}+h^{2}} = \sqrt{A^{2}+B^{2}}$
Squaring both sides:
$(a^{2}+b^{2})(c^{2}+d^{2})(e^{2}+f^{2})(g^{2}+h^{2}) = A^{2}+B^{2}$
Hence proved.

(D) Let $|z| = r$. Using the triangle inequality for complex numbers,we have $||z| - |1/z|| \leq |z - 1/z| \leq |z| + |1/z|$.
Given $|z - 1/z| = 2$,we have $|r - 1/r| \leq 2 \leq r + 1/r$.
Considering the inequality $|r - 1/r| \leq 2$,we have $-2 \leq r - 1/r \leq 2$.
Since $r > 0$,we focus on $r - 1/r \leq 2$,which implies $r^2 - 2r - 1 \leq 0$.
Solving the quadratic equation $r^2 - 2r - 1 = 0$ using the quadratic formula,we get $r = \frac{2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$.
Since $r = |z| > 0$,the range for $r$ is $0 < r \leq 1 + \sqrt{2}$.
Therefore,the maximum value of $|z|$ is $1 + \sqrt{2}$.

(C) We are given $w = \frac{z-\alpha}{1-\bar{\alpha}z}$ with $|\alpha| < 1$.
Consider the condition $|z| < 1$.
We want to check the value of $|w|^2 = w\bar{w}$.
$|w|^2 = \left(\frac{z-\alpha}{1-\bar{\alpha}z}\right) \left(\frac{\bar{z}-\bar{\alpha}}{1-\alpha\bar{z}}\right) = \frac{z\bar{z} - z\bar{\alpha} - \alpha\bar{z} + \alpha\bar{\alpha}}{1 - \alpha\bar{z} - \bar{\alpha}z + \alpha\bar{\alpha}z\bar{z}}$.
Since $|z|^2 = z\bar{z}$ and $|\alpha|^2 = \alpha\bar{\alpha}$,we have $|w|^2 = \frac{|z|^2 - z\bar{\alpha} - \bar{z}\alpha + |\alpha|^2}{1 - \bar{z}\alpha - z\bar{\alpha} + |\alpha|^2|z|^2}$.
Now,consider $|w|^2 - 1 = \frac{|z|^2 - z\bar{\alpha} - \bar{z}\alpha + |\alpha|^2 - (1 - \bar{z}\alpha - z\bar{\alpha} + |\alpha|^2|z|^2)}{1 - \bar{z}\alpha - z\bar{\alpha} + |\alpha|^2|z|^2}$.
$|w|^2 - 1 = \frac{|z|^2 + |\alpha|^2 - 1 - |\alpha|^2|z|^2}{1 - \bar{z}\alpha - z\bar{\alpha} + |\alpha|^2|z|^2} = \frac{-(1 - |z|^2)(1 - |\alpha|^2)}{|1 - \bar{\alpha}z|^2}$.
Since $|z| < 1$ and $|\alpha| < 1$,both $(1 - |z|^2) > 0$ and $(1 - |\alpha|^2) > 0$.
Thus,$|w|^2 - 1 < 0$,which implies $|w| < 1$ for all $|z| < 1$.

(B) Given $z = \frac{1}{2} - 2i$.
Substitute $z$ into the equation $|z+1| = \alpha z + \beta(1+i)$:
$|(\frac{1}{2} - 2i) + 1| = \alpha(\frac{1}{2} - 2i) + \beta(1+i)$
$|\frac{3}{2} - 2i| = (\frac{\alpha}{2} + \beta) + (\beta - 2\alpha)i$
Calculate the modulus on the left side:
$|\frac{3}{2} - 2i| = \sqrt{(\frac{3}{2})^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$
Equating the real and imaginary parts:
Imaginary part: $\beta - 2\alpha = 0 \implies \beta = 2\alpha$
Real part: $\frac{\alpha}{2} + \beta = \frac{5}{2}$
Substitute $\beta = 2\alpha$ into the real part equation:
$\frac{\alpha}{2} + 2\alpha = \frac{5}{2}$
$\frac{5\alpha}{2} = \frac{5}{2} \implies \alpha = 1$
Then $\beta = 2(1) = 2$.
Therefore,$\alpha + \beta = 1 + 2 = 3$.

