Mix Examples-Complex numbers Questions in English

(B) Given roots are $z_1 = \sqrt{3} + \sqrt{27} = \sqrt{3} + 3\sqrt{3} = 4\sqrt{3}$ and $z_2 = \sqrt{2} + 5i$.
Since the coefficients are rational,irrational roots must occur in conjugate pairs.
For $z_1 = 4\sqrt{3}$,the conjugate is $-4\sqrt{3}$.
For $z_2 = \sqrt{2} + 5i$,the conjugate is $\sqrt{2} - 5i$.
However,for a polynomial with rational coefficients,if $\sqrt{a} + \sqrt{b}$ is a root,then $\pm\sqrt{a} \pm \sqrt{b}$ are also roots.
Here,$z_1 = 4\sqrt{3}$. The minimal polynomial for $4\sqrt{3}$ over $\mathbb{Q}$ is $x^2 - 48 = 0$,which has roots $4\sqrt{3}$ and $-4\sqrt{3}$.
For $z_2 = \sqrt{2} + 5i$,the minimal polynomial over $\mathbb{Q}$ is found by $(x - \sqrt{2})^2 = (5i)^2$,which is $x^2 - 2\sqrt{2}x + 2 = -25$,or $x^2 + 27 = 2\sqrt{2}x$.
Squaring both sides gives $(x^2 + 27)^2 = 8x^2$,which is $x^4 + 54x^2 + 729 = 8x^2$,or $x^4 + 46x^2 + 729 = 0$.
This polynomial has roots $\pm\sqrt{2} \pm 5i$.
Thus,the total degree is $2 + 4 = 6$.

(C) Given that $k \sqrt{-1} = ki$ is a root of the equation $x^4+6 x^3-16 x^2+24 x-80=0$.
Since the coefficients are real,the conjugate $-ki$ must also be a root.
Thus,$(x-ki)(x+ki) = x^2+k^2$ is a factor of the polynomial.
Dividing $x^4+6 x^3-16 x^2+24 x-80$ by $x^2+k^2$:
$x^4+6 x^3-16 x^2+24 x-80 = (x^2+k^2)(x^2+6x-(16+k^2)) + (24-6k^2)x + (k^2(16+k^2)-80)$.
For $x^2+k^2$ to be a factor,the remainder must be zero.
Setting the coefficient of $x$ to zero: $24-6k^2 = 0 \implies 6k^2 = 24 \implies k^2 = 4$.
Checking the constant term: $k^2(16+k^2)-80 = 4(16+4)-80 = 4(20)-80 = 0$.
Thus,$k^2 = 4$ is the correct value.

(A) Given that $f(x)$ is a polynomial with rational coefficients.
If a complex number $a+bi$ is a root,its conjugate $a-bi$ must also be a root. Thus,$1+2i$ and $1-2i$ are roots.
If an irrational number of the form $a+\sqrt{b}$ is a root,its conjugate $a-\sqrt{b}$ must also be a root. Thus,$2-\sqrt{3}$ and $2+\sqrt{3}$ are roots.
Additionally,$5$ is given as a root.
Therefore,the roots are $1+2i, 1-2i, 2-\sqrt{3}, 2+\sqrt{3}$,and $5$.
Counting these,we have $5$ distinct roots.
Hence,the least degree $n$ of the polynomial is $5$.

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

(C) Given $x = -5 + 2 \sqrt{-4} = -5 + 4i$.
$x + 5 = 4i$
Squaring both sides:
$(x + 5)^2 = (4i)^2$
$x^2 + 10x + 25 = -16$
$x^2 + 10x + 41 = 0$
Now,divide the polynomial $P(x) = x^4 + 9x^3 + 35x^2 - x + 4$ by $x^2 + 10x + 41$:
$x^4 + 9x^3 + 35x^2 - x + 4 = (x^2 + 10x + 41)(x^2 - x + 4) - 160$
Since $x^2 + 10x + 41 = 0$,we have:
$P(x) = 0 \cdot (x^2 - x + 4) - 160 = -160$.

(A) We have,
$\frac{(1+i)^{2016}}{(1-i)^{2014}} = \frac{(1+i)^{2014} \times (1+i)^2}{(1-i)^{2014}}$
$= \left(\frac{1+i}{1-i}\right)^{2014} \times (1+i)^2$
Since $\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+2i+i^2}{1-i^2} = \frac{2i}{2} = i$
So,the expression becomes $i^{2014} \times (1+2i+i^2)$
$= i^{2014} \times (2i)$
Since $i^4 = 1$,$i^{2014} = (i^4)^{503} \times i^2 = 1^{503} \times (-1) = -1$
Therefore,$-1 \times 2i = -2i$

(D) Given $x = 3 - 2\sqrt{3}i$.
Rearranging,we get $x - 3 = -2\sqrt{3}i$.
Squaring both sides,$(x - 3)^2 = (-2\sqrt{3}i)^2$.
$x^2 - 6x + 9 = 4 \times 3 \times i^2$.
Since $i^2 = -1$,$x^2 - 6x + 9 = -12$.
$x^2 - 6x + 21 = 0$.
Now,we divide the polynomial $P(x) = x^4 - 12x^3 + 54x^2 - 108x - 54$ by $(x^2 - 6x + 21)$.
$x^4 - 12x^3 + 54x^2 - 108x - 54 = (x^2 - 6x + 21)(x^2 - 6x - 9) + 135$.
Wait,let us re-evaluate the division:
$(x^2 - 6x + 21)(x^2 - 6x) = x^4 - 6x^3 - 6x^3 + 36x^2 + 21x^2 - 126x = x^4 - 12x^3 + 57x^2 - 126x$.
Subtracting this from $P(x)$: $(x^4 - 12x^3 + 54x^2 - 108x - 54) - (x^4 - 12x^3 + 57x^2 - 126x) = -3x^2 + 18x - 54$.
$-3(x^2 - 6x + 21) + 63 - 54 = -3(0) + 9 = 9$.
Thus,the value is $9$.

