Conjugate and Modulus of complex numbers Questions in English

(A) To simplify the complex number,we multiply the numerator and denominator by the conjugate of the denominator $(1+i)$:
$\frac{1+2i}{1-i} \times \frac{1+i}{1+i} = \frac{1+i+2i+2i^2}{1^2-i^2}$
Since $i^2 = -1$,we have:
$\frac{1+3i-2}{1-(-1)} = \frac{-1+3i}{2} = -\frac{1}{2} + \frac{3}{2}i$
The real part is $-\frac{1}{2}$ (negative) and the imaginary part is $\frac{3}{2}$ (positive).
$A$ complex number with a negative real part and a positive imaginary part lies in the second quadrant.

(C) Let $z = \frac{(1+i)^2(1+3 i)}{(2-6 i)(2-2 i)}$.
Using the property of modulus $|\frac{z_1 z_2}{z_3 z_4}| = \frac{|z_1| |z_2|}{|z_3| |z_4|}$,we have:
$|z| = \frac{|1+i|^2 |1+3i|}{|2-6i| |2-2i|}$
$|1+i| = \sqrt{1^2+1^2} = \sqrt{2}$,so $|1+i|^2 = 2$.
$|1+3i| = \sqrt{1^2+3^2} = \sqrt{10}$.
$|2-6i| = \sqrt{2^2+(-6)^2} = \sqrt{4+36} = \sqrt{40} = 2\sqrt{10}$.
$|2-2i| = \sqrt{2^2+(-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$.
Substituting these values:
$|z| = \frac{2 \times \sqrt{10}}{2\sqrt{10} \times 2\sqrt{2}} = \frac{2}{4\sqrt{2}} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$.

(C) Let $z = \frac{1+2i}{1-(1-i)^{2}}$.
First,simplify the denominator: $(1-i)^{2} = 1^{2} + i^{2} - 2i = 1 - 1 - 2i = -2i$.
Substitute this back into the expression for $z$:
$z = \frac{1+2i}{1 - (-2i)} = \frac{1+2i}{1+2i} = 1$.
Thus,$z = 1 + 0i$.
The modulus is $|z| = \sqrt{1^{2} + 0^{2}} = 1$.
The amplitude (argument) is $\theta = \tan^{-1}\left(\frac{0}{1}\right) = 0$.

(B) Given that the conjugate of $(x + iy)(1 - 2i)$ is $1 + i$.
Let $z = (x + iy)(1 - 2i)$.
Then $\bar{z} = 1 + i$.
Taking the conjugate on both sides of $z = (x + iy)(1 - 2i)$,we get $\bar{z} = \overline{(x + iy)(1 - 2i)} = \overline{(x + iy)} \cdot \overline{(1 - 2i)}$.
Since $\bar{z} = 1 + i$,we have $(x - iy)(1 + 2i) = 1 + i$.
Alternatively,we can write $z = \overline{1 + i} = 1 - i$.
Thus,$(x + iy)(1 - 2i) = 1 - i$.
Dividing both sides by $(1 - 2i)$,we get $x + iy = \frac{1 - i}{1 - 2i}$.
This matches Option $(B)$.

(B) Given that $|\beta|=1$,we have $|\beta|^2 = \beta \bar{\beta} = 1$.
Consider the expression $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|$.
Substituting $1 = \beta \bar{\beta}$ in the denominator,we get:
$\left|\frac{\beta-\alpha}{\beta \bar{\beta}-\bar{\alpha} \beta}\right| = \left|\frac{\beta-\alpha}{\beta(\bar{\beta}-\bar{\alpha})}\right|$.
Using the property of modulus $|z_1/z_2| = |z_1|/|z_2|$ and $|z_1 z_2| = |z_1||z_2|$,we have:
$= \frac{|\beta-\alpha|}{ |\beta| |\bar{\beta}-\bar{\alpha}| }$.
Since $|\beta|=1$ and $|\bar{z}| = |z|$,we know $|\bar{\beta}-\bar{\alpha}| = |\overline{\beta-\alpha}| = |\beta-\alpha|$.
Therefore,the expression becomes $\frac{|\beta-\alpha|}{1 \cdot |\beta-\alpha|} = 1$.

(B) Let $z = x + iy$.
Given,$z = -\overline{z}$.
Substituting the values:
$x + iy = -(x - iy)$
$x + iy = -x + iy$
$2x = 0$
$x = 0$.
Since the real part $x$ is $0$,the complex number $z = iy$ is purely imaginary.

(D) Let $z = \frac{(1+i)^{2}}{1-i}$.
First,expand the numerator: $(1+i)^{2} = 1^{2} + i^{2} + 2i = 1 - 1 + 2i = 2i$.
So,$z = \frac{2i}{1-i}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator $(1+i)$:
$z = \frac{2i(1+i)}{(1-i)(1+i)} = \frac{2i + 2i^{2}}{1 - i^{2}} = \frac{2i - 2}{1 - (-1)} = \frac{2i - 2}{2} = i - 1$.
Thus,$z = -1 + i$.
The conjugate of a complex number $z = a + bi$ is $\bar{z} = a - bi$.
Therefore,the conjugate of $-1 + i$ is $-1 - i$.

