Geometry of complex numbers Questions in English

(C) Given $w = \frac{\sqrt{3} + i}{2} = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$.
Thus, $P = \{e^{in\pi/6} : n = 1, 2, 3, \ldots\}$.
$H_1 = \{z : \operatorname{Re}(z) > 1/2\}$. For $z = e^{in\pi/6} = \cos(n\pi/6) + i \sin(n\pi/6)$, $\operatorname{Re}(z) = \cos(n\pi/6) > 1/2$ implies $n\pi/6 \in (0, \pi/3) \cup (5\pi/3, 2\pi)$. For $n \in Z^+$, this gives $n = 1$ $(z_1 = e^{i\pi/6} = \frac{\sqrt{3}+i}{2})$ and $n = 11$ $(z_1 = e^{i11\pi/6} = \frac{\sqrt{3}-i}{2})$.
$H_2 = \{z : \operatorname{Re}(z) < -1/2\}$. $\operatorname{Re}(z) = \cos(n\pi/6) < -1/2$ implies $n\pi/6 \in (2\pi/3, 4\pi/3)$. For $n \in Z^+$, this gives $n = 5$ $(z_2 = e^{i5\pi/6} = \frac{-\sqrt{3}+i}{2})$ and $n = 7$ $(z_2 = e^{i7\pi/6} = \frac{-\sqrt{3}-i}{2})$.
The possible angles $\angle z_1 O z_2$ are the differences in arguments: $|\arg(z_1) - \arg(z_2)|$.
Possible arguments for $z_1$ are $\pm \pi/6$. Possible arguments for $z_2$ are $\pm 5\pi/6$.
The differences are $|5\pi/6 - \pi/6| = 4\pi/6 = 2\pi/3$, $|-5\pi/6 - \pi/6| = |-\pi| = \pi$, $|5\pi/6 - (-\pi/6)| = \pi$, and $|-5\pi/6 - (-\pi/6)| = |-4\pi/6| = 2\pi/3$.
Thus, the possible values are $2\pi/3$ and $\pi$. Comparing with the options, $2\pi/3$ is present.

(B,C) $1.$ $S_1$ represents the interior of a circle with radius $r=4$ centered at the origin.
$S_2: \operatorname{Im}[\frac{(x-1)+i(y+\sqrt{3})}{1-\sqrt{3} i} \cdot \frac{1+\sqrt{3} i}{1+\sqrt{3} i}] > 0 \implies \operatorname{Im}[\frac{(x-1+i(y+\sqrt{3}))(1+\sqrt{3} i)}{4}] > 0$
$\implies (x-1)\sqrt{3} + (y+\sqrt{3}) > 0 \implies \sqrt{3}x + y > 0$.
$S_3: x > 0$.
The region $S$ is the intersection of the disk $x^2+y^2 < 16$,the half-plane $y > -\sqrt{3}x$,and the half-plane $x > 0$. This forms a circular sector with angle $\theta = 150^\circ = \frac{5\pi}{6}$ radians.
Area $= \frac{1}{2} r^2 \theta = \frac{1}{2} \times 16 \times \frac{5\pi}{6} = \frac{20\pi}{3}$.
$2.$ We need the minimum distance from the point $P(1, -3)$ to the region $S$. The boundary of $S$ includes the line $y = -\sqrt{3}x$ for $x > 0$.
The distance from $(1, -3)$ to the line $\sqrt{3}x + y = 0$ is $d = \frac{|\sqrt{3}(1) + (-3)|}{\sqrt{(\sqrt{3})^2 + 1^2}} = \frac{|\sqrt{3}-3|}{2} = \frac{3-\sqrt{3}}{2}$.

(D) Given $\alpha_k = e^{i \frac{k \pi}{7}}$.
Then $|\alpha_{k+1} - \alpha_k| = |e^{i \frac{(k+1) \pi}{7}} - e^{i \frac{k \pi}{7}}| = |e^{i \frac{k \pi}{7}}| |e^{i \frac{\pi}{7}} - 1| = |e^{i \frac{\pi}{7}} - 1|$.
The numerator is $\sum_{k=1}^{12} |e^{i \frac{\pi}{7}} - 1| = 12 |e^{i \frac{\pi}{7}} - 1|$.
The denominator is $\sum_{k=1}^3 |e^{i \frac{(4k-1) \pi}{7}} - e^{i \frac{(4k-2) \pi}{7}}| = \sum_{k=1}^3 |e^{i \frac{(4k-2) \pi}{7}}| |e^{i \frac{\pi}{7}} - 1| = 3 |e^{i \frac{\pi}{7}} - 1|$.
Thus,the ratio is $\frac{12 |e^{i \frac{\pi}{7}} - 1|}{3 |e^{i \frac{\pi}{7}} - 1|} = \frac{12}{3} = 4$.