(D) Given,$|Z-1| \leq 2$ and $|\omega^2 Z-1-\omega|=a$.
Since $1+\omega+\omega^2=0$,we have $-1-\omega = \omega^2$.
Substituting this into the second equation: $|\omega^2 Z + \omega^2| = a$.
Since $|\omega^2| = 1$,this simplifies to $|Z+1| = a$.
We can rewrite this as $|(Z-1)+2| = a$.
By the triangle inequality,$|Z-1+2| \leq |Z-1| + |2|$.
Given $|Z-1| \leq 2$,we have $a = |Z+1| \leq |Z-1| + 2 \leq 2 + 2 = 4$.
Also,the modulus $a$ must be non-negative,so $a \geq 0$.
Thus,the range of possible values for $a$ is $0 \leq a \leq 4$.

(C) Given the roots are $z_1 = 2 - 3i$,$z_2 = i$,and $z_3 = \bar{z}_1 = 2 + 3i$.
The cubic equation is given by $(z - z_1)(z - z_2)(z - z_3) = 0$.
Substituting the values: $(z - i)(z - (2 - 3i))(z - (2 + 3i)) = 0$.
First,simplify the product of the last two factors: $(z - (2 - 3i))(z - (2 + 3i)) = ((z - 2) + 3i)((z - 2) - 3i) = (z - 2)^2 - (3i)^2 = z^2 - 4z + 4 + 9 = z^2 - 4z + 13$.
Now,multiply by $(z - i)$: $(z - i)(z^2 - 4z + 13) = z^3 - 4z^2 + 13z - iz^2 + 4iz - 13i = z^3 + (-4 - i)z^2 + (13 + 4i)z - 13i = 0$.
Comparing this with $z^3 + bz^2 + cz + d = 0$,we get $b = -4 - i$,$c = 13 + 4i$,and $d = -13i$.
Therefore,$b + c + d = (-4 - i) + (13 + 4i) + (-13i) = (-4 + 13) + (-i + 4i - 13i) = 9 - 10i$.

(D) Given,$(x-iy)^{1/3} = 2-i\sqrt{3}$
$\Rightarrow x-iy = (2-i\sqrt{3})^3$
Using the identity $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:
$x-iy = 2^3 - 3(2^2)(i\sqrt{3}) + 3(2)(i\sqrt{3})^2 - (i\sqrt{3})^3$
$x-iy = 8 - 12i\sqrt{3} + 6(-3) - (-3i\sqrt{3})$
$x-iy = 8 - 12i\sqrt{3} - 18 + 3i\sqrt{3}$
$x-iy = -10 - 9i\sqrt{3}$
Comparing real and imaginary parts,we get $x = -10$ and $y = 9\sqrt{3}$.
Since the point $z = (x, y)$ lies on the line $\frac{x}{2} + \frac{y}{\sqrt{3}} = k$:
$\frac{-10}{2} + \frac{9\sqrt{3}}{\sqrt{3}} = k$
$-5 + 9 = k$
$k = 4$

(D) Given the condition $(a+bi)^3 = a-bi$.
Expanding the left side: $a^3 + 3a^2(bi) + 3a(bi)^2 + (bi)^3 = a-bi$.
$a^3 + 3a^2bi - 3ab^2 - ib^3 = a-bi$.
Grouping real and imaginary parts: $(a^3 - 3ab^2) + i(3a^2b - b^3) = a - bi$.
Equating real and imaginary parts:
$1) a^3 - 3ab^2 = a \Rightarrow a(a^2 - 3b^2 - 1) = 0$.
$2) 3a^2b - b^3 = -b \Rightarrow b(3a^2 - b^2 + 1) = 0$.
Case $1$: If $a=0$,then $b(0 - b^2 + 1) = 0 \Rightarrow b(1-b^2) = 0$. So $b=0, 1, -1$. Pairs: $(0,0), (0,1), (0,-1)$.
Case $2$: If $b=0$,then $a(a^2 - 1) = 0 \Rightarrow a=0, 1, -1$. Pairs: $(0,0), (1,0), (-1,0)$.
Case $3$: If $a \neq 0$ and $b \neq 0$,then $a^2 - 3b^2 = 1$ and $3a^2 - b^2 = -1$.
From the first,$a^2 = 1 + 3b^2$. Substitute into the second: $3(1 + 3b^2) - b^2 = -1$ $\Rightarrow 3 + 9b^2 - b^2 = -1$ $\Rightarrow 8b^2 = -4$,which has no real solution for $b$.
Combining the solutions from Case $1$ and Case $2$,the distinct pairs are $(0,0), (0,1), (0,-1), (1,0), (-1,0)$.
There are $5$ such ordered pairs.

(A) Given that $x+iy = \frac{3}{2+\cos \theta + i \sin \theta}$.
Taking the modulus on both sides,we have $|x+iy| = \left| \frac{3}{2+\cos \theta + i \sin \theta} \right|$.
Since $|x+iy| = \sqrt{x^2+y^2}$,we get $\sqrt{x^2+y^2} = \frac{3}{|2+\cos \theta + i \sin \theta|}$.
Calculating the modulus of the denominator: $|2+\cos \theta + i \sin \theta| = \sqrt{(2+\cos \theta)^2 + \sin^2 \theta} = \sqrt{4 + 4\cos \theta + \cos^2 \theta + \sin^2 \theta} = \sqrt{5+4\cos \theta}$.
Thus,$\sqrt{x^2+y^2} = \frac{3}{\sqrt{5+4\cos \theta}}$,which implies $x^2+y^2 = \frac{9}{5+4\cos \theta}$.
Now,consider $4x-3$. From $x+iy = \frac{3(2+\cos \theta - i \sin \theta)}{(2+\cos \theta)^2 + \sin^2 \theta} = \frac{3(2+\cos \theta) - 3i \sin \theta}{5+4\cos \theta}$,we have $x = \frac{3(2+\cos \theta)}{5+4\cos \theta}$.
Then $4x-3 = \frac{12(2+\cos \theta)}{5+4\cos \theta} - 3 = \frac{24+12\cos \theta - 15 - 12\cos \theta}{5+4\cos \theta} = \frac{9}{5+4\cos \theta}$.
Therefore,$x^2+y^2 = 4x-3$.