(C) Given $Z = \frac{(\sqrt{3} + i)^{3}(3i + 4)^{2}}{(8 + 6i)^{2}}$.
Taking the modulus on both sides,we get $|Z| = \frac{|\sqrt{3} + i|^{3} |3i + 4|^{2}}{|8 + 6i|^{2}}$.
Calculate the modulus of each complex number:
$|\sqrt{3} + i| = \sqrt{(\sqrt{3})^{2} + 1^{2}} = \sqrt{3 + 1} = \sqrt{4} = 2$.
$|3i + 4| = \sqrt{4^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$.
$|8 + 6i| = \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10$.
Substitute these values into the expression for $|Z|$:
$|Z| = \frac{2^{3} \times 5^{2}}{10^{2}} = \frac{8 \times 25}{100} = \frac{200}{100} = 2$.
Thus,$|Z| = 2$.

(A) 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) $.
Expanding the numerator: $ (\beta-\alpha)(\bar{\beta}-\bar{\alpha}) = \beta \bar{\beta} - \beta \bar{\alpha} - \alpha \bar{\beta} + \alpha \bar{\alpha} = |\beta|^2 - \beta \bar{\alpha} - \alpha \bar{\beta} + |\alpha|^2 $.
Expanding the denominator: $ (1-\bar{\alpha} \beta)(1-\alpha \bar{\beta}) = 1 - \alpha \bar{\beta} - \bar{\alpha} \beta + \bar{\alpha} \alpha \beta \bar{\beta} = 1 - \alpha \bar{\beta} - \bar{\alpha} \beta + |\alpha|^2 |\beta|^2 $.
Since $ |\beta|=1 $,we have $ |\beta|^2 = 1 $.
Substituting $ |\beta|^2 = 1 $ into the expressions:
Numerator: $ 1 - \beta \bar{\alpha} - \alpha \bar{\beta} + |\alpha|^2 $.
Denominator: $ 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 the equation $|1-i|^x=2^x$.
First,calculate the modulus of the complex number $1-i$:
$|1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2} = 2^{\frac{1}{2}}$.
Substitute this into the equation:
$(2^{\frac{1}{2}})^x = 2^x$.
$2^{\frac{x}{2}} = 2^x$.
Since the bases are equal,equate the exponents:
$\frac{x}{2} = x$.
$x = 2x \Rightarrow x = 0$.
Thus,there is only $1$ integer solution,which is $x=0$.

(C) We know that $i^4 = 1$. Therefore,$i^{2020} = (i^4)^{505} = 1^{505} = 1$.
Similarly,$i^{2021} = i^{2020} \times i = i$,$i^{2022} = i^{2020} \times i^2 = -1$,and $i^{2023} = i^{2020} \times i^3 = -i$.
Substituting these values into the expression:
$\left|\frac{1}{1} + \frac{2}{i} + \frac{3}{-1} + \frac{4}{-i}\right|$
$= \left|1 - 2i - 3 + 4i\right|$
$= \left|-2 + 2i\right|$
$= \sqrt{(-2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.

(A) Given: $z_1 = 2 + 5i$,$z_2 = -1 + 4i$,$z_3 = i$.
First,calculate the numerator: $z_1 - z_3 = (2 + 5i) - i = 2 + 4i$.
Next,calculate the denominator: $z_3 - z_2 = i - (-1 + 4i) = i + 1 - 4i = 1 - 3i$.
Now,consider the fraction: $\frac{z_1 - z_3}{z_3 - z_2} = \frac{2 + 4i}{1 - 3i}$.
Multiply the numerator and denominator by the conjugate of the denominator $(1 + 3i)$:
$\frac{(2 + 4i)(1 + 3i)}{(1 - 3i)(1 + 3i)} = \frac{2 + 6i + 4i + 12i^2}{1^2 + 3^2} = \frac{2 + 10i - 12}{1 + 9} = \frac{-10 + 10i}{10} = -1 + i$.
Finally,find the modulus: $\left| -1 + i \right| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$.

(C) Let the multiplicative inverse of $z$ be $A$.
Since $z \cdot A = 1$,we have $A = \frac{1}{z}$.
To simplify,multiply the numerator and denominator by the conjugate $\bar{z}$:
$A = \frac{1 \cdot \bar{z}}{z \cdot \bar{z}} = \frac{\bar{z}}{|z|^2}$
Since $z \cdot \bar{z} = |z|^2$,the multiplicative inverse is $\frac{\bar{z}}{|z|^2}$.
Thus,option $C$ is correct.

(B) Let the complex number be $z = \sin \theta + i \cos \theta$.
The multiplicative inverse of $z$ is given by $\frac{1}{z}$.
$\frac{1}{z} = \frac{1}{\sin \theta + i \cos \theta}$.
To simplify,multiply the numerator and denominator by the conjugate $(\sin \theta - i \cos \theta)$:
$\frac{1}{z} = \frac{\sin \theta - i \cos \theta}{(\sin \theta + i \cos \theta)(\sin \theta - i \cos \theta)}$.
Using the identity $(a+ib)(a-ib) = a^2 + b^2$:
$\frac{1}{z} = \frac{\sin \theta - i \cos \theta}{\sin^2 \theta + \cos^2 \theta}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$\frac{1}{z} = \sin \theta - i \cos \theta$.
Thus,the multiplicative inverse is $(\sin \theta, -\cos \theta)$.

