Geometry of complex numbers Questions in English

(B) Given conditions are $|Z| \leq 3$ and $-\frac{\pi}{2} \leq \operatorname{amp}(Z) \leq \frac{\pi}{2}$.
$|Z| \leq 3$ represents a disk of radius $3$ centered at the origin.
$-\frac{\pi}{2} \leq \operatorname{amp}(Z) \leq \frac{\pi}{2}$ represents the region in the first and fourth quadrants (the right half-plane).
Thus,the locus of $Z$ is a semicircle with radius $r = 3$.
The area of the region is $\frac{1}{2} \pi r^2 = \frac{1}{2} \times \pi \times (3)^2 = \frac{9 \pi}{2}$.

(C) Let $z = x + iy$. Then $\frac{z - 3i}{z + 4} = \frac{x + i(y - 3)}{(x + 4) + iy}$.
Multiplying numerator and denominator by the conjugate of the denominator:
$\frac{x + i(y - 3)}{(x + 4) + iy} \times \frac{(x + 4) - iy}{(x + 4) - iy} = \frac{x(x + 4) - xyi + i(y - 3)(x + 4) + y(y - 3)}{(x + 4)^2 + y^2}$.
The real part is $\frac{x^2 + 4x + y^2 - 3y}{(x + 4)^2 + y^2}$ and the imaginary part is $\frac{xy - 3x + 4y - 12 - xy + 4x}{(x + 4)^2 + y^2} = \frac{4y - 3x - 12}{(x + 4)^2 + y^2}$.
Since $\operatorname{Arg}(w) = \frac{\pi}{2}$,the real part must be $0$ and the imaginary part must be positive.
Thus,$x^2 + y^2 + 4x - 3y = 0$ and $4y - 3x - 12 > 0$.
Note that the expression simplifies to $\frac{x^2 + y^2 + 4x - 3y + i(4y - 3x - 12)}{(x + 4)^2 + y^2}$.
Setting the real part to $0$ gives $x^2 + y^2 + 4x - 3y = 0$.
The condition for $\operatorname{Arg} = \frac{\pi}{2}$ is $\text{Re}(w) = 0$ and $\text{Im}(w) > 0$.
Therefore,$4y - 3x - 12 > 0$,which implies $3x - 4y < -12 < 0$.

(C) Given the equation $Z^3+i Z^2+2 i=0$.
By inspection,$Z=i$ is a root because $i^3+i(i^2)+2i = -i-i+2i = 0$.
Dividing the polynomial by $(Z-i)$,we get $(Z-i)(Z^2+2iZ-2)=0$.
Solving $Z^2+2iZ-2=0$ using the quadratic formula: $Z = \frac{-2i \pm \sqrt{(2i)^2 - 4(1)(-2)}}{2} = \frac{-2i \pm \sqrt{-4+8}}{2} = \frac{-2i \pm 2}{2} = -i \pm 1$.
The roots are $Z_1 = i$,$Z_2 = 1-i$,and $Z_3 = -1-i$.
Representing these as points in the Argand plane: $A(0, 1)$,$B(1, -1)$,and $C(-1, -1)$.
Calculating the side lengths:
$AB = \sqrt{(1-0)^2 + (-1-1)^2} = \sqrt{1+4} = \sqrt{5}$.
$BC = \sqrt{(-1-1)^2 + (-1-(-1))^2} = \sqrt{(-2)^2 + 0} = 2$.
$AC = \sqrt{(-1-0)^2 + (-1-1)^2} = \sqrt{1+4} = \sqrt{5}$.
Since $AB = AC = \sqrt{5}$,the triangle is an isosceles triangle.

(D) Given $x^3-a^3=0 \Rightarrow (x-a)(x^2+ax+a^2)=0$.
Since $a>0$,the real root is $\alpha = a$.
The other roots are $\beta = a \omega = a(-\frac{1}{2} + i\frac{\sqrt{3}}{2})$ and $\gamma = a \omega^2 = a(-\frac{1}{2} - i\frac{\sqrt{3}}{2})$.
The first curve is $|z - \beta| = \frac{\sqrt{3}a}{2}$,which represents a circle with center $\beta = (-\frac{a}{2}, \frac{a\sqrt{3}}{2})$ and radius $R = \frac{\sqrt{3}a}{2}$.
The second curve is $|z - \gamma| = \frac{\sqrt{3}a}{2}$,which represents a circle with center $\gamma = (-\frac{a}{2}, -\frac{a\sqrt{3}}{2})$ and radius $R = \frac{\sqrt{3}a}{2}$.
The distance between the centers is $d = \sqrt{(-\frac{a}{2} - (-\frac{a}{2}))^2 + (\frac{a\sqrt{3}}{2} - (-\frac{a\sqrt{3}}{2}))^2} = \sqrt{0 + (a\sqrt{3})^2} = a\sqrt{3}$.
The sum of the radii is $R_1 + R_2 = \frac{\sqrt{3}a}{2} + \frac{\sqrt{3}a}{2} = a\sqrt{3}$.
Since the distance between the centers $d$ is equal to the sum of the radii $R_1 + R_2$,the two circles touch each other externally at exactly one point.
Thus,the number of common points is $1$.