(A) Given,$z=x-iy$ and $z^{1/3}=a+ib$.
Taking the cube on both sides,we get:
$(z^{1/3})^3 = (a+ib)^3$
$z = a^3 + (ib)^3 + 3a^2(ib) + 3a(ib)^2$
$z = a^3 - ib^3 + 3a^2bi - 3ab^2$
$z = (a^3 - 3ab^2) - i(b^3 - 3a^2b)$
Since $z = x - iy$,we compare the real and imaginary parts:
$x = a^3 - 3ab^2$ and $y = b^3 - 3a^2b$.
Now,substitute these into the expression:
$\frac{(x/a + y/b)}{a^2+b^2} = \frac{((a^3 - 3ab^2)/a + (b^3 - 3a^2b)/b)}{a^2+b^2}$
$= \frac{(a^2 - 3b^2 + b^2 - 3a^2)}{a^2+b^2}$
$= \frac{-2a^2 - 2b^2}{a^2+b^2} = \frac{-2(a^2+b^2)}{a^2+b^2} = -2$.

(C) Let $z = x + iy$. Then $\bar{z} = x - iy$.
$z + \bar{z} = 2x$ and $z - \bar{z} = 2iy$.
Substituting these into the given equation:
$(7 + i)(2x) - (4 + i)(2iy) + 116i = 0$
$(14x + 2ix) - (8iy - 2y) + 116i = 0$
$(14x + 2y) + i(2x - 8y + 116) = 0$
Equating real and imaginary parts to zero:
$14x + 2y = 0 \Rightarrow y = -7x$
$2x - 8y + 116 = 0$
Substitute $y = -7x$ into the second equation:
$2x - 8(-7x) + 116 = 0$
$2x + 56x + 116 = 0$
$58x = -116 \Rightarrow x = -2$
Then $y = -7(-2) = 14$.
Thus,$z\bar{z} = |z|^2 = x^2 + y^2 = (-2)^2 + 14^2 = 4 + 196 = 200$.

(D) Given: $z_1=1-2 i, z_2=1+i, z_3=3+4 i$.
We need to evaluate $\left(\frac{1}{z_1}+\frac{3}{z_2}\right) \frac{z_3}{z_2}$.
First,calculate the term inside the bracket:
$\frac{1}{1-2 i}+\frac{3}{1+i} = \frac{(1+i)+3(1-2 i)}{(1-2 i)(1+i)} = \frac{1+i+3-6 i}{1+i-2 i-2 i^2} = \frac{4-5 i}{1-i+2} = \frac{4-5 i}{3-i}$.
Now,multiply by $\frac{z_3}{z_2}$:
$\left(\frac{4-5 i}{3-i}\right) \left(\frac{3+4 i}{1+i}\right) = \frac{(4-5 i)(3+4 i)}{(3-i)(1+i)} = \frac{12+16 i-15 i-20 i^2}{3+3 i-i-i^2} = \frac{12+i+20}{3+2 i+1} = \frac{32+i}{4+2 i}$.
Rationalize the denominator:
$\frac{32+i}{4+2 i} \times \frac{4-2 i}{4-2 i} = \frac{128-64 i+4 i-2 i^2}{16-4 i^2} = \frac{128-60 i+2}{16+4} = \frac{130-60 i}{20} = \frac{13}{2}-3 i$.

(D) Let $z = \frac{2+3i \sin \theta}{1-2i \sin \theta}$.
To make $z$ purely imaginary,the real part of $z$ must be zero.
Multiply the numerator and denominator by the conjugate of the denominator: $1+2i \sin \theta$.
$z = \frac{(2+3i \sin \theta)(1+2i \sin \theta)}{(1-2i \sin \theta)(1+2i \sin \theta)} = \frac{2 + 4i \sin \theta + 3i \sin \theta + 6i^2 \sin^2 \theta}{1 + 4 \sin^2 \theta}$.
Since $i^2 = -1$,$z = \frac{2 - 6 \sin^2 \theta + 7i \sin \theta}{1 + 4 \sin^2 \theta}$.
The real part is $\frac{2 - 6 \sin^2 \theta}{1 + 4 \sin^2 \theta}$.
Setting the real part to $0$: $2 - 6 \sin^2 \theta = 0 \implies 6 \sin^2 \theta = 2 \implies \sin^2 \theta = \frac{1}{3}$.
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$,we get $\cos^2 \theta = 1 - \frac{1}{3} = \frac{2}{3}$.

(B) Let $t = z^4$. The equation becomes $t^2 - 20t + 64 = 0$.
Factoring the quadratic equation,we get $(t-16)(t-4) = 0$.
So,$z^4 = 16$ or $z^4 = 4$.
For $z^4 = 16$,the roots are $z = 2, -2, 2i, -2i$.
The imaginary roots are $2i$ and $-2i$.
For $z^4 = 4$,the roots are $z = \sqrt{2}, -\sqrt{2}, \sqrt{2}i, -\sqrt{2}i$.
The imaginary roots are $\sqrt{2}i$ and $-\sqrt{2}i$.
The squares of the imaginary roots are $(2i)^2 = -4$,$(-2i)^2 = -4$,$(\sqrt{2}i)^2 = -2$,and $(-\sqrt{2}i)^2 = -2$.
The sum of the squares of the imaginary roots is $(-4) + (-4) + (-2) + (-2) = -12$.