(C) Given,$\alpha - i\beta = \left(\frac{3-4ix}{3+4ix}\right)$.
Taking the modulus on both sides,we get:
$|\alpha - i\beta| = \left|\frac{3-4ix}{3+4ix}\right|$.
Since the modulus of a quotient is the quotient of the moduli,we have:
$|\alpha - i\beta| = \frac{|3-4ix|}{|3+4ix|}$.
We know that for any complex number $z = a + ib$,$|z| = \sqrt{a^2 + b^2}$.
Thus,$\sqrt{\alpha^2 + (-\beta)^2} = \frac{\sqrt{3^2 + (-4x)^2}}{\sqrt{3^2 + (4x)^2}}$.
$\sqrt{\alpha^2 + \beta^2} = \frac{\sqrt{9 + 16x^2}}{\sqrt{9 + 16x^2}}$.
Since $x$ is a real value,$9 + 16x^2 \neq 0$,so the ratio is $1$.
$\sqrt{\alpha^2 + \beta^2} = 1$.
Squaring both sides,we get $\alpha^2 + \beta^2 = 1$.

(C) Given,$z = x+iy = \frac{(3+2i)(4-7i)(12+13i)}{(13-12i)(2-3i)(11+3i)}$.
Taking the modulus on both sides,we have $|z| = |x+iy| = \sqrt{x^2+y^2}$.
Using the property $|\frac{z_1 z_2 z_3}{z_4 z_5 z_6}| = \frac{|z_1| |z_2| |z_3|}{|z_4| |z_5| |z_6|}$,we get:
$|z| = \frac{|3+2i| \cdot |4-7i| \cdot |12+13i|}{|13-12i| \cdot |2-3i| \cdot |11+3i|}$.
Note that $|3+2i| = |2-3i| = \sqrt{3^2+2^2} = \sqrt{13}$.
Also,$|12+13i| = |13-12i| = \sqrt{12^2+13^2} = \sqrt{313}$.
Substituting these values,we get:
$|z| = \frac{\sqrt{13} \cdot |4-7i| \cdot \sqrt{313}}{\sqrt{313} \cdot \sqrt{13} \cdot |11+3i|} = \frac{|4-7i|}{|11+3i|}$.
$|z| = \frac{\sqrt{4^2+(-7)^2}}{\sqrt{11^2+3^2}} = \frac{\sqrt{16+49}}{\sqrt{121+9}} = \frac{\sqrt{65}}{\sqrt{130}} = \frac{\sqrt{65}}{\sqrt{2 \times 65}} = \frac{1}{\sqrt{2}}$.
Since $|z| = \sqrt{x^2+y^2}$,we have $\sqrt{x^2+y^2} = \frac{1}{\sqrt{2}}$.
Squaring both sides,$x^2+y^2 = \frac{1}{2}$.

(D) Given that $|z_1|=1, |z_2|=2, |z_3|=3$ and $|9 z_1 z_2 + 4 z_1 z_3 + z_2 z_3| = 12$.
We know that $|z|^2 = z \bar{z}$,which implies $\bar{z} = \frac{|z|^2}{z}$.
Substituting $\bar{z}_1 = \frac{1}{z_1}$,$\bar{z}_2 = \frac{4}{z_2}$,and $\bar{z}_3 = \frac{9}{z_3}$ is not directly helpful here.
Instead,divide the given expression by $|z_1 z_2 z_3|$:
$|9 z_1 z_2 + 4 z_1 z_3 + z_2 z_3| = 12$
$|z_1 z_2 z_3| \left| \frac{9}{z_3} + \frac{4}{z_2} + \frac{1}{z_1} \right| = 12$
Since $|z_1|=1, |z_2|=2, |z_3|=3$,we have $|z_1 z_2 z_3| = 1 \times 2 \times 3 = 6$.
$6 \left| \frac{9}{z_3} + \frac{4}{z_2} + \frac{1}{z_1} \right| = 12 \implies \left| \frac{9}{z_3} + \frac{4}{z_2} + \frac{1}{z_1} \right| = 2$.
Using $\bar{z} = \frac{|z|^2}{z}$,we have $\bar{z}_1 = \frac{1}{z_1}$,$\bar{z}_2 = \frac{4}{z_2}$,and $\bar{z}_3 = \frac{9}{z_3}$.
Thus,$|\bar{z}_1 + \bar{z}_2 + \bar{z}_3| = 2$.
Since $|\bar{z}| = |z|$,we get $|z_1 + z_2 + z_3| = 2$.

(C) First,calculate the product of the complex numbers:
$(4-3i)(2+3i) = 8 + 12i - 6i - 9i^2 = 8 + 6i + 9 = 17 + 6i$.
Then,multiply by $(1+4i)$:
$(17+6i)(1+4i) = 17 + 68i + 6i + 24i^2 = 17 + 74i - 24 = -7 + 74i$.
The complex conjugate of a complex number $z = a + bi$ is $\bar{z} = a - bi$.
Therefore,the complex conjugate of $-7 + 74i$ is $-7 - 74i$.