(A) $z_3 = \frac{z_1+z_2}{2} = \frac{(2+3i)+(4-5i)}{2} = \frac{6-2i}{2} = 3-i$.
Given $5z_1+xz_2+yz_3=0$.
Substituting the values: $5(2+3i) + x(4-5i) + y(3-i) = 0$.
$(10+15i) + (4x-5xi) + (3y-yi) = 0$.
Grouping real and imaginary parts: $(10+4x+3y) + i(15-5x-y) = 0$.
Equating real and imaginary parts to zero:
$4x+3y = -10$ $(i)$
$5x+y = 15$ (ii)
From (ii),$y = 15-5x$.
Substituting into $(i)$: $4x + 3(15-5x) = -10$.
$4x + 45 - 15x = -10$.
$-11x = -55 \Rightarrow x = 5$.
$y = 15 - 5(5) = 15-25 = -10$.
Therefore,$x+y = 5-10 = -5$.

(C) Since $\triangle ABC$ is an isosceles right-angled triangle with $\angle C = 90^{\circ}$,we have $AC = BC$.
Using the property of rotation,the vector $\vec{CA}$ is obtained by rotating $\vec{CB}$ by $90^{\circ}$ ($i.e., \frac{\pi}{2}$ radians) counter-clockwise.
Thus,$z_1 - z_3 = i(z_2 - z_3)$.
Also,since it is isosceles,$|z_1 - z_3| = |z_2 - z_3|$.
Now,consider the vector $\vec{BA} = z_1 - z_2$ and $\vec{BC} = z_3 - z_2$.
In $\triangle ABC$,$\angle B = 45^{\circ}$ and $AC = BC$.
Using the rotation formula,$\frac{z_1 - z_2}{z_3 - z_2} = \sqrt{2} e^{i\pi/4} = \sqrt{2}(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}) = \sqrt{2}(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}) = 1 + i$.
Similarly,$\frac{z_2 - z_1}{z_3 - z_1} = \sqrt{2} e^{-i\pi/4} = 1 - i$.
Multiplying these two,we get $\frac{(z_1 - z_2)(z_2 - z_1)}{(z_3 - z_2)(z_3 - z_1)} = (1+i)(1-i) = 1 - i^2 = 2$.
$-(z_1 - z_2)^2 = 2(z_3 - z_2)(z_3 - z_1)$.
Since $z_3 - z_1 = -(z_1 - z_3)$,we have $-(z_1 - z_2)^2 = 2(z_3 - z_2)(-(z_1 - z_3))$.
Therefore,$(z_1 - z_2)^2 = 2(z_1 - z_3)(z_3 - z_2)$.

(D) The condition for three complex numbers $z_1, z_2, z_3$ to form an equilateral triangle is given by the relation:
$z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$
Given that the vertices are $z_1, z_2$ and $0$,we substitute $z_3 = 0$ into the equation:
$z_1^2 + z_2^2 + 0^2 = z_1 z_2 + z_2(0) + (0)z_1$
Simplifying this,we get:
$z_1^2 + z_2^2 = z_1 z_2$

(A) Given equation: $\bar{z} z^3+z \bar{z}^3=350$
$\Rightarrow z \bar{z}(z^2+\bar{z}^2)=350$
Let $z=x+iy$,then $\bar{z}=x-iy$.
Substituting these into the equation:
$(x+iy)(x-iy)[(x+iy)^2+(x-iy)^2]=350$
$(x^2+y^2)[(x^2-y^2+2ixy)+(x^2-y^2-2ixy)]=350$
$(x^2+y^2) \cdot 2(x^2-y^2)=350$
$(x^2+y^2)(x^2-y^2)=175$
Since $x, y \in \mathbb{Z}$,we have $x^2+y^2=25$ and $x^2-y^2=7$.
Adding these equations: $2x^2=32$ $\Rightarrow x^2=16$ $\Rightarrow x=\pm 4$.
Subtracting these equations: $2y^2=18$ $\Rightarrow y^2=9$ $\Rightarrow y=\pm 3$.
The vertices of the rectangle are $(4,3), (4,-3), (-4,-3),$ and $(-4,3)$.
The length of the sides are $6$ and $8$.
Area $= 6 \times 8 = 48$ sq units.

(A) Let $z = x + iy$. Then $iz = -y + ix$ and $z + iz = (x - y) + i(x + y)$.
The vertices of the triangle are $(x, y)$,$(-y, x)$,and $(x - y, x + y)$.
The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the coordinates:
Area $= \frac{1}{2} |x(x - (x + y)) + (-y)((x + y) - y) + (x - y)(y - x)|$
Area $= \frac{1}{2} |x(-y) - y(x) + (x - y)(-(x - y))|$
Area $= \frac{1}{2} |-xy - xy - (x - y)^2|$
Area $= \frac{1}{2} |-2xy - (x^2 - 2xy + y^2)|$
Area $= \frac{1}{2} |-2xy - x^2 + 2xy - y^2|$
Area $= \frac{1}{2} |-x^2 - y^2| = \frac{1}{2} (x^2 + y^2) = \frac{1}{2} |z|^2$.

(C) Given the equations of the curves:
$|z|=1 \Rightarrow x^2+y^2=1$ $(i)$
$|z-2|=1 \Rightarrow (x-2)^2+y^2=1$ $(ii)$
$|z-1|=0 \Rightarrow (x-1)^2+y^2=0$ $(iii)$
Solving equations $(i)$ and $(ii)$:
$(x-2)^2+y^2 = x^2+y^2$
$(x-2)^2 = x^2$
$x^2-4x+4 = x^2$
$4x = 4 \Rightarrow x=1$
Substituting $x=1$ into equation $(i)$:
$1^2+y^2=1$ $\Rightarrow y^2=0$ $\Rightarrow y=0$
So,the intersection point of $(i)$ and $(ii)$ is $(1,0)$.
Now,check if $(1,0)$ satisfies equation $(iii)$:
$(1-1)^2+0^2 = 0^2+0^2 = 0$
Since it satisfies all three equations,the common point is $(1,0)$.