(D) Given the equation $z^3+\bar{z}=0$,where $z=x+iy$ is a complex number.
Taking the modulus on both sides: $|z^3| = |-\bar{z}|$.
Since $|z^3| = |z|^3$ and $|-\bar{z}| = |z|$,we have $|z|^3 = |z|$.
This implies $|z|(|z|^2-1) = 0$.
Case $1$: $|z|=0$,which gives the solution $z=0$.
Case $2$: $|z|^2=1$,which implies $z\bar{z}=1$,so $\bar{z}=\frac{1}{z}$.
Substituting $\bar{z}=\frac{1}{z}$ into the original equation: $z^3 + \frac{1}{z} = 0$,which simplifies to $z^4+1=0$.
The equation $z^4 = -1$ has $4$ distinct roots.
Including the solution $z=0$ from Case $1$,the total number of distinct solutions is $1 + 4 = 5$.

(C) Let $z = \frac{3-2 i \sin \theta}{1+2 i \sin \theta}$.
To rationalize,multiply the numerator and denominator by the conjugate of the denominator $(1-2 i \sin \theta)$:
$z = \frac{(3-2 i \sin \theta)(1-2 i \sin \theta)}{(1+2 i \sin \theta)(1-2 i \sin \theta)}$
$z = \frac{3 - 6 i \sin \theta - 2 i \sin \theta + 4 i^2 \sin^2 \theta}{1 + 4 \sin^2 \theta}$
Since $i^2 = -1$,we have:
$z = \frac{3 - 4 \sin^2 \theta - 8 i \sin \theta}{1 + 4 \sin^2 \theta}$
For $z$ to be a purely imaginary number,the real part must be zero:
$\frac{3 - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta} = 0$
$3 - 4 \sin^2 \theta = 0$
$\sin^2 \theta = \frac{3}{4} = \left(\frac{\sqrt{3}}{2}\right)^2$
$\sin^2 \theta = \sin^2 \left(\frac{\pi}{3}\right)$
Therefore,the general solution is $\theta = n \pi \pm \frac{\pi}{3}$.

(B) Given the equation $z^3+\bar{z}=0$.
Taking the modulus on both sides,we get $|z^3| = |-\bar{z}|$,which implies $|z|^3 = |z|$.
This gives $|z|(|z|^2 - 1) = 0$.
Case $1$: $|z| = 0$,which implies $z = 0$. This is $1$ solution.
Case $2$: $|z|^2 = 1$,which implies $z\bar{z} = 1$,so $\bar{z} = \frac{1}{z}$.
Substituting this into the original equation: $z^3 + \frac{1}{z} = 0$,which gives $z^4 + 1 = 0$.
The equation $z^4 = -1$ has $4$ distinct roots.
Thus,the total number of solutions is $1 + 4 = 5$.

(C) Given $\left|z-\frac{2}{z}\right|=2$.
Using the triangle inequality property $||z_1|-|z_2|| \leq |z_1-z_2|$,we have:
$||z|-\left|\frac{2}{z}\right|| \leq \left|z-\frac{2}{z}\right| = 2$.
Let $|z| = r$. Then $|r - \frac{2}{r}| \leq 2$.
This implies $-2 \leq r - \frac{2}{r} \leq 2$.
Considering the right inequality: $r - \frac{2}{r} \leq 2
\Rightarrow r^2 - 2r - 2 \leq 0$.
Solving the quadratic equation $r^2 - 2r - 2 = 0$ using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$r = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$.
Since $r = |z| > 0$,we have $r \leq 1 + \sqrt{3}$.
Thus,the greatest value of $|z|$ is $\sqrt{3} + 1$.

(D) Let $z = -3-4i$. We are given $\sqrt{z} = re^{i\theta}$.
Squaring both sides,we get $z = r^2 e^{i2\theta} = r^2(\cos 2\theta + i \sin 2\theta)$.
Comparing the real and imaginary parts with $z = -3-4i$:
$r^2 \cos 2\theta = -3$ and $r^2 \sin 2\theta = -4$.
We know that $r^2 = |z| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9+16} = 5$.
Using the double angle formula for tangent: $\tan 2\theta = \frac{r^2 \sin 2\theta}{r^2 \cos 2\theta} = \frac{-4}{-3} = \frac{4}{3}$.
Using $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{4}{3}$,we get $6 \tan \theta = 4 - 4 \tan^2 \theta$,or $4 \tan^2 \theta + 6 \tan \theta - 4 = 0$.
Dividing by $2$,$2 \tan^2 \theta + 3 \tan \theta - 2 = 0$.
Factoring gives $(2 \tan \theta - 1)(\tan \theta + 2) = 0$,so $\tan \theta = \frac{1}{2}$ or $\tan \theta = -2$.
Since $z$ is in the third quadrant,its square root $\sqrt{z}$ lies in the second or fourth quadrant. For $\sqrt{z} = x+iy$,$xy = -2$,so $x$ and $y$ have opposite signs.
Testing $\tan \theta = -2$,$r^2 \tan \theta = 5 \times (-2) = -10$.

(C) Given equation: $i z^3+z^2-z+i=0$
Factorizing the equation:
$i z^2(z-i) - 1(z-i) = 0$
$(z-i)(i z^2-1) = 0$
This gives two cases:
Case $1$: $z-i=0 \Rightarrow z=i$. Then $|z| = |i| = 1$.
Case $2$: $i z^2-1=0 \Rightarrow i z^2=1$.
Taking the modulus on both sides:
$|i z^2| = |1|$
$|i| \cdot |z|^2 = 1$
$1 \cdot |z|^2 = 1$ $\Rightarrow |z|^2 = 1$ $\Rightarrow |z| = 1$.
In both cases,$|z| = 1$.