(B) Given $z_1 = 4+3 i$.
The modulus $\alpha = |z_1| = \sqrt{4^2+3^2} = \sqrt{16+9} = 5$.
The region is defined by $|z - \overline{z_1}| \leq 5$.
Since $\overline{z_1} = 4-3 i$,the inequality is $|z - (4-3 i)| \leq 5$.
We test the options by checking if they satisfy $|z - (4-3 i)| \leq 5$:
For option $B$,$z = z_1 = 4+3 i$:
$|(4+3 i) - (4-3 i)| = |6 i| = 6$.
Since $6 > 5$,the point $z_1$ does not satisfy the inequality.
Thus,$z_1$ does not lie in the region.

(C) Given the equation: $4a + i(3a - b) = b - 6i$
Equating the real and imaginary parts on both sides:
Real part: $4a = b$
Imaginary part: $3a - b = -6$
Substitute $b = 4a$ into the second equation:
$3a - 4a = -6$ $\Rightarrow -a = -6$ $\Rightarrow a = 6$
Then,$b = 4(6) = 24$
Now,substitute $a$ and $b$ into $z$:
$z = 6 + \frac{24}{4}i = 6 + 6i$
The modulus $|z|$ is:
$|z| = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$
Finally,calculate $\frac{|z|}{a}$:
$\frac{|z|}{a} = \frac{6\sqrt{2}}{6} = \sqrt{2}$

(B) Given $z = \frac{-2+i}{(1-2i)^2}$.
First,simplify the denominator: $(1-2i)^2 = 1^2 + (2i)^2 - 2(1)(2i) = 1 - 4 - 4i = -3 - 4i$.
So,$z = \frac{-2+i}{-3-4i} = \frac{2-i}{3+4i}$.
Multiply the numerator and denominator by the conjugate of the denominator $(3-4i)$:
$z = \frac{(2-i)(3-4i)}{(3+4i)(3-4i)} = \frac{6 - 8i - 3i + 4i^2}{3^2 + 4^2} = \frac{6 - 11i - 4}{9 + 16} = \frac{2 - 11i}{25}$.
The conjugate is $\bar{z} = \frac{2 + 11i}{25}$.
The modulus is $|\bar{z}| = |z| = \sqrt{(\frac{2}{25})^2 + (-\frac{11}{25})^2} = \sqrt{\frac{4 + 121}{625}} = \sqrt{\frac{125}{625}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}$.

(D) Given $z = 1 + \cos \theta - i \sin \theta$.
$z - 1 = \cos \theta - i \sin \theta$.
$|z - 1| = \sqrt{\cos^2 \theta + \sin^2 \theta} = 1$.
$|z|^2 = (1 + \cos \theta)^2 + (-\sin \theta)^2 = 1 + 2 \cos \theta + \cos^2 \theta + \sin^2 \theta = 2 + 2 \cos \theta$.
Now,substitute these into the expression:
$\left[|z - 1|^2 - \frac{|z|^2}{4}\right]^{1/2} = \left[1^2 - \frac{2 + 2 \cos \theta}{4}\right]^{1/2} = \left[1 - \frac{1 + \cos \theta}{2}\right]^{1/2} = \left[\frac{2 - 1 - \cos \theta}{2}\right]^{1/2} = \left[\frac{1 - \cos \theta}{2}\right]^{1/2}$.
Using the identity $1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)$,we get:
$\left[\frac{2 \sin^2 \left(\frac{\theta}{2}\right)}{2}\right]^{1/2} = \sqrt{\sin^2 \left(\frac{\theta}{2}\right)} = \sin \left(\frac{\theta}{2}\right)$ (since $0 < \frac{\theta}{2} < \frac{\pi}{2}$,$\sin \left(\frac{\theta}{2}\right) > 0$).

(A) Given,$\left|z+\frac{2}{z}\right|=2$.
Using the triangle inequality,$|z| = |(z+\frac{2}{z}) - \frac{2}{z}| \leq |z+\frac{2}{z}| + |\frac{2}{z}|$.
Substituting the given value,$|z| \leq 2 + \frac{2}{|z|}$.
Multiplying by $|z|$ (since $|z| > 0$),we get $|z|^2 \leq 2|z| + 2$,which implies $|z|^2 - 2|z| - 2 \leq 0$.
Solving the quadratic inequality $x^2 - 2x - 2 \leq 0$ for $x = |z|$,the roots are $x = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = 1 \pm \sqrt{3}$.
Since $|z| \geq 0$,we have $0 \leq |z| \leq 1+\sqrt{3}$.
Thus,the maximum value of $|z|$ is $1+\sqrt{3}$.

(A) Given $Z_1$ and $Z_2$ are conjugate complex numbers. Let $Z_1 = a + ib$, then $Z_2 = a - ib$.
$(A) Z_1 Z_2 = (a + ib)(a - ib) = a^2 + b^2 = |Z_1|^2$. Thus, $A-3$.
$(B) Z_1 + Z_2 = (a + ib) + (a - ib) = 2a$. If $Z_1 + Z_2 = 0$, then $2a = 0 \Rightarrow a = 0$. This represents the imaginary axis. Thus, $B-1$.
$(C) \text{Im}(Z_1) = b$. Also, $\text{Im}(-Z_2) = \text{Im}(-(a - ib)) = \text{Im}(-a + ib) = b$. Thus, $\text{Im}(Z_1) = \text{Im}(-Z_2)$. So, $C-2$.
$(D) \text{Re}(Z_1) = a$ and $\text{Re}(Z_2) = a$. Thus, $\text{Re}(Z_1) = \text{Re}(Z_2)$. So, $D-4$.
The correct matching is $A-3, B-1, C-2, D-4$.