(B) Given the equation $z^2 + i\bar{z} = 0$,we have $z^2 = -i\bar{z}$.
Taking the modulus on both sides,we get $|z^2| = |-i\bar{z}|$.
Since $|-i| = 1$ and $|\bar{z}| = |z|$,this simplifies to $|z|^2 = |z|$.
This implies $|z|^2 - |z| = 0$,or $|z|(|z| - 1) = 0$.
Since $xy \neq 0$,$z \neq 0$,so $|z| \neq 0$. Thus,$|z| = 1$.
Therefore,$|z^2| = |z|^2 = 1^2 = 1$.

(C) Let $z = 6+8i$. We want to find the modulus of $\sqrt{z}$.
Using the property of the modulus of complex numbers,$|\sqrt{z}| = \sqrt{|z|}$.
First,calculate the modulus of $z = 6+8i$:
$|z| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.
Therefore,the modulus of the square root is $\sqrt{|z|} = \sqrt{10}$.

(B) Given $z = \frac{(1+i)^2}{a+i} = \frac{1+i^2+2i}{a+i} = \frac{2i}{a+i}$.
Magnitude $|z| = \left| \frac{2i}{a+i} \right| = \frac{|2i|}{|a+i|} = \frac{2}{\sqrt{a^2+1}}$.
Given $|z| = \frac{2}{\sqrt{5}}$,so $\frac{2}{\sqrt{a^2+1}} = \frac{2}{\sqrt{5}}$ $\Rightarrow a^2+1 = 5$ $\Rightarrow a^2 = 4$.
Since $a > 0$,we have $a = 2$.
Substituting $a = 2$ into $z$,we get $z = \frac{2i}{2+i} = \frac{2i(2-i)}{(2+i)(2-i)} = \frac{4i - 2i^2}{4+1} = \frac{2+4i}{5} = \frac{2}{5} + \frac{4}{5}i$.
Therefore,$\bar{z} = \frac{2}{5} - \frac{4}{5}i$.

(D) Let $z = x + iy$,where $x, y \in \mathbb{R}$.
Given $|z| + z = 3 + i$.
Substituting $z = x + iy$,we get $\sqrt{x^2 + y^2} + x + iy = 3 + i$.
Equating the real and imaginary parts:
$y = 1$ and $\sqrt{x^2 + y^2} + x = 3$.
Substituting $y = 1$ into the second equation:
$\sqrt{x^2 + 1} = 3 - x$.
Squaring both sides:
$x^2 + 1 = (3 - x)^2 = 9 - 6x + x^2$.
$1 = 9 - 6x$ $\Rightarrow 6x = 8$ $\Rightarrow x = \frac{4}{3}$.
Now,$|z| = \sqrt{x^2 + y^2} = \sqrt{(\frac{4}{3})^2 + 1^2} = \sqrt{\frac{16}{9} + 1} = \sqrt{\frac{25}{9}} = \frac{5}{3}$.

(D) Let $z = -7 + 24i$,where $i = \sqrt{-1}$.
The conjugate of $z$ is $\bar{z} = -7 - 24i$.
We need to find the modulus of the square root of $\bar{z}$,which is $|\sqrt{\bar{z}}|$.
Using the property of modulus,$|\sqrt{\bar{z}}| = \sqrt{|\bar{z}|}$.
The modulus of $\bar{z}$ is $|\bar{z}| = \sqrt{(-7)^2 + (-24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25$.
Therefore,$|\sqrt{\bar{z}}| = \sqrt{25} = 5$.