(A) Let $z=x+iy$. Substituting this into the equation of the curve:
$(x+iy)(1+i)+(x-iy)(1-i)=4$
$x+ix+iy-y+x-ix-iy-y=4$
$2x-2y=4 \implies x-y=2$.
The region $|z-3| \leq 1$ represents a circle with center $(3,0)$ and radius $r=1$,given by $(x-3)^2+y^2 \leq 1$.
The line $x-y=2$ intersects the circle at points where $(x-3)^2+(x-2)^2=1$,which simplifies to $x^2-6x+9+x^2-4x+4=1$,or $2x^2-10x+12=0$,so $x^2-5x+6=0$. The intersection points are $(2,0)$ and $(3,1)$.
The distance from the center $(3,0)$ to the line $x-y-2=0$ is $d = \frac{|3-0-2|}{\sqrt{1^2+(-1)^2}} = \frac{1}{\sqrt{2}}$.
The area of the circular segment cut by the chord is $A_{segment} = r^2 \cos^{-1}(\frac{d}{r}) - d\sqrt{r^2-d^2} = 1^2 \cos^{-1}(\frac{1}{\sqrt{2}}) - \frac{1}{\sqrt{2}}\sqrt{1-\frac{1}{2}} = \frac{\pi}{4} - \frac{1}{2}$.
Let $\alpha = \frac{\pi}{4} - \frac{1}{2}$ be the smaller area. The total area of the circle is $\pi r^2 = \pi$. The larger area is $\beta = \pi - (\frac{\pi}{4} - \frac{1}{2}) = \frac{3\pi}{4} + \frac{1}{2}$.
Then $|\alpha-\beta| = |(\frac{\pi}{4} - \frac{1}{2}) - (\frac{3\pi}{4} + \frac{1}{2})| = |-\frac{2\pi}{4} - 1| = \frac{\pi}{2} + 1$.

(A) Given $\left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3}$.
Dividing numerator and denominator by $2$ inside the modulus: $\left|\frac{\bar{z}-i}{2(\bar{z}+i/2)}\right|=\frac{1}{3}$ $\Rightarrow \left|\frac{\bar{z}-i}{\bar{z}+i/2}\right|=\frac{2}{3}$.
Let $z = x+iy$,then $\bar{z} = x-iy$. Substituting this:
$3|x-iy-i| = 2|x-iy+i/2|$
$9(x^2 + (-y-1)^2) = 4(x^2 + (-y+1/2)^2)$
$9(x^2 + y^2 + 2y + 1) = 4(x^2 + y^2 - y + 1/4)$
$9x^2 + 9y^2 + 18y + 9 = 4x^2 + 4y^2 - 4y + 1$
$5x^2 + 5y^2 + 22y + 8 = 0$
$x^2 + y^2 + \frac{22}{5}y + \frac{8}{5} = 0$.
The center $C$ is $(0, -11/5)$.
The area of the triangle with vertices $(0,0)$,$(0, -11/5)$,and $(\alpha, 0)$ is given by $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| = 11$.
$\frac{1}{2} |0(-11/5 - 0) + 0(0 - 0) + \alpha(0 - (-11/5))| = 11$.
$\frac{1}{2} |\alpha \cdot \frac{11}{5}| = 11$.
$|\alpha| = 10$.
Therefore,$\alpha^2 = 100$.

(D) Given $|z|=1$,we can write $z = e^{i\theta}$ where $\theta \in [0, 2\pi)$.
Then $\bar{z} = e^{-i\theta}$.
Thus,$\frac{z}{\bar{z}} = \frac{e^{i\theta}}{e^{-i\theta}} = e^{i2\theta}$ and $\frac{\bar{z}}{z} = e^{-i2\theta}$.
The given equation is $\left|e^{i2\theta} + e^{-i2\theta}\right| = 1$.
Using Euler's formula,$e^{i2\theta} + e^{-i2\theta} = 2\cos(2\theta)$.
So,$|2\cos(2\theta)| = 1$,which implies $|\cos(2\theta)| = \frac{1}{2}$.
This means $\cos(2\theta) = \pm \frac{1}{2}$.
For $\theta \in [0, 2\pi)$,$2\theta \in [0, 4\pi)$.
If $\cos(2\theta) = \frac{1}{2}$,then $2\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}$.
If $\cos(2\theta) = -\frac{1}{2}$,then $2\theta = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}$.
There are $8$ distinct values for $\theta$,hence there are $8$ such complex numbers $z$.

(D) Given $z_1 = \sqrt{3} + 2\sqrt{2}i$. The modulus is $|z_1| = \sqrt{(\sqrt{3})^2 + (2\sqrt{2})^2} = \sqrt{3 + 8} = \sqrt{11}$.
Given $\sqrt{3}|z_2| = |z_1|$,so $|z_2| = \frac{|z_1|}{\sqrt{3}} = \sqrt{\frac{11}{3}}$.
Given $\arg(z_2) - \arg(z_1) = \frac{\pi}{6}$,the angle $\angle AOB = \frac{\pi}{6}$.
The area of $\triangle ABO = \frac{1}{2} |z_1| |z_2| \sin(\angle AOB) = \frac{1}{2} \cdot \sqrt{11} \cdot \sqrt{\frac{11}{3}} \cdot \sin(\frac{\pi}{6}) = \frac{1}{2} \cdot \frac{11}{\sqrt{3}} \cdot \frac{1}{2} = \frac{11}{4\sqrt{3}}$.
Using the Law of Cosines in $\triangle ABO$,$AB^2 = |z_1|^2 + |z_2|^2 - 2|z_1||z_2| \cos(\frac{\pi}{6}) = 11 + \frac{11}{3} - 2 \cdot \sqrt{11} \cdot \sqrt{\frac{11}{3}} \cdot \frac{\sqrt{3}}{2} = \frac{44}{3} - 11 = \frac{11}{3}$.
Since $|z_2|^2 = \frac{11}{3}$ and $AB^2 = \frac{11}{3}$,we have $|z_2| = AB$. Thus,$\triangle ABO$ is an isosceles triangle with $\angle OAB = \angle AOB = \frac{\pi}{6}$.
The third angle $\angle ABO = \pi - (\frac{\pi}{6} + \frac{\pi}{6}) = \frac{2\pi}{3}$.
Since $\angle ABO = \frac{2\pi}{3} > \frac{\pi}{2}$,the triangle is obtuse angled isosceles.