(D) Given $f(z) = |z|$.
$(a)$ $f(-z) = |-z| = |z| = f(z)$,which is true.
$(b)$ $f(\bar{z}) = |\bar{z}| = |z| = f(z)$,which is true.
$(c)$ $f(z^2) = |z^2| = |z|^2 = (f(z))^2$,which is true.
$(d)$ $f(z_1^2 + z_2^2) = |z_1^2 + z_2^2|$ and $f(z_1^2) + f(z_2^2) = |z_1^2| + |z_2^2| = |z_1|^2 + |z_2|^2$.
By the triangle inequality,$|z_1^2 + z_2^2| \leq |z_1^2| + |z_2^2| = |z_1|^2 + |z_2|^2$. Equality does not hold in general.
Therefore,$f(z_1^2 + z_2^2) = f(z_1^2) + f(z_2^2)$ is false.

(B) Let $z = (1+2i)(-2+i)$.
Using the property of the modulus,$|z_1 z_2| = |z_1| |z_2|$,we have:
$|z| = |1+2i| \times |-2+i|$
$|z| = \sqrt{1^2 + 2^2} \times \sqrt{(-2)^2 + 1^2}$
$|z| = \sqrt{1+4} \times \sqrt{4+1}$
$|z| = \sqrt{5} \times \sqrt{5}$
$|z| = 5$

(A) To find the conjugate of $z = \frac{5i}{7+i}$,we first simplify the expression by multiplying the numerator and denominator by the conjugate of the denominator,which is $(7-i)$.
$z = \frac{5i}{7+i} \times \frac{7-i}{7-i}$
$z = \frac{35i - 5i^2}{7^2 - i^2}$
Since $i^2 = -1$,we have:
$z = \frac{35i - 5(-1)}{49 - (-1)} = \frac{5 + 35i}{50}$
Dividing both terms by $5$,we get:
$z = \frac{1 + 7i}{10} = \frac{1}{10} + \frac{7}{10}i$
The conjugate of a complex number $a + bi$ is $a - bi$.
Therefore,the conjugate of $\frac{1}{10} + \frac{7}{10}i$ is $\frac{1}{10} - \frac{7}{10}i = \frac{1}{10}(1-7i)$.
Thus,option $A$ is correct.

(A) Given $\left|\frac{z-25}{z-1}\right|=5$.
Squaring both sides,we get $\left|\frac{z-25}{z-1}\right|^2 = 25$.
This implies $\frac{(z-25)(\bar{z}-25)}{(z-1)(\bar{z}-1)} = 25$.
Expanding the terms: $(z-25)(\bar{z}-25) = 25(z-1)(\bar{z}-1)$.
$z\bar{z} - 25z - 25\bar{z} + 625 = 25(z\bar{z} - z - \bar{z} + 1)$.
$|z|^2 - 25(z+\bar{z}) + 625 = 25|z|^2 - 25(z+\bar{z}) + 25$.
$|z|^2 + 625 = 25|z|^2 + 25$.
$24|z|^2 = 600$.
$|z|^2 = 25$.
Therefore,$|z| = 5$.

(C) Given that $x^2+y+4 i$ and $-3+x^2 y i$ are conjugates to each other.
Therefore,$x^2+y+4 i = -3 - x^2 y i$.
Comparing the real and imaginary parts on both sides:
$x^2+y = -3$ $(i)$
$4 = -x^2 y \Rightarrow y = -\frac{4}{x^2}$ $(ii)$
Substituting $(ii)$ into $(i)$:
$x^2 - \frac{4}{x^2} = -3$
$x^4 - 4 = -3x^2$
$x^4 + 3x^2 - 4 = 0$
$(x^2+4)(x^2-1) = 0$
Since $x \in R$,$x^2$ must be non-negative,so $x^2+4 \neq 0$.
Thus,$x^2-1 = 0$ $\Rightarrow x^2 = 1$ $\Rightarrow x = \pm 1$.
Substituting $x^2=1$ into $(ii)$:
$y = -\frac{4}{1} = -4$.
Now,$(|x|+|y|)^2 = (|\pm 1| + |-4|)^2 = (1+4)^2 = 5^2 = 25$.

(A) Given,$13 e^{i \tan ^{-1} \frac{5}{12}} = a + i b$.
Using Euler's formula $e^{i \theta} = \cos \theta + i \sin \theta$,we have:
$13 [\cos(\tan ^{-1} \frac{5}{12}) + i \sin(\tan ^{-1} \frac{5}{12})] = a + i b$.
Let $\theta = \tan ^{-1} \frac{5}{12}$,then $\tan \theta = \frac{5}{12}$.
In a right-angled triangle with opposite side $5$ and adjacent side $12$,the hypotenuse is $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.
Thus,$\cos \theta = \frac{12}{13}$ and $\sin \theta = \frac{5}{13}$.
Substituting these values:
$13 [\frac{12}{13} + i \frac{5}{13}] = a + i b$.
$12 + 5i = a + i b$.
Comparing real and imaginary parts,we get $a = 12$ and $b = 5$.
Therefore,the ordered pair $(a, b) = (12, 5)$.