(A) The general equation of a circle in the Cartesian plane is given by $x^2 + y^2 + 2gx + 2fy + c = 0$ ... $(i)$
Let $z = x + iy$ and $\bar{z} = x - iy$.
Then $z + \bar{z} = 2x$ and $z \bar{z} = x^2 + y^2$.
Also,$y = \frac{z - \bar{z}}{2i} = -\frac{i}{2}(z - \bar{z})$.
Substituting these into equation $(i)$:
$z \bar{z} + 2g(\frac{z + \bar{z}}{2}) + 2f(\frac{z - \bar{z}}{2i}) + c = 0$
$z \bar{z} + g(z + \bar{z}) - if(z - \bar{z}) + c = 0$
$z \bar{z} + (g - if)z + (g + if)\bar{z} + c = 0$
Let $b = g + if$,then $\bar{b} = g - if$.
Substituting these,we get $z \bar{z} + \bar{b}z + b\bar{z} + c = 0$.
This represents a circle in the complex plane.
Hence,option $A$ is correct.

(B) The given inequality is $|z - (2 + 2i)| \leq 1$.
Let $z = x + iy$,where $x, y \in \mathbb{R}$.
Substituting $z$ into the inequality,we get $|(x - 2) + i(y - 2)| \leq 1$.
Squaring both sides,we obtain $(x - 2)^2 + (y - 2)^2 \leq 1^2$.
This represents a closed circular disc in the complex plane with center at $(2, 2)$ and radius $r = 1$.
Therefore,option $B$ is correct.

(D) For condition $(i)$, $\frac{2z+i}{z-2} = \frac{2(x+iy)+i}{(x-2)+iy} = \frac{2x + i(2y+1)}{(x-2)+iy}$.
For this to be purely imaginary, the real part must be zero:
$\operatorname{Re}\left(\frac{2x + i(2y+1)}{(x-2)+iy} \cdot \frac{(x-2)-iy}{(x-2)-iy}\right) = 0 \implies 2x(x-2) + y(2y+1) = 0$.
$2x^2 - 4x + 2y^2 + y = 0 \implies x^2 + y^2 - 2x + \frac{1}{2}y = 0$ (Circle $C_1$).
For condition $(ii)$, $\operatorname{Arg}\left(\frac{z+i}{z+1}\right) = \frac{\pi}{2}$ implies $\frac{z+i}{z+1}$ is purely imaginary with a positive imaginary part.
Let $z+i = x+i(y+1)$ and $z+1 = (x+1)+iy$.
$\frac{z+i}{z+1} = \frac{(x+i(y+1))((x+1)-iy)}{(x+1)^2+y^2} = \frac{x(x+1) + y(y+1) + i((x+1)(y+1) - xy)}{(x+1)^2+y^2}$.
Real part $x(x+1) + y(y+1) = 0 \implies x^2 + x + y^2 + y = 0$ (Circle $C_2$).
Subtracting the equations of $C_1$ and $C_2$:
$(x^2 + y^2 - 2x + \frac{1}{2}y) - (x^2 + y^2 + x + y) = 0 \implies -3x - \frac{1}{2}y = 0 \implies y = -6x$.
Substituting $y = -6x$ into $x^2 + y^2 + x + y = 0$:
$x^2 + 36x^2 + x - 6x = 0 \implies 37x^2 - 5x = 0$.
$x(37x - 5) = 0 \implies x = 0$ or $x = \frac{5}{37}$.
For $x = \frac{5}{37}$, $y = -6(\frac{5}{37}) = -\frac{30}{37}$.
The intersection point other than the origin is $\left(\frac{5}{37}, -\frac{30}{37}\right)$.

(C) Let the complex numbers be $z_1 = z$,$z_2 = z e^{i \pi / 3}$,and $z_3 = z(1 + e^{i \pi / 3})$.
$PQ = |z_2 - z_1| = |z e^{i \pi / 3} - z| = |z| |e^{i \pi / 3} - 1|$.
Using $e^{i \theta} = \cos \theta + i \sin \theta$,we have $|e^{i \pi / 3} - 1| = |(\cos \frac{\pi}{3} - 1) + i \sin \frac{\pi}{3}| = |-\frac{1}{2} + i \frac{\sqrt{3}}{2}| = \sqrt{(-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = 1$.
Thus,$PQ = |z|$.
$QR = |z_3 - z_2| = |z(1 + e^{i \pi / 3}) - z e^{i \pi / 3}| = |z|$.
$PR = |z_3 - z_1| = |z(1 + e^{i \pi / 3}) - z| = |z e^{i \pi / 3}| = |z|$.
Since $PQ = QR = PR = |z|$,the triangle $PQR$ is an equilateral triangle with side length $s = |z|$.
The area of an equilateral triangle is given by $\frac{\sqrt{3}}{4} s^2$.
Therefore,the area of $\triangle PQR = \frac{\sqrt{3}}{4} |z|^2$.