(B) The given inequalities represent two disks in the complex plane:
$|z_1 - (8 + 2i)| \leq 1$ is a disk centered at $A(8, 2)$ with radius $r_1 = 1$.
$|z_2 - (2 - 6i)| \leq 2$ is a disk centered at $B(2, -6)$ with radius $r_2 = 2$.
The distance between the centers $A(8, 2)$ and $B(2, -6)$ is given by the distance formula:
$d = \sqrt{(8 - 2)^2 + (2 - (-6))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.
The minimum distance between two points $z_1$ and $z_2$ in these disks is given by $d - r_1 - r_2$.
$|z_1 - z_2|_{\text{min}} = 10 - 1 - 2 = 7$.

(A) For an equilateral triangle with vertices $z_1, z_2, z_3$ and centroid $z_0$,we have the property $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.
Also,the centroid is given by $z_0 = \frac{z_1 + z_2 + z_3}{3}$,which implies $z_1 + z_2 + z_3 = 3z_0$.
We want to evaluate $\sum_{k=1}^3 (z_k - z_0)^2 = (z_1 - z_0)^2 + (z_2 - z_0)^2 + (z_3 - z_0)^2$.
Expanding this,we get $(z_1^2 + z_2^2 + z_3^2) - 2z_0(z_1 + z_2 + z_3) + 3z_0^2$.
Substituting $z_1 + z_2 + z_3 = 3z_0$,the expression becomes $(z_1^2 + z_2^2 + z_3^2) - 2z_0(3z_0) + 3z_0^2 = (z_1^2 + z_2^2 + z_3^2) - 3z_0^2$.
Since $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$,and using the identity $(z_1 + z_2 + z_3)^2 = z_1^2 + z_2^2 + z_3^2 + 2(z_1 z_2 + z_2 z_3 + z_3 z_1)$,we have $(3z_0)^2 = 3(z_1^2 + z_2^2 + z_3^2)$,so $z_1^2 + z_2^2 + z_3^2 = 3z_0^2$.
Thus,the expression is $3z_0^2 - 3z_0^2 = 0$.

(C) Let $z = x + iy$.
For set $A$, $|(x - 2) + i(y - 1)| = 3$, which implies $(x - 2)^2 + (y - 1)^2 = 9$.
For set $B$, $\operatorname{Re}((x + iy) - i(x + iy)) = \operatorname{Re}((x + y) + i(y - x)) = x + y = 2$.
Substituting $y = 2 - x$ into the equation for $A$:
$(x - 2)^2 + (2 - x - 1)^2 = 9$
$(x - 2)^2 + (1 - x)^2 = 9$
$x^2 - 4x + 4 + 1 - 2x + x^2 = 9$
$2x^2 - 6x - 4 = 0 \implies x^2 - 3x - 2 = 0$.
The roots are $x = \frac{3 \pm \sqrt{17}}{2}$.
Since $y = 2 - x$, the points are $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$.
$|z|^2 = x^2 + y^2 = x^2 + (2 - x)^2 = 2x^2 - 4x + 4$.
Using $x^2 = 3x + 2$, we get $|z|^2 = 2(3x + 2) - 4x + 4 = 2x + 8$.
Sum $= (2x_1 + 8) + (2x_2 + 8) = 2(x_1 + x_2) + 16$.
Since $x_1 + x_2 = 3$, Sum $= 2(3) + 16 = 22$.

(C) Given $\operatorname{Re}\left(\frac{z-1}{2z+i}\right) + \operatorname{Re}\left(\frac{\bar{z}-1}{2\bar{z}-i}\right)=2$
Since $\operatorname{Re}(w)=\operatorname{Re}(\bar{w})$, we have
$\operatorname{Re}\left(\frac{\bar{z}-1}{2\bar{z}-i}\right)=\operatorname{Re}\left(\overline{\left(\frac{\bar{z}-1}{2\bar{z}-i}\right)}\right)=\operatorname{Re}\left(\frac{z-1}{2z+i}\right)$
Thus, $2\operatorname{Re}\left(\frac{z-1}{2z+i}\right)=2 \Rightarrow \operatorname{Re}\left(\frac{z-1}{2z+i}\right)=1$
Let $z=x+iy$. Then
$\frac{z-1}{2z+i}=\frac{(x-1)+iy}{2x+i(2y+1)}=\frac{((x-1)+iy)(2x-i(2y+1))}{4x^2+(2y+1)^2}$
Real part: $\frac{2x(x-1)+y(2y+1)}{4x^2+(2y+1)^2}=1$
$2x^2-2x+2y^2+y=4x^2+4y^2+4y+1$
$2x^2+2y^2+2x+3y+1=0$
$x^2+y^2+x+\frac{3}{2}y+\frac{1}{2}=0$
Comparing with $x^2+y^2-2ax-2by+c=0$, center is
$(a,b)=\left(-\frac{1}{2},-\frac{3}{4}\right)$
$r^2=a^2+b^2-c=\frac{1}{4}+\frac{9}{16}-\frac{1}{2}=\frac{5}{16}$
Now, $\frac{15ab}{r^2}=15\cdot\left(-\frac{1}{2}\right)\cdot\left(-\frac{3}{4}\right)\cdot\frac{16}{5}=18$