(B) 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 $z_2 = \cos x - i \sin 2x$.
For these to be conjugates,we must have $\sin x = \cos x$ and $\cos 2x = -(-\sin 2x) = \sin 2x$.
From $\sin x = \cos x$,we get $\tan x = 1$,which implies $x = n\pi + \frac{\pi}{4}$.
From $\cos 2x = \sin 2x$,we get $\tan 2x = 1$,which implies $2x = m\pi + \frac{\pi}{4}$,or $x = \frac{m\pi}{2} + \frac{\pi}{8}$.
Since there is no value of $x$ that satisfies both conditions simultaneously,there is no solution.

(B) Given equation is $\overline{z} = i z^2 \dots (i)$
Taking modulus on both sides,$|\overline{z}| = |i z^2|$ $\Rightarrow |z| = |i| |z|^2$ $\Rightarrow |z| = |z|^2$.
This implies $|z|(|z| - 1) = 0$,so $|z| = 0$ or $|z| = 1$.
Case $1$: If $|z| = 0$,then $z = 0$. This is one solution.
Case $2$: If $|z| = 1$,then $\overline{z} = 1/z$. Substituting into $(i)$,$1/z = i z^2 \Rightarrow z^3 = 1/i = -i$.
We know $-i = e^{i(3\pi/2 + 2k\pi)}$ for $k = 0, 1, 2$.
Thus,$z = e^{i(\pi/2 + 2k\pi/3)}$.
For $k = 0$,$z = e^{i\pi/2} = i$.
For $k = 1$,$z = e^{i(7\pi/6)} = -\frac{\sqrt{3}}{2} - \frac{i}{2}$.
For $k = 2$,$z = e^{i(11\pi/6)} = \frac{\sqrt{3}}{2} - \frac{i}{2}$.
Including $z = 0$,there are $1 + 3 = 4$ solutions.

(C) Given the equation: $|z|^2 w - |w|^2 z = z - w$
Rearranging the terms: $|z|^2 w + w = |w|^2 z + z$
$(|z|^2 + 1) w = (|w|^2 + 1) z$
Since $z$ and $w$ are non-zero,we can write: $\frac{z}{|z|^2 + 1} = \frac{w}{|w|^2 + 1}$
Let $\frac{z}{|z|^2 + 1} = \frac{w}{|w|^2 + 1} = k$
Then $z = k(|z|^2 + 1)$ and $w = k(|w|^2 + 1)$
Since $z$ and $w$ are complex numbers,$k$ must be a real number.
If $k$ is real,then $\frac{z}{w} = \frac{|z|^2 + 1}{|w|^2 + 1}$,which is a real number.
Thus,$z = cw$ for some real constant $c \neq 1$.
Substituting $z = cw$ into the original equation:
$|cw|^2 w - |w|^2 (cw) = cw - w$
$c^2 |w|^2 w - c |w|^2 w = (c - 1) w$
Since $w \neq 0$,we divide by $w$:
$c^2 |w|^2 - c |w|^2 = c - 1$
$c |w|^2 (c - 1) = c - 1$
Since $z \neq w$,$c \neq 1$,so we divide by $(c - 1)$:
$c |w|^2 = 1 \Rightarrow c = \frac{1}{|w|^2}$
Therefore,$z = \frac{w}{|w|^2} = \frac{w}{w \bar{w}} = \frac{1}{\bar{w}}$
This implies $z \bar{w} = 1$.

(A) We want to find the minimum value of $f(z) = |z| + |2z - 3| + |z - 1|$.
Using the property of modulus $|a| + |b| \geq |a + b|$,we can rewrite the expression as:
$|z| + |z - 1| + |3 - 2z| \geq |z + z - 1 + 3 - 2z|$
$|z| + |z - 1| + |3 - 2z| \geq |2|$
$|z| + |z - 1| + |3 - 2z| \geq 2$
Thus,the minimum value is $2$.

(A) Given that $(2+i)$ is a root of the polynomial equation $x^3-5x^2+9x-5=0$ with real coefficients,its complex conjugate $(2-i)$ must also be a root.
Let the three roots be $r_1 = 2+i$,$r_2 = 2-i$,and $r_3 = \alpha$.
According to the properties of roots of a cubic equation $ax^3+bx^2+cx+d=0$,the product of the roots is given by $r_1 r_2 r_3 = -d/a$.
Here,$a=1$ and $d=-5$,so $r_1 r_2 r_3 = -(-5)/1 = 5$.
Substituting the known roots: $(2+i)(2-i) \alpha = 5$.
Since $(2+i)(2-i) = 2^2 - i^2 = 4 - (-1) = 5$,we have $5 \alpha = 5$,which implies $\alpha = 1$.
Thus,the other roots are $1$ and $(2-i)$.

(C) Given equation: $i x^2 - 3 x - 2 i = 0$
Since $i^2 = -1$,we can write $-3x$ as $i^2 x - 4x$ or manipulate the middle term.
Alternatively,using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
Here $a = i$,$b = -3$,$c = -2i$.
$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(i)(-2i)}}{2(i)}$
$x = \frac{3 \pm \sqrt{9 + 8i^2}}{2i}$
$x = \frac{3 \pm \sqrt{9 - 8}}{2i}$
$x = \frac{3 \pm \sqrt{1}}{2i}$
$x = \frac{3 \pm 1}{2i}$
Case $1$: $x = \frac{3+1}{2i} = \frac{4}{2i} = \frac{2}{i} = -2i$
Case $2$: $x = \frac{3-1}{2i} = \frac{2}{2i} = \frac{1}{i} = -i$
Thus,the solutions are $x = -i$ and $x = -2i$.