(D) The condition $\arg \left(\frac{z-z_1}{z-z_2}\right)=0$ or $\pi$ implies that the vectors $(z-z_1)$ and $(z-z_2)$ are collinear.
This means that the point $z$ lies on the line passing through the points $z_1$ and $z_2$.
Specifically,if $\arg \left(\frac{z-z_1}{z-z_2}\right)=0$,the point $z$ lies on the line outside the segment $\overline{AB}$.
If $\arg \left(\frac{z-z_1}{z-z_2}\right)=\pi$,the point $z$ lies on the line segment $\overline{AB}$.
Therefore,the locus of $z$ is the straight line passing through the points $A$ and $B$,excluding the segment $\overline{AB}$ for the $0$ case and including it for the $\pi$ case,effectively covering the entire line.

(A) Let $Z = x + iy$.
Then $\frac{Z-1}{Z+1} = \frac{(x-1) + iy}{(x+1) + iy}$.
Multiplying the numerator and denominator by the conjugate of the denominator $(x+1) - iy$,we get:
$\frac{Z-1}{Z+1} = \frac{((x-1) + iy)((x+1) - iy)}{(x+1)^2 + y^2} = \frac{(x^2 - 1 + y^2) + i(y(x+1) - y(x-1))}{(x+1)^2 + y^2} = \frac{(x^2 + y^2 - 1) + 2iy}{(x+1)^2 + y^2}$.
Given $\operatorname{Arg}\left(\frac{Z-1}{Z+1}\right) = \frac{\pi}{4}$,we have $\tan\left(\frac{\pi}{4}\right) = \frac{\text{Imaginary part}}{\text{Real part}} = \frac{2y}{x^2 + y^2 - 1}$.
Since $\tan\left(\frac{\pi}{4}\right) = 1$,we have $1 = \frac{2y}{x^2 + y^2 - 1}$.
This simplifies to $x^2 + y^2 - 1 = 2y$,or $x^2 + y^2 - 2y - 1 = 0$.

(C) Given points are $A(3-i)$ and $B(2+i)$.
In the Argand plane,these correspond to coordinates $A(3, -1)$ and $B(2, 1)$.
The given equation is $|z-3+i|=|z-2-i|$.
This can be rewritten as $|z-(3-i)|=|z-(2+i)|$.
Let $P$ be the point representing the complex number $z$. Then the equation represents the set of points $P$ such that the distance from $P$ to $A$ is equal to the distance from $P$ to $B$,i.e.,$PA = PB$.
The locus of points equidistant from two fixed points $A$ and $B$ is the perpendicular bisector of the line segment $AB$.

(B) Given $\left|\frac{z-i}{z i}\right|=2$.
Since $z=x iy$,we have $\left|\frac{x i(y-1)}{x i(y 1)}\right|=2$.
Squaring both sides,we get $\frac{x^2 (y-1)^2}{x^2 (y 1)^2}=4$.
$x^2 y^2-2y 1=4(x^2 y^2 2y 1)$.
$x^2 y^2-2y 1=4x^2 4y^2 8y 4$.
Rearranging the terms,we get $3x^2 3y^2 10y 3=0$.

(C) Given that $|z|=1$,we can write $z = x + iy$ where $x^2 + y^2 = 1$.
Then,$\frac{1+z^2}{2iz} = \frac{1}{2i} \left(\frac{1}{z} + z\right)$.
Since $|z|=1$,we have $\frac{1}{z} = \bar{z} = x - iy$.
Substituting this,we get $\frac{1}{2i} (x - iy + x + iy) = \frac{1}{2i} (2x) = \frac{x}{i} = -ix$.
Thus,$\frac{1+z^2}{2iz} = -ix$.
Taking the imaginary part,$a = \operatorname{Im}(-ix) = -x$.
Since $x = \operatorname{Re}(z)$,we have $a = -\operatorname{Re}(z)$.

(D) Given $|a \omega_1 + b \omega_2| = |a \omega_1 - b \omega_2|$.
Squaring both sides,we get $|a \omega_1 + b \omega_2|^2 = |a \omega_1 - b \omega_2|^2$.
Using the property $|z|^2 = z \bar{z}$,we have $(a \omega_1 + b \omega_2)(a \bar{\omega}_1 + b \bar{\omega}_2) = (a \omega_1 - b \omega_2)(a \bar{\omega}_1 - b \bar{\omega}_2)$.
Expanding both sides: $a^2 |\omega_1|^2 + ab \omega_1 \bar{\omega}_2 + ab \bar{\omega}_1 \omega_2 + b^2 |\omega_2|^2 = a^2 |\omega_1|^2 - ab \omega_1 \bar{\omega}_2 - ab \bar{\omega}_1 \omega_2 + b^2 |\omega_2|^2$.
Canceling common terms $a^2 |\omega_1|^2$ and $b^2 |\omega_2|^2$,we get $2ab \omega_1 \bar{\omega}_2 + 2ab \bar{\omega}_1 \omega_2 = 0$.
Since $a, b \neq 0$,we have $\omega_1 \bar{\omega}_2 + \bar{\omega}_1 \omega_2 = 0$.
This implies $\omega_1 \bar{\omega}_2 + \overline{\omega_1 \bar{\omega}_2} = 0$.
Let $z = \frac{\omega_1}{\omega_2}$. Then $\omega_1 = z \omega_2$. Substituting this,we get $z \omega_2 \bar{\omega}_2 + \bar{z} \bar{\omega}_2 \omega_2 = 0$.
Since $\omega_2 \neq 0$,$|\omega_2|^2 \neq 0$,so $z + \bar{z} = 0$.
This means $2 \text{Re}(z) = 0$,so $\text{Re}(\frac{\omega_1}{\omega_2}) = 0$.
Thus,$\frac{\omega_1}{\omega_2}$ is a purely imaginary number.