(B) Given $|x + iy - (1 + 2i)| = 2|x + iy - 3i|$.
Squaring both sides,we get $(x - 1)^2 + (y - 2)^2 = 4(x^2 + (y - 3)^2)$.
Expanding the terms: $x^2 - 2x + 1 + y^2 - 4y + 4 = 4(x^2 + y^2 - 6y + 9)$.
$x^2 + y^2 - 2x - 4y + 5 = 4x^2 + 4y^2 - 24y + 36$.
Rearranging terms: $3x^2 + 3y^2 + 2x - 20y + 31 = 0$.
Dividing by $3$: $x^2 + y^2 + \frac{2}{3}x - \frac{20}{3}y + \frac{31}{3} = 0$.
The centre is $\left(-\frac{1}{3}, \frac{10}{3}\right)$.
The radius is $\sqrt{\left(-\frac{1}{3}\right)^2 + \left(\frac{10}{3}\right)^2 - \frac{31}{3}} = \sqrt{\frac{1}{9} + \frac{100}{9} - \frac{93}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.
Thus,statements $A$ and $D$ are true.

(A) Given $w = \frac{z-1}{z+1}$.
Since $|z|=1$,we can write $z = x+iy$ where $x^2+y^2=1$.
Then $w = \frac{(x-1)+iy}{(x+1)+iy}$.
To find the real part,multiply the numerator and denominator by the conjugate of the denominator: $(x+1)-iy$.
$w = \frac{((x-1)+iy)((x+1)-iy)}{(x+1)^2+y^2} = \frac{(x^2-1) + y^2 + i(y(x+1) - y(x-1))}{(x+1)^2+y^2}$.
$w = \frac{(x^2+y^2-1) + 2iy}{(x+1)^2+y^2}$.
Since $|z|=1$,we have $x^2+y^2=1$,so $x^2+y^2-1=0$.
Therefore,$\operatorname{Re}(w) = \frac{0}{(x+1)^2+y^2} = 0$.

(B) Given Cartesian coordinates are $(x, y) = (-2 \sqrt{3}, 2)$.
First,calculate the modulus $r$:
$r = \sqrt{x^2 + y^2} = \sqrt{(-2 \sqrt{3})^2 + (2)^2} = \sqrt{12 + 4} = \sqrt{16} = 4$.
Since the point lies in the second quadrant $(x < 0, y > 0)$,the argument $\theta$ is given by:
$\theta = \pi - \tan^{-1} \left| \frac{y}{x} \right| = \pi - \tan^{-1} \left| \frac{2}{-2 \sqrt{3}} \right| = \pi - \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \pi - \frac{\pi}{6} = \frac{5 \pi}{6}$.
Thus,the polar coordinates are $(r, \theta) = (4, \frac{5 \pi}{6})$.

(D) The vertices of the triangle are $z_1 = i$,$z_2 = \omega$,and $z_3 = \omega^2$.
We know that $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $\omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$.
The area of a triangle with vertices $z_1, z_2, z_3$ in the Argand plane is given by $\frac{1}{2} |\text{Im}(\bar{z_1}z_2 + \bar{z_2}z_3 + \bar{z_3}z_1)|$.
Substituting the values:
$z_1 = 0 + i$,$z_2 = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$,$z_3 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$.
$\bar{z_1}z_2 = (-i)(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{2} + i\frac{1}{2}$.
$\bar{z_2}z_3 = (-\frac{1}{2} - i\frac{\sqrt{3}}{2})(-\frac{1}{2} - i\frac{\sqrt{3}}{2}) = \frac{1}{4} - \frac{3}{4} + i\frac{\sqrt{3}}{2} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$.
$\bar{z_3}z_1 = (-\frac{1}{2} + i\frac{\sqrt{3}}{2})(i) = -\frac{\sqrt{3}}{2} - i\frac{1}{2}$.
Sum $= (\frac{\sqrt{3}}{2} - \frac{1}{2} - \frac{\sqrt{3}}{2}) + i(\frac{1}{2} + \frac{\sqrt{3}}{2} - \frac{1}{2}) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$.
The imaginary part is $\frac{\sqrt{3}}{2}$.
Area $= \frac{1}{2} |\frac{\sqrt{3}}{2}| = \frac{\sqrt{3}}{4}$ sq.units.

(A) Let $z = x + iy$.
Substituting this into the equation: $|(x+1) + i(y-1)| = |(x-1) + i(y+1)|$.
Squaring both sides: $(x+1)^2 + (y-1)^2 = (x-1)^2 + (y+1)^2$.
Expanding the terms: $x^2 + 2x + 1 + y^2 - 2y + 1 = x^2 - 2x + 1 + y^2 + 2y + 1$.
Simplifying: $2x - 2y = -2x + 2y$.
$4x = 4y$,which implies $y = x$.
The equation $y = x$ represents a straight line passing through the origin and the first and third quadrants.

(D) Given the equation $|z+3|-|z-3|=6$.
Let $z = x + iy$. The points $z_1 = -3$ and $z_2 = 3$ are the foci of the hyperbola.
The distance between the foci is $2c = |3 - (-3)| = 6$.
The definition of a hyperbola is $| |z - z_1| - |z - z_2| | = 2a$.
Here,$2a = 6$,so $a = 3$.
Since $2a = 2c$,the hyperbola degenerates into the line segment connecting the foci.
Specifically,for $|z+3|-|z-3|=6$,the condition $|z+3| = |z-3| + 6$ implies that $z$ must lie on the $x$-axis to the right of $3$ (i.e.,$x \ge 3$).
However,looking at the options provided,the most appropriate description for the locus defined by $|z+3|-|z-3|=6$ is a ray on the $x$-axis starting from $3$ to $\infty$.