(C) Given the equation $\bar{z} = i z^2$.
Taking the conjugate on both sides,we get $z = \bar{i} \bar{z}^2 = -i \bar{z}^2$.
Substitute $\bar{z} = i z^2$ into the equation:
$z = -i (i z^2)^2$
$z = -i (i^2 z^4)$
$z = -i (-1) z^4$
$z = i z^4$
$z^4 - i z = 0$
$z (z^3 - i) = 0$
Since we are looking for non-zero solutions,$z \neq 0$,so $z^3 - i = 0$.

(C) Given $Z = a + ib$,we have $\hat{Z} = b + ia = i(a - ib) = i\bar{Z}$.
Let $Z_1 = a + ib$ and $Z_2 = c + id$.
Then $Z_1 Z_2 = (ac - bd) + i(ad + bc)$.
Applying the definition of $\hat{Z}$,we get $\widehat{Z_1 Z_2} = (ad + bc) + i(ac - bd)$.
Now,calculate $\hat{Z}_1 \hat{Z}_2 = (b + ia)(d + ic) = (bd + ibc + iad + i^2 ac) = (bd - ac) + i(bc + ad)$.
Comparing the two,we see $\widehat{Z_1 Z_2} = -i \hat{Z}_1 \hat{Z}_2$ or similar relations depending on the definition.
However,evaluating the provided options,there is a discrepancy. Given the standard form,the correct identity is $\widehat{Z_1 Z_2} = \frac{\hat{Z}_1 \hat{Z}_2}{i}$.

(A) Given that,$\alpha = \frac{a+ib}{1-c}$.
Taking the modulus squared on both sides,we get $|\alpha|^2 = \frac{a^2+b^2}{(1-c)^2} \quad \dots(i)$.
Given $a^2+b^2+c^2=c$,we have $a^2+b^2 = c-c^2 = c(1-c) \quad \dots(ii)$.
Substituting $(ii)$ into $(i)$,we get $|\alpha|^2 = \frac{c(1-c)}{(1-c)^2} = \frac{c}{1-c}$.
This implies $1+|\alpha|^2 = 1 + \frac{c}{1-c} = \frac{1-c+c}{1-c} = \frac{1}{1-c}$.
Thus,$1-c = \frac{1}{1+|\alpha|^2}$.
From $(ii)$,$a^2+b^2 = c(1-c) = (1-(1-c))(1-c) = (1-c) - (1-c)^2$.
Substituting $1-c = \frac{1}{1+|\alpha|^2}$,we get $a^2+b^2 = \frac{1}{1+|\alpha|^2} - \left(\frac{1}{1+|\alpha|^2}\right)^2 = \frac{1+|\alpha|^2-1}{(1+|\alpha|^2)^2} = \frac{|\alpha|^2}{(1+|\alpha|^2)^2}$.

(D) Given the equation $z^4+z^2+1=0$.
Dividing by $z^2$ (since $z \neq 0$),we get $z^2+1+\frac{1}{z^2}=0$,which implies $z^2+\frac{1}{z^2}=-1$.
Also,$(z+\frac{1}{z})^2 = z^2+\frac{1}{z^2}+2 = -1+2 = 1$,so $z+\frac{1}{z} = \pm 1$.
For $z^3+\frac{1}{z^3}$,we use the identity $z^3+\frac{1}{z^3} = (z+\frac{1}{z})(z^2-1+\frac{1}{z^2}) = (z+\frac{1}{z})(-1-1) = -2(z+\frac{1}{z})$.
If $z+\frac{1}{z} = 1$,then $z^3+\frac{1}{z^3} = -2(1) = -2$.
The expression becomes $(1)^3+(-1)^2+(-2)^3 = 1+1-8 = -6$.
If $z+\frac{1}{z} = -1$,then $z^3+\frac{1}{z^3} = -2(-1) = 2$.
The expression becomes $(-1)^3+(-1)^2+(2)^3 = -1+1+8 = 8$.
Since the question implies a unique value,we check the roots of $z^4+z^2+1=0$,which are $e^{i2\pi/3}, e^{i4\pi/3}, e^{i8\pi/3}, e^{i10\pi/3}$. For these,$z+\frac{1}{z} = 2\cos(2\pi/3) = -1$. Thus,the value is $8$.

(A) Given,$(a+ib)^2=(c+id)^2(x+iy)$
Taking the modulus on both sides,we get:
$|a+ib|^2 = |c+id|^2 |x+iy|$
Since $|z|^2 = a^2+b^2$ for $z = a+ib$,we have:
$a^2+b^2 = (c^2+d^2) \sqrt{x^2+y^2}$
Substituting the given values $a^2+b^2=4$ and $c^2+d^2=2$:
$4 = 2 \sqrt{x^2+y^2}$
$\sqrt{x^2+y^2} = 2$
Squaring both sides,we get:
$x^2+y^2 = 4$

(D) Given,$\left|z-\frac{4}{z}\right|=2$.
Using the triangle inequality $|a+b| \leq |a|+|b|$,we have:
$|z| = \left|z-\frac{4}{z} + \frac{4}{z}\right| \leq \left|z-\frac{4}{z}\right| + \left|\frac{4}{z}\right|$.
Substituting the given value: $|z| \leq 2 + \frac{4}{|z|}$.
Multiplying by $|z|$ (since $|z| > 0$): $|z|^2 \leq 2|z| + 4$.
$|z|^2 - 2|z| - 4 \leq 0$.
Completing the square: $(|z|-1)^2 - 1 - 4 \leq 0 \Rightarrow (|z|-1)^2 \leq 5$.
Taking the square root: $|z|-1 \leq \sqrt{5}$.
Therefore,$|z| \leq \sqrt{5}+1$.
The greatest value of $|z|$ is $\sqrt{5}+1$.