(A) Given $z = \frac{1}{1 - \cos \theta + i \sin \theta}$.
Using trigonometric identities $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$,we get:
$z = \frac{1}{2 \sin^2 \frac{\theta}{2} + i (2 \sin \frac{\theta}{2} \cos \frac{\theta}{2})}$
$z = \frac{1}{2 \sin \frac{\theta}{2} (\sin \frac{\theta}{2} + i \cos \frac{\theta}{2})}$
Since $\sin \frac{\theta}{2} + i \cos \frac{\theta}{2} = i (\cos \frac{\theta}{2} - i \sin \frac{\theta}{2}) = i e^{-i \theta/2}$,we have:
$z = \frac{1}{2 \sin \frac{\theta}{2} \cdot i e^{-i \theta/2}} = \frac{1}{2 \sin \frac{\theta}{2}} \cdot \frac{1}{i} e^{i \theta/2}$
Since $\frac{1}{i} = -i = e^{-i \pi/2}$,we get:
$z = \frac{1}{2 \sin \frac{\theta}{2}} e^{-i \pi/2} e^{i \theta/2} = \frac{1}{2} \operatorname{cosec} \frac{\theta}{2} e^{i (\theta/2 - \pi/2)}$
Thus,the modulus is $\frac{1}{2} \operatorname{cosec} \frac{\theta}{2}$ and the amplitude is $-(\frac{\pi}{2} - \frac{\theta}{2})$.

(C) We are given $|z| \geq 5$.
Using the triangle inequality,we know that $|z + w| \geq ||z| - |w||$.
Therefore,$\left|z + \frac{2}{z}\right| \geq ||z| - \left|\frac{2}{z}\right|| = ||z| - \frac{2}{|z|}||$.
Let $f(t) = t - \frac{2}{t}$ where $t = |z| \geq 5$.
Since $f(t)$ is an increasing function for $t > 0$,the minimum value occurs at the smallest value of $t$,which is $t = 5$.
Thus,$\left|z + \frac{2}{z}\right| \geq 5 - \frac{2}{5} = \frac{25-2}{5} = \frac{23}{5}$.
The least value is $\frac{23}{5}$.

(C) Given complex numbers are $z_1 = 1+4i, z_2 = 3+i, z_3 = 1-i, z_4 = 2-3i$.
The modulus of a complex number $z = a+bi$ is given by $|z| = \sqrt{a^2+b^2}$.
Calculating the moduli:
$m_1 = |1+4i| = \sqrt{1^2+4^2} = \sqrt{1+16} = \sqrt{17}$
$m_2 = |3+i| = \sqrt{3^2+1^2} = \sqrt{9+1} = \sqrt{10}$
$m_3 = |1-i| = \sqrt{1^2+(-1)^2} = \sqrt{1+1} = \sqrt{2}$
$m_4 = |2-3i| = \sqrt{2^2+(-3)^2} = \sqrt{4+9} = \sqrt{13}$
Comparing the values: $\sqrt{2} < \sqrt{10} < \sqrt{13} < \sqrt{17}$
Therefore,$m_3 < m_2 < m_4 < m_1$.

(C) Given the complex numbers:
$\alpha_1 = |-i| = 1$
$\alpha_2 = |\frac{1}{3}(1+i)| = \frac{1}{3} \sqrt{1^2 + 1^2} = \frac{\sqrt{2}}{3} \approx 0.471$
$\alpha_3 = |-1+i| = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \approx 1.414$
Comparing the values: $\frac{\sqrt{2}}{3} < 1 < \sqrt{2}$.
Therefore,the increasing order is $\alpha_2 < \alpha_1 < \alpha_3$.

(B) Given the equation $(1+i)(1+3i)(1+7i) = x+iy$.
First,multiply the first two complex numbers: $(1+i)(1+3i) = 1 + 3i + i + 3i^2 = 1 + 4i - 3 = -2 + 4i$.
Now,multiply the result by the third complex number: $(-2+4i)(1+7i) = -2 - 14i + 4i + 28i^2 = -2 - 10i - 28 = -30 - 10i$.
Thus,$x = -30$ and $y = -10$.
The radius of the circle represented by the complex number $z = x+iy$ in the Argand plane is given by $|z| = \sqrt{x^2 + y^2}$.
$|z| = \sqrt{(-30)^2 + (-10)^2} = \sqrt{900 + 100} = \sqrt{1000} = 10\sqrt{10}$.

(B) Given the equation $(3+4i)^{2025} = 5^{2023}(x+iy)$.
Taking the modulus on both sides,we get $|(3+4i)^{2025}| = |5^{2023}(x+iy)|$.
Since $|z^n| = |z|^n$,we have $|3+4i|^{2025} = 5^{2023} \cdot |x+iy|$.
We know that $|3+4i| = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$.
Substituting this into the equation,we get $5^{2025} = 5^{2023} \cdot \sqrt{x^2+y^2}$.
Dividing both sides by $5^{2023}$,we get $\sqrt{x^2+y^2} = \frac{5^{2025}}{5^{2023}} = 5^{2025-2023} = 5^2 = 25$.