(A) Given: $\cos \alpha+4 \cos \beta+9 \cos \gamma=0$ and $\sin \alpha+4 \sin \beta+9 \sin \gamma=0$.
Let $z_1 = e^{i\alpha}$,$z_2 = e^{i\beta}$,$z_3 = e^{i\gamma}$.
The equations can be written as $z_1 + 4z_2 + 9z_3 = 0$.
Thus,$4z_2 = -(z_1 + 9z_3)$.
Squaring both sides: $16|z_2|^2 = |z_1 + 9z_3|^2 = |z_1|^2 + 81|z_3|^2 + 18 \text{Re}(z_1 \bar{z_3})$.
Since $|z_1|=|z_2|=|z_3|=1$,we have $16 = 1 + 81 + 18 \cos(\alpha - \gamma)$ $\Rightarrow 18 \cos(\alpha - \gamma) = -66$ $\Rightarrow \cos(\alpha - \gamma) = -\frac{11}{3}$.
Similarly,$9z_3 = -(z_1 + 4z_2)$ $\Rightarrow 81 = 1 + 16 + 8 \cos(\alpha - \beta)$ $\Rightarrow 8 \cos(\alpha - \beta) = 64$ $\Rightarrow \cos(\alpha - \beta) = 8$.
Now,$81 \cos(2\gamma - 2\alpha) = 81(2 \cos^2(\gamma - \alpha) - 1) = 81(2(-\frac{11}{3})^2 - 1) = 81(2 \cdot \frac{121}{9} - 1) = 81(\frac{242-9}{9}) = 9(233) = 2097$.
And $16 \cos(2\beta - 2\alpha) = 16(2 \cos^2(\beta - \alpha) - 1) = 16(2(8)^2 - 1) = 16(128 - 1) = 16(127) = 2032$.
Thus,$2097 - 2032 = 65$.
Checking the options: $1 + 8 \cos(\beta - \alpha) = 1 + 8(8) = 1 + 64 = 65$.
Therefore,the correct option is $A$.

(D) Let $z_1 = \cos A + i \sin A$,$z_2 = \cos B + i \sin B$,and $z_3 = \cos C + i \sin C$.
Given $\cos A + \cos B + \cos C = 0$ and $\sin A + \sin B + \sin C = 0$,we have $z_1 + z_2 + z_3 = 0$.
Taking the conjugate,$\bar{z}_1 + \bar{z}_2 + \bar{z}_3 = 0$.
Since $\bar{z} = \frac{1}{z}$ for $|z|=1$,we have $\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} = 0$.
This implies $\frac{z_2 z_3 + z_3 z_1 + z_1 z_2}{z_1 z_2 z_3} = 0$,so $z_1 z_2 + z_2 z_3 + z_3 z_1 = 0$.
Substituting the polar forms:
$\sum (\cos A + i \sin A)(\cos B + i \sin B) = 0$
$\sum (\cos A \cos B - \sin A \sin B) + i \sum (\sin A \cos B + \cos A \sin B) = 0$
$\sum \cos (A+B) + i \sum \sin (A+B) = 0$.
Equating the real parts,we get $\cos (A+B) + \cos (B+C) + \cos (C+A) = 0$.

(B) Let $w = \frac{z-2}{z-4i}$. The given conditions are $\operatorname{Re}(w) = 0$ and $\operatorname{Im}(w) = 1$.
This implies $w = 0 + 1i = i$.
So, $\frac{z-2}{z-4i} = i$.
Multiplying both sides by $(z-4i)$, we get $z-2 = i(z-4i)$.
$z-2 = iz - 4i^2$.
Since $i^2 = -1$, we have $z-2 = iz + 4$.
Rearranging the terms, $z - iz = 4 + 2$.
$z(1-i) = 6$.
$z = \frac{6}{1-i} = \frac{6(1+i)}{(1-i)(1+i)} = \frac{6(1+i)}{1^2 - i^2} = \frac{6(1+i)}{1+1} = \frac{6(1+i)}{2} = 3(1+i)$.
Since there is a unique value for $z$, the number of points is $1$.

(B) Given $\left(\frac{1+iz}{1-iz}\right)^4 = P$ where $|P| = 1$.
Let $w = \frac{1+iz}{1-iz}$. Then $w^4 = P$.
Since $|P| = 1$,we have $|w|^4 = 1$,which implies $|w| = 1$.
So,$\left|\frac{1+iz}{1-iz}\right| = 1$.
This implies $|1+iz| = |1-iz|$.
Let $z = x+iy$. Then $|1+i(x+iy)| = |1-i(x+iy)|$.
$|1-y+ix| = |1+y-ix|$.
Squaring both sides: $(1-y)^2 + x^2 = (1+y)^2 + x^2$.
$1 - 2y + y^2 + x^2 = 1 + 2y + y^2 + x^2$.
$-2y = 2y$ $\Rightarrow 4y = 0$ $\Rightarrow y = 0$.
Thus,$z$ must be a real number. Since $z$ is real,the roots of the equation are real.