(A) Given $\left|\frac{z+i}{z-i}\right| = \sqrt{3}$.
Squaring both sides,we get $\frac{|z+i|^2}{|z-i|^2} = 3$.
Substituting $z = x + iy$,we have $|x + i(y+1)|^2 = 3|x + i(y-1)|^2$.
This simplifies to $x^2 + (y+1)^2 = 3(x^2 + (y-1)^2)$.
$x^2 + y^2 + 2y + 1 = 3(x^2 + y^2 - 2y + 1)$.
$x^2 + y^2 + 2y + 1 = 3x^2 + 3y^2 - 6y + 3$.
$2x^2 + 2y^2 - 8y + 2 = 0$.
Dividing by $2$,we get $x^2 + y^2 - 4y + 1 = 0$.
Completing the square for $y$: $x^2 + (y-2)^2 - 4 + 1 = 0$.
$x^2 + (y-2)^2 = 3$.
This is a circle with centre $(0, 2)$ and radius $\sqrt{3}$.

(D) Given $\left|\frac{z-1}{z+2i}\right| = 1$.
This implies $|z-1| = |z+2i|$.
Substituting $z = x + iy$:
$|x + iy - 1| = |x + iy + 2i|$
$|(x-1) + iy| = |x + i(y+2)|$
Squaring both sides:
$(x-1)^2 + y^2 = x^2 + (y+2)^2$
$x^2 - 2x + 1 + y^2 = x^2 + y^2 + 4y + 4$
$-2x + 1 = 4y + 4$
$2x + 4y + 3 = 0$.
This is the equation of a straight line.

(B) Given $z=x+iy$ and $|z+1|=1$.
Substituting $z=x+iy$ into the equation:
$|(x+1)+iy|=1$.
Squaring both sides,we get:
$(x+1)^2+y^2=1^2$.
This is the standard equation of a circle $(x-h)^2+(y-k)^2=r^2$,where the centre is $(h,k)=(-1,0)$ and the radius is $r=1$.

(B) Let $w = \frac{z-1}{2z+1}$. Since $w$ is purely imaginary,$w + \overline{w} = 0$.
Substituting $w$,we get $\frac{z-1}{2z+1} + \frac{\overline{z}-1}{2\overline{z}+1} = 0$.
$(z-1)(2\overline{z}+1) + (\overline{z}-1)(2z+1) = 0$.
$2z\overline{z} + z - 2\overline{z} - 1 + 2z\overline{z} + \overline{z} - 2z - 1 = 0$.
$4z\overline{z} - z - \overline{z} - 2 = 0$.
Dividing by $4$,we get $z\overline{z} - \frac{1}{4}z - \frac{1}{4}\overline{z} - \frac{1}{2} = 0$.
This is the equation of a circle in the form $z\overline{z} - \overline{\alpha}z - \alpha\overline{z} + k = 0$,where the center is $\alpha = \frac{1}{8}$ and the radius $r = \sqrt{|\alpha|^2 - k}$.
Here,$\alpha = \frac{1}{8}$ and $k = -\frac{1}{2}$.
$r = \sqrt{(\frac{1}{8})^2 - (-\frac{1}{2})} = \sqrt{\frac{1}{64} + \frac{1}{2}} = \sqrt{\frac{1+32}{64}} = \sqrt{\frac{33}{64}} = \frac{\sqrt{33}}{8}$.
Wait,re-evaluating the standard form: $z\overline{z} - \frac{1}{4}z - \frac{1}{4}\overline{z} = \frac{1}{2}$.
$(z - \frac{1}{4})(\overline{z} - \frac{1}{4}) = \frac{1}{2} + \frac{1}{16} = \frac{9}{16}$.
$|z - \frac{1}{4}|^2 = (\frac{3}{4})^2$.
Thus,the radius is $\frac{3}{4}$ units.

(C) The initial position is $Z_0 = 1 + 2i$.
Moving $5$ units horizontally away from the origin means adding $5$ to the real part: $Z_{temp} = (1 + 5) + 2i = 6 + 2i$.
Moving $3$ units vertically upwards means adding $3$ to the imaginary part: $Z_1 = 6 + (2 + 3)i = 6 + 5i$.
From $Z_1$,the particle moves $\sqrt{2}$ units in the direction of $\hat{i} + \hat{j}$. The unit vector in this direction is $\frac{1}{\sqrt{2}}(\hat{i} + \hat{j}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$. Thus,the displacement is $\sqrt{2} \times (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i) = 1 + i$.
The new position is $Z_{1.5} = (6 + 5i) + (1 + i) = 7 + 6i$.
Finally,moving through an angle $\frac{\pi}{2}$ anticlockwise on a circle centered at the origin is equivalent to multiplying by $e^{i\pi/2} = i$.
$Z_2 = (7 + 6i) \times i = 7i + 6i^2 = 7i - 6 = -6 + 7i$.