(A) Given $x_n = \cos \left(\frac{\pi}{4^n}\right) + i \sin \left(\frac{\pi}{4^n}\right) = e^{i \frac{\pi}{4^n}}$.
The product $P = x_1 x_2 x_3 \ldots \infty$ is given by:
$P = e^{i \frac{\pi}{4^1}} \cdot e^{i \frac{\pi}{4^2}} \cdot e^{i \frac{\pi}{4^3}} \ldots = e^{i \pi \left(\frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} + \ldots \right)}$.
The exponent is a geometric series with first term $a = \frac{1}{4}$ and common ratio $r = \frac{1}{4}$.
The sum of the infinite geometric series is $S = \frac{a}{1 - r} = \frac{1/4}{1 - 1/4} = \frac{1/4}{3/4} = \frac{1}{3}$.
Therefore,$P = e^{i \pi \left(\frac{1}{3}\right)} = e^{i \frac{\pi}{3}}$.
$P = \cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right) = \frac{1}{2} + i \frac{\sqrt{3}}{2} = \frac{1 + i \sqrt{3}}{2}$.

(D) Given $z = 3+5i$,then the conjugate is $\bar{z} = 3-5i$.
First,calculate $z^3$:
$z^2 = (3+5i)^2 = 9 + 25i^2 + 30i = 9 - 25 + 30i = -16 + 30i$.
$z^3 = z^2 \cdot z = (-16 + 30i)(3 + 5i) = -48 - 80i + 90i + 150i^2$.
Since $i^2 = -1$,$z^3 = -48 + 10i - 150 = -198 + 10i$.
Now,substitute these into the expression:
$z^3 + \bar{z} + 198 = (-198 + 10i) + (3 - 5i) + 198$.
$= (-198 + 198) + (10i - 5i) + 3 = 0 + 5i + 3 = 3 + 5i$.

(A) Given that $(r, \theta) = r(\cos \theta + i \sin \theta) = re^{i\theta}$.
Thus,$x = e^{i\alpha}$,$y = e^{i\beta}$,and $z = e^{i\gamma}$.
Given $x + y + z = 0$,we know that for complex numbers,if $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.
Dividing by $xyz$,we get $\frac{x^2}{yz} + \frac{y^2}{xz} + \frac{z^2}{xy} = 3$.
Substituting the exponential forms:
$\frac{e^{2i\alpha}}{e^{i\beta}e^{i\gamma}} + \frac{e^{2i\beta}}{e^{i\alpha}e^{i\gamma}} + \frac{e^{2i\gamma}}{e^{i\alpha}e^{i\beta}} = 3$.
$e^{i(2\alpha - \beta - \gamma)} + e^{i(2\beta - \alpha - \gamma)} + e^{i(2\gamma - \alpha - \beta)} = 3$.
Taking the real part of both sides:
$\cos(2\alpha - \beta - \gamma) + \cos(2\beta - \alpha - \gamma) + \cos(2\gamma - \alpha - \beta) = 3$.

(C) Given $|z|=1$ and $\operatorname{Arg}(z)=\frac{\pi}{6}$.
$z = |z|e^{i \operatorname{Arg}(z)} = 1 \cdot e^{i \frac{\pi}{6}} = e^{i \frac{\pi}{6}}$.
Then $z^6 = (e^{i \frac{\pi}{6}})^6 = e^{i \pi} = -1$.
And $z^{12} = (e^{i \frac{\pi}{6}})^{12} = e^{2 \pi i} = 1$.
Substituting these values into the expression:
$\frac{z^{12}+1-z^6}{z^{12}+i z^6-1} = \frac{1+1-(-1)}{1+i(-1)-1} = \frac{3}{-i} = \frac{3}{-i} \cdot \frac{i}{i} = \frac{3i}{-i^2} = \frac{3i}{1} = 3i$.

(C) Let $C = \sum_{n=0}^{\infty} \frac{\cos (n \theta)}{2^n} = 1 + \frac{\cos \theta}{2} + \frac{\cos (2 \theta)}{2^2} + \dots$ and $S = \sum_{n=0}^{\infty} \frac{\sin (n \theta)}{2^n} = 0 + \frac{\sin \theta}{2} + \frac{\sin (2 \theta)}{2^2} + \dots$
Consider $C + iS = \sum_{n=0}^{\infty} \frac{e^{in \theta}}{2^n} = \sum_{n=0}^{\infty} \left(\frac{e^{i \theta}}{2}\right)^n$.
This is an infinite geometric series with first term $a = 1$ and common ratio $r = \frac{e^{i \theta}}{2}$.
Since $|r| = |\frac{e^{i \theta}}{2}| = \frac{1}{2} < 1$, the sum is $C + iS = \frac{1}{1 - \frac{e^{i \theta}}{2}} = \frac{2}{2 - e^{i \theta}}$.
Substituting $e^{i \theta} = \cos \theta + i \sin \theta$, we get $C + iS = \frac{2}{2 - \cos \theta - i \sin \theta}$.
Multiply the numerator and denominator by the conjugate $(2 - \cos \theta + i \sin \theta)$:
$C + iS = \frac{2(2 - \cos \theta + i \sin \theta)}{(2 - \cos \theta)^2 + \sin^2 \theta} = \frac{4 - 2 \cos \theta + 2i \sin \theta}{4 - 4 \cos \theta + \cos^2 \theta + \sin^2 \theta}$.
Since $\cos^2 \theta + \sin^2 \theta = 1$, the denominator is $4 - 4 \cos \theta + 1 = 5 - 4 \cos \theta$.
Thus, $C + iS = \frac{4 - 2 \cos \theta}{5 - 4 \cos \theta} + i \frac{2 \sin \theta}{5 - 4 \cos \theta}$.
Comparing the real parts, $C = \frac{4 - 2 \cos \theta}{5 - 4 \cos \theta}$.