(C) Given $|1 + iz| = |1 - iz|$.
Substitute $z = x + iy$:
$|1 + i(x + iy)| = |1 - i(x + iy)|$
$|1 + ix - y| = |1 - ix + y|$
$|(1 - y) + ix| = |(1 + y) - ix|$
Squaring both sides:
$(1 - y)^2 + x^2 = (1 + y)^2 + x^2$
$1 - 2y + y^2 + x^2 = 1 + 2y + y^2 + x^2$
$-2y = 2y$
$4y = 0 \Rightarrow y = 0$.
Since $z = x + iy$ and $y = 0$, we have $z = x$.
Also, $\bar{z} = x - iy = x - i(0) = x$.
Therefore, $z = \bar{z}$.

(A) Let $z = x+iy$. The given expression is $\frac{(x-2)+i(y-1)}{(x+1)+i(y-2)}$.
To make this purely real,the imaginary part of the product of the numerator and the conjugate of the denominator must be zero.
Multiplying by the conjugate of the denominator: $\frac{[(x-2)+i(y-1)][(x+1)-i(y-2)]}{(x+1)^2+(y-2)^2}$.
The imaginary part is $(x+1)(y-1) - (x-2)(y-2) = 0$.
Expanding this: $(xy - x + y - 1) - (xy - 2x - 2y + 4) = 0$.
$xy - x + y - 1 - xy + 2x + 2y - 4 = 0$.
$x + 3y - 5 = 0$.
Since the denominator cannot be zero,$z \neq -(1-2i)$,which means $x \neq -1$ and $y \neq 2$.
Thus,the locus is the line $x+3y-5=0$ excluding the point $(-1,2)$.

(A) Let $z = x + iy$.
Then,the vertices of the rectangle in the Argand plane are $(x, y), (x, -y), (-x, -y),$ and $(-x, y)$.
The length of the sides of the rectangle are $|x - (-x)| = |2x|$ and $|y - (-y)| = |2y|$.
Since $x$ and $y$ are coordinates,the side lengths are $2|x|$ and $2|y|$.
The area of the rectangle is $(2|x|)(2|y|) = 4|xy|$.
Given that the area is $2\sqrt{3}$,we have $4|xy| = 2\sqrt{3}$,which implies $|xy| = \frac{\sqrt{3}}{2}$.
For option $A$,$z = \frac{1}{2} + \sqrt{3}i$,so $x = \frac{1}{2}$ and $y = \sqrt{3}$.
Then $|xy| = |\frac{1}{2} \times \sqrt{3}| = \frac{\sqrt{3}}{2}$.
Thus,$z = \frac{1}{2} + \sqrt{3}i$ is a valid solution.

(A) Given $A = \{z : |z| \leq 2\}$,which implies $\sqrt{x^2 + y^2} \leq 2$,or $x^2 + y^2 \leq 4$.
Given $B = \{z : (1-i)z + (1+i)\bar{z} \geq 4\}$.
Substituting $z = x + iy$ and $\bar{z} = x - iy$:
$(1-i)(x+iy) + (1+i)(x-iy) \geq 4$
$(x + iy - ix - i^2y) + (x - iy + ix - i^2y) \geq 4$
$(x + iy - ix + y) + (x - iy + ix + y) \geq 4$
$2x + 2y \geq 4 \implies x + y \geq 2$.
Thus,$A \cap B = \{z : x^2 + y^2 \leq 4 \text{ and } x + y \geq 2\}$.
Checking the options:
For $A$,$z = \sqrt{3} + \frac{1}{2}i$: $|z|^2 = 3 + \frac{1}{4} = 3.25 \leq 4$ (True). $x+y = \sqrt{3} + 0.5 \approx 1.732 + 0.5 = 2.232 \geq 2$ (True).
Thus,$\sqrt{3} + \frac{1}{2}i$ belongs to $A \cap B$.

(D) Given equation is $z^2(1-z^2)=16$,where $z \in \mathbb{C}$.
This can be rewritten as $z^2 - z^4 = 16$,or $z^4 - z^2 + 16 = 0$.
Let $z^2 = w$. Then $w^2 - w + 16 = 0$.
Using the quadratic formula,$w = \frac{1 \pm \sqrt{1 - 64}}{2} = \frac{1 \pm i\sqrt{63}}{2} = \frac{1 \pm 3i\sqrt{7}}{2}$.
Since $w = z^2$,we have $|w| = |z^2| = |z|^2$.
Calculating the modulus of $w$:
$|w| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{3\sqrt{7}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{63}{4}} = \sqrt{\frac{64}{4}} = \sqrt{16} = 4$.
Therefore,$|z|^2 = 4$,which implies $|z| = 2$.

(A) We know that the definition of the hyperbolic sine function is $\sinh(z) = \frac{e^z - e^{-z}}{2}$.
Substituting $z = ix$,we get:
$\sinh(ix) = \frac{e^{ix} - e^{-ix}}{2}$.
Using Euler's formula,$\sin x = \frac{e^{ix} - e^{-ix}}{2i}$,which implies $\frac{e^{ix} - e^{-ix}}{2} = i \sin x$.
Therefore,$\sinh(ix) = i \sin x$.