(D) Given $\left|\frac{z}{1+i}\right|=2$.
Since $|1+i| = \sqrt{1^2+1^2} = \sqrt{2}$,we have $|z| = 2|1+i| = 2\sqrt{2}$.
Substituting $z=x+iy$,we get $|x+iy| = 2\sqrt{2}$.
Squaring both sides,we get $x^2+y^2 = (2\sqrt{2})^2 = 8$.
This represents a circle $x^2+y^2 = 8$ with centre $C \equiv (0,0)$ and radius $r = \sqrt{8} = 2\sqrt{2}$.

(C) Given the condition $|z+1|=1$.
Substitute $z=x+iy$ into the equation:
$|x+iy+1|=1$
$|(x+1)+iy|=1$
Using the definition of the modulus of a complex number $|a+ib|=\sqrt{a^2+b^2}$,we get:
$\sqrt{(x+1)^2+y^2}=1$
Squaring both sides:
$(x+1)^2+y^2=1^2$
$(x+1)^2+y^2=1$
This is the standard equation of a circle $(x-h)^2+(y-k)^2=r^2$,where the centre is $(h,k)$ and the radius is $r$.
Comparing the equations,the centre is $(-1,0)$ and the radius is $1$ unit.

(A) Given,$z\bar{z}^3+\bar{z}z^3=350$
$\Rightarrow z\bar{z}(\bar{z}^2+z^2)=350$
$\Rightarrow |z|^2(x-iy)^2+(x+iy)^2=350$
$\Rightarrow (x^2+y^2)(x^2-y^2-2ixy+x^2-y^2+2ixy)=350$
$\Rightarrow (x^2+y^2)(2x^2-2y^2)=350$
$\Rightarrow 2(x^2+y^2)(x^2-y^2)=350$
$\Rightarrow x^4-y^4=175$
Since $x$ and $y$ are integers,we test values: $x^4-y^4=(x^2-y^2)(x^2+y^2)=175$.
For $x=4, y=3$: $4^4-3^4=256-81=175$.
Thus,the vertices are $(4,3), (-4,3), (-4,-3), (4,-3)$.
The length of the rectangle is $|4-(-4)|=8$ and the breadth is $|3-(-3)|=6$.
Area $= 8 \times 6 = 48 \text{ sq. units}$.

(B) Given $w = \frac{z}{z - \frac{1}{3}i}$.
Multiplying numerator and denominator by $3$,we get $w = \frac{3z}{3z - i}$.
Since $|w| = 1$,we have $|\frac{3z}{3z - i}| = 1$,which implies $3|z| = |3z - i|$.
Let $z = x + iy$. Then $3|x + iy| = |3x + i(3y - 1)|$.
Squaring both sides,we get $9(x^2 + y^2) = (3x)^2 + (3y - 1)^2$.
$9x^2 + 9y^2 = 9x^2 + 9y^2 - 6y + 1$.
$6y - 1 = 0$,which is the equation of a straight line.

(C) Given the inequality $|z-(2-i)| \leq 2$,this represents a disk in the complex plane centered at $2-i$ with radius $r=2$.
Let $z_0 = 2-i$. The distance of the center from the origin is $|z_0| = |2-i| = \sqrt{2^2+(-1)^2} = \sqrt{5}$.
The maximum value of $|z|$ is $|z_0| + r = \sqrt{5} + 2$.
The minimum value of $|z|$ is $|z_0| - r = \sqrt{5} - 2$.
The difference between the greatest and least value is $(\sqrt{5} + 2) - (\sqrt{5} - 2) = 4$.

(B) Given the amplitude of $(z-2-3i)$ is $\frac{3\pi}{4}$ and $z=x+iy$.
Substituting $z$,we get $(x-2) + i(y-3)$.
Since $\text{arg}((x-2) + i(y-3)) = \frac{3\pi}{4}$,we have $\tan^{-1}\left(\frac{y-3}{x-2}\right) = \frac{3\pi}{4}$.
Taking tangent on both sides,$\frac{y-3}{x-2} = \tan\left(\frac{3\pi}{4}\right) = -1$.
This implies $y-3 = -(x-2)$,which simplifies to $y-3 = -x+2$.
Therefore,$x+y=5$.

(D) Let $z = \frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}$.
Multiply the numerator and denominator by the conjugate of the denominator $(\sqrt{3}-i)$:
$z = \frac{(1-i \sqrt{3})(1+i)(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)}$
$z = \frac{(1+i-i\sqrt{3}+\sqrt{3})(\sqrt{3}-i)}{3+1}$
$z = \frac{(1+\sqrt{3}) + i(1-\sqrt{3})}{4} \cdot (\sqrt{3}-i)$
$z = \frac{1}{4} [\sqrt{3}(1+\sqrt{3}) - i(1+\sqrt{3}) + i\sqrt{3}(1-\sqrt{3}) - i^2(1-\sqrt{3})]$
$z = \frac{1}{4} [\sqrt{3} + 3 - i - i\sqrt{3} + i\sqrt{3} - 3i + 1 - \sqrt{3}]$
$z = \frac{1}{4} [4 - 4i] = 1 - i$
The point $(1, -1)$ corresponds to $1-i$ in the Argand plane.
Since the $x$-coordinate is positive and the $y$-coordinate is negative,the point lies in the $IV^{th}$ quadrant.