(D) Given $z = \cos 6^{\circ} + i \sin 6^{\circ} = e^{i 6^{\circ}}$.
We need to find $\sum_{n=1}^{20} \operatorname{Im}(z^{2n-1}) = \operatorname{Im} \left( \sum_{n=1}^{20} z^{2n-1} \right)$.
The sum is a geometric progression: $S = z + z^3 + z^5 + \dots + z^{39}$.
Here,the first term $a = z$ and the common ratio $r = z^2 = e^{i 12^{\circ}}$.
The sum of $20$ terms is $S = z \frac{(z^2)^{20} - 1}{z^2 - 1} = z \frac{z^{40} - 1}{z^2 - 1}$.
Substituting $z = e^{i 6^{\circ}}$,we get $S = e^{i 6^{\circ}} \frac{e^{i 240^{\circ}} - 1}{e^{i 12^{\circ}} - 1}$.
Using the identity $e^{i \theta} - 1 = e^{i \theta/2} (e^{i \theta/2} - e^{-i \theta/2}) = e^{i \theta/2} (2i \sin(\theta/2))$,we have:
$S = e^{i 6^{\circ}} \frac{e^{i 120^{\circ}} (2i \sin 120^{\circ})}{e^{i 6^{\circ}} (2i \sin 6^{\circ})} = e^{i 120^{\circ}} \frac{\sin 120^{\circ}}{\sin 6^{\circ}}$.
$S = (\cos 120^{\circ} + i \sin 120^{\circ}) \frac{\sin 120^{\circ}}{\sin 6^{\circ}} = \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) \frac{\sqrt{3}/2}{\sin 6^{\circ}}$.
$S = -\frac{\sqrt{3}}{4 \sin 6^{\circ}} + i \frac{3}{4 \sin 6^{\circ}}$.
Taking the imaginary part,$\operatorname{Im}(S) = \frac{3}{4 \sin 6^{\circ}}$.

(C) Given $\theta = \frac{\pi}{2}$,we have $\cos \frac{\pi}{2} = 0$ and $\sin \frac{\pi}{2} = 1$.
Substituting these values into the first term: $\left(\frac{0+i(1)}{1+i(0)}\right)^{2024} = (i)^{2024} = (i^4)^{506} = 1^{506} = 1$.
For the second term: $\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta+i \sin \theta} = \frac{1+0+i(1)}{1-0+i(1)} = \frac{1+i}{1+i} = 1$.
Thus,the expression becomes $1^{2024} + 1^{2025} = 1 + 1 = 2$.
Since $x+iy = 2$,we have $x=2$ and $y=0$.
Therefore,$x+y = 2+0 = 2$.

(C) Using the property of complex numbers in polar form,$e^{i\theta} = \cos \theta + i \sin \theta$,the given expression can be written as:
$e^{i \frac{\pi}{2}} \cdot e^{i \frac{\pi}{4}} \cdot e^{i \frac{\pi}{8}} \ldots \infty$
$= e^{i(\frac{\pi}{2} + \frac{\pi}{4} + \frac{\pi}{8} + \ldots \infty)}$
The exponent is an infinite geometric series with first term $a = \frac{\pi}{2}$ and common ratio $r = \frac{1}{2}$.
The sum of an infinite geometric series is $S_{\infty} = \frac{a}{1-r}$.
$S_{\infty} = \frac{\frac{\pi}{2}}{1 - \frac{1}{2}} = \frac{\frac{\pi}{2}}{\frac{1}{2}} = \pi$.
Thus,the expression becomes $e^{i\pi}$.
Using Euler's formula,$e^{i\pi} = \cos \pi + i \sin \pi = -1 + i(0) = -1$.
Hence,option $C$ is correct.

(D) Given,$x^2-\sqrt{3} x+1=0$.
Dividing by $x$,we get $x+\frac{1}{x}=\sqrt{3}$.
We know that $\left(x^n-\frac{1}{x^n}\right)^2 = \left(x^n+\frac{1}{x^n}\right)^2 - 4$.
Let $x = e^{i\theta} = \cos \theta + i \sin \theta$. Then $x+\frac{1}{x} = 2 \cos \theta = \sqrt{3}$,so $\cos \theta = \frac{\sqrt{3}}{2}$,which implies $\theta = \frac{\pi}{6}$.
Thus,$x^n = \cos \frac{n\pi}{6} + i \sin \frac{n\pi}{6}$ and $\frac{1}{x^n} = \cos \frac{n\pi}{6} - i \sin \frac{n\pi}{6}$.
Then $x^n - \frac{1}{x^n} = 2i \sin \frac{n\pi}{6}$.
Therefore,$\left(x^n - \frac{1}{x^n}\right)^2 = (2i \sin \frac{n\pi}{6})^2 = -4 \sin^2 \frac{n\pi}{6}$.
Now,$\sum_{n=1}^{24} -4 \sin^2 \frac{n\pi}{6} = -4 \sum_{n=1}^{24} \sin^2 \frac{n\pi}{6}$.
Since $\sin^2 \frac{n\pi}{6}$ has a period of $6$,the sum over $24$ terms is $4 \times \sum_{n=1}^{6} \sin^2 \frac{n\pi}{6}$.
$\sum_{n=1}^{6} \sin^2 \frac{n\pi}{6} = \sin^2 \frac{\pi}{6} + \sin^2 \frac{2\pi}{6} + \sin^2 \frac{3\pi}{6} + \sin^2 \frac{4\pi}{6} + \sin^2 \frac{5\pi}{6} + \sin^2 \frac{6\pi}{6} = (\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2 + (1)^2 + (\frac{\sqrt{3}}{2})^2 + (\frac{1}{2})^2 + 0^2 = \frac{1}{4} + \frac{3}{4} + 1 + \frac{3}{4} + \frac{1}{4} + 0 = 3$.
Thus,the total sum is $-4 \times (4 \times 3) = -48$.