(B) Let $z = x + iy$,where $x, y \in \mathbb{R}$.
Given $|z| + z = 3 + i$.
Substituting $z = x + iy$,we get $\sqrt{x^2 + y^2} + x + iy = 3 + i$.
Equating the real and imaginary parts:
$1) \sqrt{x^2 + y^2} + x = 3$
$2) y = 1$
Substitute $y = 1$ into the first equation:
$\sqrt{x^2 + 1} = 3 - x$
Squaring both sides:
$x^2 + 1 = (3 - x)^2$
$x^2 + 1 = 9 - 6x + x^2$
$6x = 8 \implies x = \frac{8}{6} = \frac{4}{3}$.
Now,$|z| = \sqrt{x^2 + y^2} = 3 - x = 3 - \frac{4}{3} = \frac{9 - 4}{3} = \frac{5}{3}$.
Thus,$|z| = \frac{5}{3}$.

(B) Given $z = \frac{(1 + i)^2}{a - i}$.
Since $(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$,we have $z = \frac{2i}{a - i}$.
Multiply numerator and denominator by the conjugate $(a + i)$:
$z = \frac{2i(a + i)}{a^2 + 1} = \frac{2ai - 2}{a^2 + 1} = \frac{-2 + 2ai}{a^2 + 1}$.
The magnitude is $|z| = \sqrt{\frac{(-2)^2 + (2a)^2}{(a^2 + 1)^2}} = \sqrt{\frac{4(1 + a^2)}{(a^2 + 1)^2}} = \sqrt{\frac{4}{a^2 + 1}} = \frac{2}{\sqrt{a^2 + 1}}$.
Given $|z| = \sqrt{\frac{2}{5}}$,so $\frac{2}{\sqrt{a^2 + 1}} = \sqrt{\frac{2}{5}}$.
Squaring both sides: $\frac{4}{a^2 + 1} = \frac{2}{5}$ $\Rightarrow 20 = 2(a^2 + 1)$ $\Rightarrow 10 = a^2 + 1$ $\Rightarrow a^2 = 9$.
Since $a > 0$,$a = 3$.
Substituting $a = 3$ into $z$: $z = \frac{-2 + 2(3)i}{3^2 + 1} = \frac{-2 + 6i}{10} = -\frac{1}{5} + \frac{3}{5}i$.
Therefore,the conjugate $\overline{z} = -\frac{1}{5} - \frac{3}{5}i$.

(D) Let $Z = a + ib$.
Then $|Z| = \sqrt{a^2 + b^2}$.
Given $|Z| + Z = 2 + i$.
Substituting the values,we get $\sqrt{a^2 + b^2} + a + ib = 2 + i$.
Comparing the real and imaginary parts,we get $b = 1$ and $\sqrt{a^2 + b^2} + a = 2$.
Since $b = 1$,$\sqrt{a^2 + 1} = 2 - a$.
Squaring both sides,$a^2 + 1 = (2 - a)^2$.
$a^2 + 1 = 4 - 4a + a^2$.
$4a = 3$,so $a = \frac{3}{4}$.
Now,$|Z| = \sqrt{a^2 + b^2} = \sqrt{(\frac{3}{4})^2 + 1^2} = \sqrt{\frac{9}{16} + 1} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

(B) Given $|z|+z=2+i$.
Let $z=x+iy$,where $x, y \in \mathbb{R}$.
Then $|z|=\sqrt{x^2+y^2}$.
Substituting these into the equation: $\sqrt{x^2+y^2}+x+iy=2+i$.
Equating real and imaginary parts,we get:
$y=1$ and $\sqrt{x^2+y^2}+x=2$.
Substituting $y=1$ into the second equation: $\sqrt{x^2+1}=2-x$.
Squaring both sides: $x^2+1=(2-x)^2 = 4-4x+x^2$.
$1=4-4x$ $\Rightarrow 4x=3$ $\Rightarrow x=\frac{3}{4}$.
Thus,$z=\frac{3}{4}+i$.
Therefore,$|z|=\sqrt{(\frac{3}{4})^2+1^2} = \sqrt{\frac{9}{16}+1} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