(B) Let $z = x + iy$.
Then $z^{2} = x^{2} - y^{2} + 2ixy$.
Given $z^{2} + \overline{z} = 0$,we have $(x^{2} - y^{2} + 2ixy) + (x - iy) = 0$.
Grouping real and imaginary parts: $(x^{2} + x - y^{2}) + i(2xy - y) = 0$.
Equating real and imaginary parts to zero:
$x^{2} + x - y^{2} = 0$ $(i)$
$y(2x - 1) = 0$ (ii)
From (ii),either $y = 0$ or $x = 1/2$.
Case $1$: If $y = 0$,then $x^{2} + x = 0 \Rightarrow x(x + 1) = 0$,so $x = 0$ or $x = -1$. This gives solutions $z = 0$ and $z = -1$.
Case $2$: If $x = 1/2$,then $(1/2)^{2} + 1/2 - y^{2} = 0$ $\Rightarrow 1/4 + 1/2 = y^{2}$ $\Rightarrow y^{2} = 3/4$ $\Rightarrow y = \pm \sqrt{3}/2$. This gives solutions $z = 1/2 + i\sqrt{3}/2$ and $z = 1/2 - i\sqrt{3}/2$.
Thus,there are $4$ solutions in total.

(D) Given,$z = x + iy$ and $\left|\frac{z-1}{z+2i}\right| = 1$.
Taking the modulus on both sides,we have $|z-1| = |z+2i|$.
Substituting $z = x + iy$,we get $|(x-1) + iy| = |x + i(y+2)|$.
Squaring both sides,we get $(x-1)^2 + y^2 = x^2 + (y+2)^2$.
Expanding the squares,$x^2 - 2x + 1 + y^2 = x^2 + y^2 + 4y + 4$.
Simplifying the equation,$-2x + 1 = 4y + 4$,which leads to $2x + 4y + 3 = 0$.
This is the equation of a straight line.

(A) Let $z = \sqrt{3}+i$. The modulus is $|z| = \sqrt{(\sqrt{3})^2 + 1^2} = 2$ and the argument is $\arg(z) = \tan^{-1}(\frac{1}{\sqrt{3}}) = 30^{\circ}$.
Since $OPQ$ is an isosceles right-angled triangle with $\angle POQ = 90^{\circ}$ and $OP = OQ$,the point $Q$ can be obtained by rotating $P$ by $90^{\circ}$ in either the clockwise or counter-clockwise direction.
If we rotate by $+90^{\circ}$,the argument of $Q$ is $30^{\circ} + 90^{\circ} = 120^{\circ}$. The complex number is $2(\cos 120^{\circ} + i\sin 120^{\circ}) = 2(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) = -1 + i\sqrt{3}$.
If we rotate by $-90^{\circ}$,the argument of $Q$ is $30^{\circ} - 90^{\circ} = -60^{\circ}$. The complex number is $2(\cos(-60^{\circ}) + i\sin(-60^{\circ})) = 2(\frac{1}{2} - i\frac{\sqrt{3}}{2}) = 1 - i\sqrt{3}$.
Thus,$Q$ represents $-1+i\sqrt{3}$ or $1-i\sqrt{3}$.

(D) Given,$z=x+iy$.
The equation is $|z+1|=|z-1|$.
Squaring both sides,we get $|z+1|^2 = |z-1|^2$.
Substituting $z=x+iy$,we have $|(x+1)+iy|^2 = |(x-1)+iy|^2$.
This implies $(x+1)^2 + y^2 = (x-1)^2 + y^2$.
Simplifying,$(x+1)^2 - (x-1)^2 = 0$.
Using the identity $a^2 - b^2 = (a-b)(a+b)$,we get $(x+1-x+1)(x+1+x-1) = 0$.
$(2)(2x) = 0$,which gives $4x = 0$,so $x = 0$.
The equation $x=0$ represents the $Y$-axis in the complex plane.

(B) Let $z_1 = e^{i\alpha} = \cos \alpha + i \sin \alpha$,$z_2 = e^{i\beta} = \cos \beta + i \sin \beta$,and $z_3 = e^{i\gamma} = \cos \gamma + i \sin \gamma$.
Given that $z_1 + 2z_2 + 3z_3 = (\cos \alpha + 2 \cos \beta + 3 \cos \gamma) + i(\sin \alpha + 2 \sin \beta + 3 \sin \gamma) = 0 + i(0) = 0$.
Let $x = z_1$,$y = 2z_2$,and $z = 3z_3$. Then $x + y + z = 0$.
Using the identity if $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.
Thus,$z_1^3 + (2z_2)^3 + (3z_3)^3 = 3(z_1)(2z_2)(3z_3) = 18z_1 z_2 z_3$.
Substituting the exponential forms: $e^{i3\alpha} + 8e^{i3\beta} + 27e^{i3\gamma} = 18e^{i(\alpha + \beta + \gamma)}$.
Since $\alpha + \beta + \gamma = \pi$,we have $e^{i3\alpha} + 8e^{i3\beta} + 27e^{i3\gamma} = 18e^{i\pi} = 18(\cos \pi + i \sin \pi) = -18 + 0i$.
Equating the imaginary parts,we get $\sin 3 \alpha + 8 \sin 3 \beta + 27 \sin 3 \gamma = 0$.