(C) Given $|Z-1| \leq 2$,this represents a disk centered at $1$ with radius $2$.
We are given $|\omega Z - 1 - \omega^2| = r$.
Since $\omega^2 + \omega + 1 = 0$,we have $1 + \omega^2 = -\omega$.
Substituting this,the equation becomes $|\omega Z + \omega| = r$.
Since $|\omega| = 1$,we can factor it out: $|\omega| |Z + 1| = r$,which simplifies to $|Z - (-1)| = r$.
This represents a circle centered at $-1$ with radius $r$.
The distance between the centers of the disk $(C_1 = 1)$ and the circle $(C_2 = -1)$ is $d = |1 - (-1)| = 2$.
For the disk $|Z-1| \leq 2$ and the circle $|Z+1| = r$ to have no common solution,the circle must lie entirely outside the disk.
This occurs if the distance between centers $d$ is greater than the sum of the radius of the disk $(R=2)$ and the radius of the circle $(r)$.
Thus,$d > R + r \implies 2 > 2 + r \implies r < 0$.
Alternatively,if the circle is entirely inside the disk,they intersect.
If the circle is outside,$r > d + R$ is not possible here as $r$ must be positive.
Re-evaluating: The circle $|Z+1|=r$ and disk $|Z-1| \leq 2$ have no common points if the circle is completely outside the disk.
The distance between centers is $2$. The disk covers the region from $x = -1$ to $x = 3$ on the real axis.
The circle $|Z+1|=r$ covers $x = -1-r$ to $x = -1+r$.
For no common points,$-1+r < -1$,which implies $r < 0$. Since $r$ is a radius,$r > 0$.
Thus,there are no positive values of $r$ for which they have no common solution.
However,checking the options,$r > 4$ is the standard condition for disjoint circles where $d > R+r$.

(C) Given $z = x + iy$.
The equation is $|z - 1| + |z + i| = 2$.
Substituting $z = x + iy$,we get $|(x - 1) + iy| + |x + i(y + 1)| = 2$.
This implies $\sqrt{(x - 1)^2 + y^2} + \sqrt{x^2 + (y + 1)^2} = 2$.
Rearranging,$\sqrt{(x - 1)^2 + y^2} = 2 - \sqrt{x^2 + (y + 1)^2}$.
Squaring both sides: $(x - 1)^2 + y^2 = 4 + x^2 + (y + 1)^2 - 4\sqrt{x^2 + (y + 1)^2}$.
$x^2 - 2x + 1 + y^2 = 4 + x^2 + y^2 + 2y + 1 - 4\sqrt{x^2 + (y + 1)^2}$.
$-2x - 2y - 4 = -4\sqrt{x^2 + (y + 1)^2}$.
Dividing by $-2$: $x + y + 2 = 2\sqrt{x^2 + (y + 1)^2}$.
Squaring both sides again: $(x + y + 2)^2 = 4(x^2 + y^2 + 2y + 1)$.
$x^2 + y^2 + 4 + 2xy + 4x + 4y = 4x^2 + 4y^2 + 8y + 4$.
Simplifying: $3x^2 - 2xy + 3y^2 - 4x + 4y = 0$.

(D) The condition $\operatorname{Arg}\left(\frac{z - z_1}{z - z_2}\right) = \frac{\pi}{2}$ represents a semicircular arc connecting $z_1$ and $z_2$. \\ Here,$z_1 = 3$ and $z_2 = -2i$. \\ The locus is a semicircle passing through $(3, 0)$ and $(0, -2)$. \\ To check if the origin $(0, 0)$ lies on this arc,we substitute $z = 0$ into the expression: \\ $\operatorname{Arg}\left(\frac{0 - 3}{0 + 2i}\right) = \operatorname{Arg}\left(\frac{-3}{2i}\right) = \operatorname{Arg}\left(\frac{3i}{2}\right) = \frac{\pi}{2}$. \\ Since the condition is satisfied at $z = 0$,the locus is a semicircular arc containing the origin.

(C) Given $z=x+iy$. The equation is $|z-2|+|z-2i|=4$.
Substituting $z=x+iy$,we get $|(x-2)+iy|+|x+(y-2)i|=4$.
This represents $\sqrt{(x-2)^2+y^2} + \sqrt{x^2+(y-2)^2} = 4$.
Squaring both sides: $(x-2)^2+y^2 = 16 + x^2+(y-2)^2 - 8\sqrt{x^2+(y-2)^2}$.
Simplifying: $x^2-4x+4+y^2 = 16+x^2+y^2-4y+4 - 8\sqrt{x^2+(y-2)^2}$.
$-4x+4y-16 = -8\sqrt{x^2+(y-2)^2}$.
Dividing by $-4$: $x-y+4 = 2\sqrt{x^2+(y-2)^2}$.
Squaring again: $(x-y+4)^2 = 4(x^2+y^2-4y+4)$.
$x^2+y^2+16-2xy+8x-8y = 4x^2+4y^2-16y+16$.
Rearranging terms: $3x^2+3y^2+2xy-8x-8y=0$.

(A) Given $\left|\frac{z-i}{z+i}\right|=2$.
Squaring both sides,we get $\left|\frac{z-i}{z+i}\right|^2=4$.
Substituting $z=x+iy$,we have $\left|\frac{x+i(y-1)}{x+i(y+1)}\right|^2=4$.
$\frac{x^2+(y-1)^2}{x^2+(y+1)^2}=4$.
$x^2+y^2-2y+1=4(x^2+y^2+2y+1)$.
$x^2+y^2-2y+1=4x^2+4y^2+8y+4$.
Rearranging the terms: $3x^2+3y^2+10y+3=0$.