(D) Given equation: $(x^2+\frac{1}{x^2})-5(x+\frac{1}{x})+6=0$
Let $t = x+\frac{1}{x}$. Then $x^2+\frac{1}{x^2} = t^2-2$.
The equation becomes $(t^2-2)-5t+6=0$,which simplifies to $t^2-5t+4=0$.
Factoring gives $(t-4)(t-1)=0$,so $t=4$ or $t=1$.
Case $1$: $x+\frac{1}{x}=4 \Rightarrow x^2-4x+1=0$. The discriminant $D = 16-4 = 12 > 0$,so roots are real.
Case $2$: $x+\frac{1}{x}=1 \Rightarrow x^2-x+1=0$. The discriminant $D = 1-4 = -3 < 0$,so roots are complex.
The roots are $x = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2}$.
Let the roots be $\alpha = \frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $\beta = \frac{1}{2} - i\frac{\sqrt{3}}{2}$.
The modulus of each complex root is $|\alpha| = \sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$.
Similarly,$|\beta| = 1$.
The sum of the moduli is $|\alpha| + |\beta| = 1 + 1 = 2$.

(A) Given the equation $z^2+az+a^2=0$ where $a \in R$.
This is a quadratic equation in $z$. The roots are given by the quadratic formula:
$z = \frac{-a \pm \sqrt{a^2 - 4(1)(a^2)}}{2} = \frac{-a \pm \sqrt{-3a^2}}{2} = \frac{-a \pm i a \sqrt{3}}{2}$.
Now,we find the modulus of $z$:
$|z| = \left| \frac{-a}{2} \pm i \frac{a \sqrt{3}}{2} \right|$.
$|z| = \sqrt{\left( \frac{-a}{2} \right)^2 + \left( \frac{a \sqrt{3}}{2} \right)^2}$.
$|z| = \sqrt{\frac{a^2}{4} + \frac{3a^2}{4}} = \sqrt{\frac{4a^2}{4}} = \sqrt{a^2} = |a|$.
Thus,$|z|=|a|$.

(B) Given $z = (\alpha + i\beta)(2 + 7i) = (2\alpha - 7\beta) + i(7\alpha + 2\beta)$.
Since $z$ is purely imaginary,the real part must be zero:
$2\alpha - 7\beta = 0 \Rightarrow 2\alpha = 7\beta$.
Since $\alpha, \beta$ are integers,let $\alpha = 7k$ and $\beta = 2k$ for some non-zero integer $k$.
Then $|z|^2 = (\text{Re}(z))^2 + (\text{Im}(z))^2 = 0^2 + (7\alpha + 2\beta)^2$.
Substituting $\alpha = 7k$ and $\beta = 2k$:
$|z|^2 = (7(7k) + 2(2k))^2 = (49k + 4k)^2 = (53k)^2 = 2809k^2$.
For the minimum non-zero value,we take $k = 1$ (or $k = -1$):
$|z|^2 = 2809(1)^2 = 2809$.

(B) Given $x+iy=\sqrt{\frac{3+i}{1+3i}}$.
Taking the modulus on both sides,we have $|x+iy| = \left|\sqrt{\frac{3+i}{1+3i}}\right|$.
$|x+iy| = \sqrt{\left|\frac{3+i}{1+3i}\right|}$.
Since $|z_1/z_2| = |z_1|/|z_2|$,we get $|x+iy| = \sqrt{\frac{|3+i|}{|1+3i|}} = \sqrt{\frac{\sqrt{3^2+1^2}}{\sqrt{1^2+3^2}}} = \sqrt{\frac{\sqrt{10}}{\sqrt{10}}} = \sqrt{1} = 1$.
We know that $|x+iy| = \sqrt{x^2+y^2}$,so $\sqrt{x^2+y^2} = 1$.
Squaring both sides,$x^2+y^2 = 1$.
Therefore,$\left(x^2+y^2\right)^2 = 1^2 = 1$.

(B) Given $Z = \alpha + i \beta$.
Since $|Z| - Z = 1 + 2i$,we have $\sqrt{\alpha^2 + \beta^2} - (\alpha + i \beta) = 1 + 2i$.
Equating real and imaginary parts:
$\sqrt{\alpha^2 + \beta^2} - \alpha = 1$ and $-\beta = 2 \Rightarrow \beta = -2$.
Substitute $\beta = -2$ into the real part equation:
$\sqrt{\alpha^2 + (-2)^2} - \alpha = 1
\Rightarrow \sqrt{\alpha^2 + 4} = \alpha + 1$.
Squaring both sides:
$\alpha^2 + 4 = (\alpha + 1)^2
$ $\Rightarrow \alpha^2 + 4 = \alpha^2 + 2\alpha + 1
$ $\Rightarrow 2\alpha = 3
$ $\Rightarrow \alpha = \frac{3}{2}$.
We need to find $Z \bar{Z} = |Z|^2 = \alpha^2 + \beta^2$.
$Z \bar{Z} = (\frac{3}{2})^2 + (-2)^2 = \frac{9}{4} + 4 = \frac{9 + 16}{4} = \frac{25}{4}$.