(B) Given $Z_1=2+i$ and $Z_2=3-4i$.
The conjugates are $\overline{Z_1}=2-i$ and $\overline{Z_2}=3+4i$.
We need to evaluate $\frac{\overline{Z_1}}{\overline{Z_2}} = \frac{2-i}{3+4i}$.
Multiply the numerator and denominator by the conjugate of the denominator $(3-4i)$:
$\frac{2-i}{3+4i} \times \frac{3-4i}{3-4i} = \frac{6-8i-3i+4i^2}{3^2-(4i)^2}$.
Since $i^2=-1$,we have $\frac{6-11i-4}{9+16} = \frac{2-11i}{25} = \frac{2}{25} - \frac{11}{25}i$.
Comparing this with $a+bi$,we get $a=\frac{2}{25}$ and $b=\frac{-11}{25}$.
Now,calculate $-7a+b = -7(\frac{2}{25}) - \frac{11}{25} = \frac{-14-11}{25} = \frac{-25}{25} = -1$.

(A) Given $z = \frac{(1+i)^2}{a-i} = \frac{1 + 2i + i^2}{a-i} = \frac{2i}{a-i}$.
Taking the modulus on both sides: $|z| = \left| \frac{2i}{a-i} \right| = \frac{|2i|}{|a-i|} = \frac{2}{\sqrt{a^2 + 1}}$.
Given $|z| = \frac{2}{\sqrt{5}}$,we have $\frac{2}{\sqrt{a^2 + 1}} = \frac{2}{\sqrt{5}}$.
Thus,$a^2 + 1 = 5$,which implies $a^2 = 4$. Since $a > 0$,we get $a = 2$.
Substituting $a = 2$ into $z$: $z = \frac{2i}{2-i} = \frac{2i(2+i)}{(2-i)(2+i)} = \frac{4i + 2i^2}{4+1} = \frac{-2 + 4i}{5} = -\frac{2}{5} + \frac{4}{5}i$.
Therefore,the conjugate $\bar{z} = -\frac{2}{5} - \frac{4}{5}i$.

(A) Given the complex number in polar form: $z = 4(\cos 300^{\circ} + i \sin 300^{\circ})$.
We know that $\cos 300^{\circ} = \cos(360^{\circ} - 60^{\circ}) = \cos 60^{\circ} = \frac{1}{2}$.
And $\sin 300^{\circ} = \sin(360^{\circ} - 60^{\circ}) = -\sin 60^{\circ} = -\frac{\sqrt{3}}{2}$.
Substituting these values into the expression:
$z = 4\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)$.
$z = 4 \times \frac{1}{2} - 4 \times i\frac{\sqrt{3}}{2}$.
$z = 2 - 2\sqrt{3}i$.

(D) Let $z = \frac{1+i \sqrt{3}}{\left(1+\frac{1}{i+1}\right)^{2}}$.
First,simplify the denominator: $1+\frac{1}{i+1} = \frac{i+1+1}{i+1} = \frac{i+2}{i+1}$.
Then,$z = \frac{1+i \sqrt{3}}{\left(\frac{i+2}{i+1}\right)^{2}} = \frac{(1+i \sqrt{3})(i+1)^{2}}{(i+2)^{2}}$.
Since $(i+1)^{2} = i^{2}+1+2i = 2i$ and $(i+2)^{2} = i^{2}+4+4i = 3+4i$,we have $z = \frac{(1+i \sqrt{3})(2i)}{3+4i} = \frac{2i-2\sqrt{3}}{3+4i}$.
Now,find the modulus: $|z| = \frac{|2i-2\sqrt{3}|}{|3+4i|} = \frac{\sqrt{(-2\sqrt{3})^{2} + 2^{2}}}{\sqrt{3^{2}+4^{2}}}$.
$|z| = \frac{\sqrt{12+4}}{\sqrt{9+16}} = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5}$.