(C) Given $z=x+iy$ and $\operatorname{Im}\left(\frac{z-3i}{iz+4}\right)=0$.
Substituting $z=x+iy$, we get $\frac{x+i(y-3)}{i(x+iy)+4} = \frac{x+i(y-3)}{(4-y)+ix}$.
To find the imaginary part, multiply the numerator and denominator by the conjugate of the denominator: $(4-y)-ix$.
The denominator becomes $(4-y)^2+x^2$.
The numerator becomes $[x+i(y-3)][(4-y)-ix] = x(4-y)-ix^2+i(y-3)(4-y)+x(y-3)$.
Simplifying the numerator: $4x-xy-ix^2+i(-y^2+7y-12)+xy-3x = x - i(x^2+y^2-7y+12)$.
For the imaginary part to be zero, the coefficient of $i$ must be zero: $-(x^2+y^2-7y+12) = 0$, which implies $x^2+y^2-7y+12=0$.
Also, the denominator must not be zero, so $(4-y)^2+x^2 \neq 0$, which means $(x,y) \neq (0,4)$.

(A) Given $z = x + iy$,then $\bar{z} = x - iy$.
Substituting into the expression: $\frac{\bar{z}-1}{z-i} = \frac{(x-1) - iy}{x + i(y-1)}$.
To find the real part,multiply the numerator and denominator by the conjugate of the denominator: $x - i(y-1)$.
$\frac{((x-1) - iy)(x - i(y-1))}{x^2 + (y-1)^2} = \frac{x(x-1) - y(y-1) + i(\dots)}{x^2 + (y-1)^2}$.
The real part is $\frac{x^2 - x - y^2 + y}{x^2 + (y-1)^2} = 2$.
$x^2 - x - y^2 + y = 2(x^2 + y^2 - 2y + 1)$.
$x^2 + 3y^2 + x - 5y + 2 = 0$.
This equation represents a pair of straight lines passing through the point $(-1, 1)$.

(D) Given,$\arg(z-2-3i) = \frac{\pi}{4}$.
Let $z = x+iy$.
Then $z-2-3i = (x-2) + i(y-3)$.
Since $\arg(z-2-3i) = \frac{\pi}{4}$,we have $\tan^{-1}\left(\frac{y-3}{x-2}\right) = \frac{\pi}{4}$.
This implies $\frac{y-3}{x-2} = \tan\left(\frac{\pi}{4}\right) = 1$.
Therefore,$y-3 = x-2$,which simplifies to $x-y+1 = 0$.
Thus,the locus of $z$ is $x-y+1 = 0$.

(D) Given $\left|\frac{z+3i}{3z+i}\right| < 1$,where $z = x + iy$.
Squaring both sides,we get $\frac{|z+3i|^2}{|3z+i|^2} < 1$.
This implies $|x + i(y+3)|^2 < |3x + i(3y+1)|^2$.
$x^2 + (y+3)^2 < (3x)^2 + (3y+1)^2$.
$x^2 + y^2 + 6y + 9 < 9x^2 + 9y^2 + 6y + 1$.
$8x^2 + 8y^2 - 8 > 0$,which simplifies to $x^2 + y^2 > 1$.
Since the point $\left(\frac{k-1}{k}, \frac{k-2}{k}\right)$ satisfies this inequality:
$\left(\frac{k-1}{k}\right)^2 + \left(\frac{k-2}{k}\right)^2 > 1$.
$\frac{k^2 - 2k + 1 + k^2 - 4k + 4}{k^2} > 1$.
$2k^2 - 6k + 5 > k^2$.
$k^2 - 6k + 5 > 0$.
$(k-1)(k-5) > 0$.
Thus,$k \in (-\infty, 1) \cup (5, \infty)$.

(C) Given the equation: $\left|\frac{z-2i}{z+2i}\right|=2$.
Substituting $z=x+iy$: $\left|\frac{x+i(y-2)}{x+i(y+2)} \right|=2$.
Squaring both sides: $\frac{x^2+(y-2)^2}{x^2+(y+2)^2}=4$.
$x^2+y^2-4y+4 = 4(x^2+y^2+4y+4)$.
$x^2+y^2-4y+4 = 4x^2+4y^2+16y+16$.
$3x^2+3y^2+20y+12=0$.
Dividing by $3$: $x^2+y^2+\frac{20}{3}y+4=0$.
Completing the square for $y$: $x^2+(y+\frac{10}{3})^2 = \frac{100}{9}-4 = \frac{100-36}{9} = \frac{64}{9}$.
Thus,$x^2+(y+\frac{10}{3})^2 = (\frac{8}{3})^2$.
The radius of the circle is $\frac{8}{3}$.

(C) Let $z = x + iy$.
Given,$\left|\frac{z + 2i}{2z + i}\right| < 1$.
This implies $|z + 2i| < |2z + i|$.
Substituting $z = x + iy$,we get $|x + i(y + 2)| < |2x + i(2y + 1)|$.
Squaring both sides,we get $x^2 + (y + 2)^2 < (2x)^2 + (2y + 1)^2$.
$x^2 + y^2 + 4y + 4 < 4x^2 + 4y^2 + 4y + 1$.
$3 < 3x^2 + 3y^2$.
Dividing by $3$,we get $x^2 + y^2 > 1$.

(A) Given,$\left|\frac{z-2i}{z+2i}\right|=1$
$\Rightarrow |z-2i| = |z+2i|$
Substituting $z=x+iy$:
$|x+i(y-2)| = |x+i(y+2)|$
Squaring both sides:
$x^2+(y-2)^2 = x^2+(y+2)^2$
$x^2+y^2-4y+4 = x^2+y^2+4y+4$
$-4y = 4y$
$8y = 0$
$y=0$
This represents the $x$-axis.