(B) Given that $a = |\bar{a}|$ and $b = |\bar{b}|$. Since $|\bar{a}| = |a|$,we have $a = |a|$ and $b = |b|$.
We know that for any complex number $z$,$z \bar{z} = |z|^2$. Thus,$\bar{a} = \frac{|a|^2}{a} = \frac{a^2}{a} = a$ is not necessarily true,but rather $\frac{\bar{a}}{a^2} = \frac{\bar{a}}{a \bar{a} \cdot a} = \frac{1}{a \bar{a}} \cdot \bar{a} = \frac{1}{a}$.
Actually,using $z \bar{z} = |z|^2$,we have $\frac{\bar{a}}{a^2} = \frac{\bar{a}}{|a|^2} = \frac{\bar{a}}{a \bar{a}} = \frac{1}{a}$.
Similarly,$\frac{\bar{b}}{b^2} = \frac{1}{b}$.
Therefore,$\left(\frac{\bar{a}}{a^2} - \frac{\bar{b}}{b^2}\right)^2 = \left(\frac{1}{a} - \frac{1}{b}\right)^2 = \left(\frac{b-a}{ab}\right)^2 = \left(\frac{a-b}{ab}\right)^2$.

(D) Let $z_1 = \sin x + i \cos 2x$ and $z_2 = \cos x - i \sin 2x$.
For $z_1$ and $z_2$ to be conjugates,we must have $z_1 = \overline{z_2}$.
This implies $\sin x + i \cos 2x = \overline{\cos x - i \sin 2x} = \cos x + i \sin 2x$.
Equating real and imaginary parts,we get:
$1) \sin x = \cos x \implies \tan x = 1 \implies x = \frac{\pi}{4}, \frac{5\pi}{4}$.
$2) \cos 2x = \sin 2x \implies \tan 2x = 1 \implies 2x = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4} \implies x = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}$.
Since there is no common value of $x$ that satisfies both equations,the set is empty.
Thus,the correct option is $D$.

(C) Given $\left|\left(\frac{1}{z_1}+\frac{2}{z_2}\right) \frac{z_3}{z_2}\right| = \left|\frac{1}{z_1}+\frac{2}{z_2}\right| \cdot \frac{|z_3|}{|z_2|}$
Substituting the values: $\left|\frac{1}{1-2 i}+\frac{2}{1+i}\right| \cdot \frac{|3+4 i|}{|1+i|}$
$= \left|\frac{1+2 i}{1^2+2^2} + \frac{2(1-i)}{1^2+1^2}\right| \cdot \frac{\sqrt{3^2+4^2}}{\sqrt{1^2+1^2}}$
$= \left|\frac{1+2 i}{5} + \frac{2-2 i}{2}\right| \cdot \frac{5}{\sqrt{2}}$
$= \left|\frac{2+4 i + 10-10 i}{10}\right| \cdot \frac{5}{\sqrt{2}}$
$= \left|\frac{12-6 i}{10}\right| \cdot \frac{5}{\sqrt{2}} = \frac{6}{10} |2-i| \cdot \frac{5}{\sqrt{2}}$
$= \frac{3}{5} \sqrt{2^2+(-1)^2} \cdot \frac{5}{\sqrt{2}} = \frac{3 \sqrt{5}}{\sqrt{2}} = \sqrt{\frac{9 \times 5}{2}} = \sqrt{\frac{45}{2}}$

(D) We have,$|z|-z=2+i$.
Let $z=x+iy$. Then $\sqrt{x^2+y^2}-(x+iy)=2+i$.
Equating the real and imaginary parts,we get:
$\sqrt{x^2+y^2}-x=2$ and $-y=1$.
Thus,$y=-1$.
Substituting $y=-1$ into the first equation: $\sqrt{x^2+(-1)^2}-x=2$.
$\sqrt{x^2+1}=x+2$.
Squaring both sides: $x^2+1=(x+2)^2 = x^2+4x+4$.
Solving for $x$: $1=4x+4$ $\Rightarrow 4x=-3$ $\Rightarrow x=-\frac{3}{4}$.
Therefore,$z=-\frac{3}{4}-i$.
The modulus is $|z|=\sqrt{(-\frac{3}{4})^2+(-1)^2} = \sqrt{\frac{9}{16}+1} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

(B) Given that the conjugate of $(x+iy)(1-2i)$ is $(1+i)$.
Let $z = (x+iy)(1-2i)$.
Then $\bar{z} = 1+i$.
Since $\bar{z} = (x-iy)(1+2i)$,we have $(x-iy)(1+2i) = 1+i$.
Therefore,$x-iy = \frac{1+i}{1+2i}$.
Taking the conjugate of both sides,we get $\overline{x-iy} = \overline{\left(\frac{1+i}{1+2i}\right)}$.
This simplifies to $x+iy = \frac{1-i}{1-2i}$.

(D) Given,$\left|z-\frac{4}{z}\right|=2$.
Using the triangle inequality,$|z| = \left|z-\frac{4}{z}+\frac{4}{z}\right| \leq \left|z-\frac{4}{z}\right| + \left|\frac{4}{z}\right|$.
Substituting the given value,$|z| \leq 2 + \frac{4}{|z|}$.
Multiplying by $|z|$ (since $|z| > 0$),we get $|z|^2 - 2|z| - 4 \leq 0$.
Solving the quadratic inequality $|z|^2 - 2|z| - 4 = 0$ using the quadratic formula,$|z| = \frac{2 \pm \sqrt{4 - 4(1)(-4)}}{2} = \frac{2 \pm \sqrt{20}}{2} = 1 \pm \sqrt{5}$.
Since $|z| > 0$,we have $0 < |z| \leq 1 + \sqrt{5}$.
Therefore,the greatest value of $|z|$ is $1 + \sqrt{5}$.