(C) Let $\alpha = \sqrt{2}+\sqrt{3} i$. Since the coefficients are integers,the conjugate $\bar{\alpha} = \sqrt{2}-\sqrt{3} i$ must also be a root. Furthermore,since $\sqrt{2}$ is irrational,$-\sqrt{2}+\sqrt{3} i$ and $-\sqrt{2}-\sqrt{3} i$ must also be roots to ensure integer coefficients.
The polynomial is given by:
$(x-(\sqrt{2}+\sqrt{3} i))(x-(\sqrt{2}-\sqrt{3} i))(x-(-\sqrt{2}+\sqrt{3} i))(x-(-\sqrt{2}-\sqrt{3} i)) = 0$
Grouping the terms:
$((x-\sqrt{2})-\sqrt{3} i)((x-\sqrt{2})+\sqrt{3} i) \times ((x+\sqrt{2})-\sqrt{3} i)((x+\sqrt{2})+\sqrt{3} i) = 0$
$((x-\sqrt{2})^2 + 3)((x+\sqrt{2})^2 + 3) = 0$
$(x^2 - 2\sqrt{2}x + 2 + 3)(x^2 + 2\sqrt{2}x + 2 + 3) = 0$
$(x^2 + 5 - 2\sqrt{2}x)(x^2 + 5 + 2\sqrt{2}x) = 0$
$(x^2+5)^2 - (2\sqrt{2}x)^2 = 0$
$x^4 + 10x^2 + 25 - 8x^2 = 0$
$x^4 + 2x^2 + 25 = 0$

(A) Let $Z = x + iy$.
Then $Z - 2 = (x - 2) + iy$.
The amplitude (argument) of a complex number $w = a + ib$ is $\theta = \tan^{-1}(\frac{b}{a})$.
Given $\arg(Z - 2) = \frac{\pi}{2}$,this implies that the real part must be $0$ and the imaginary part must be positive.
Therefore,$x - 2 = 0 \Rightarrow x = 2$ and $y > 0$.
Thus,the locus of $Z$ is the vertical line $x = 2$ where $y > 0$.

(D) Let $z=x+iy$, then $z^2=(x+iy)^2 = x^2-y^2+i(2xy)$.
For $C_1$, $\operatorname{Re}(z^2)=x^2-y^2=4$ $(i)$.
For $C_2$, $\operatorname{Im}(z^2)=2xy=4 \Rightarrow xy=2$ $(ii)$.
From $(ii)$, $y=\frac{2}{x}$. Substituting into $(i)$:
$x^2 - (\frac{2}{x})^2 = 4 \Rightarrow x^2 - \frac{4}{x^2} = 4$.
Let $t=x^2$ (where $t > 0$): $t - \frac{4}{t} = 4 \Rightarrow t^2 - 4t - 4 = 0$.
Using the quadratic formula, $t = \frac{4 \pm \sqrt{16 - 4(1)(-4)}}{2} = \frac{4 \pm \sqrt{32}}{2} = 2 \pm 2\sqrt{2}$.
Since $t=x^2 > 0$, we must have $t = 2+2\sqrt{2}$.
Thus, $x^2 = 2+2\sqrt{2}$, which gives two real values for $x$ $(x = \pm \sqrt{2+2\sqrt{2}})$.
For each $x$, $y = \frac{2}{x}$ gives a unique real value for $y$.
Therefore, there are $2$ common points.

(C) As point $\alpha$ lies on the circle $\left(x-x_0\right)^2+\left(y-y_0\right)^2=r^2$,we have $|\alpha-z_0|^2=r^2$,where $z_0=x_0+iy_0$.
Expanding this,we get $|\alpha|^2+|z_0|^2-(\alpha\bar{z}_0+\bar{\alpha}z_0)=r^2 \quad \ldots (i)$
Since $\frac{1}{\bar{\alpha}}$ lies on the circle $\left(x-x_0\right)^2+\left(y-y_0\right)^2=4r^2$,we have $|\frac{1}{\bar{\alpha}}-z_0|^2=4r^2$.
Expanding this,we get $\frac{1}{|\alpha|^2}+|z_0|^2-(\frac{\alpha\bar{z}_0}{|\alpha|^2}+\frac{\bar{\alpha}z_0}{|\alpha|^2})=4r^2$.
Multiplying by $|\alpha|^2$,we get $1+|z_0|^2|\alpha|^2-(\alpha\bar{z}_0+\bar{\alpha}z_0)=4r^2|\alpha|^2 \quad \ldots (ii)$
Subtracting $(i)$ from $(ii)$,we get $(|\alpha|^2-1)|z_0|^2 - (|\alpha|^2-1) = r^2(4|\alpha|^2-1)$.
$(|\alpha|^2-1)(|z_0|^2-1) = r^2(4|\alpha|^2-1)$.
Given $2|z_0|^2=r^2+2$,we have $|z_0|^2-1 = \frac{r^2}{2}$.
Substituting this,$(|\alpha|^2-1)\frac{r^2}{2} = r^2(4|\alpha|^2-1)$.
Dividing by $r^2$ (assuming $r \neq 0$),we get $\frac{|\alpha|^2-1}{2} = 4|\alpha|^2-1$.
$|\alpha|^2-1 = 8|\alpha|^2-2$.
$7|\alpha|^2=1 \Rightarrow |\alpha|=\frac{1}{\sqrt{7}}$.

(C) Given the equation $\bar{z}z^3+z(\bar{z})^3=700$.
Since $z=x+iy$,we have $\bar{z}=x-iy$ and $z\bar{z}=x^2+y^2$.
The equation can be written as $\bar{z}z(z^2+(\bar{z})^2)=700$.
Substituting $z=x+iy$ and $\bar{z}=x-iy$,we get $(x^2+y^2)((x+iy)^2+(x-iy)^2)=700$.
$(x^2+y^2)(x^2-y^2+2ixy+x^2-y^2-2ixy)=700$.
$(x^2+y^2)(2(x^2-y^2))=700$.
$(x^2+y^2)(x^2-y^2)=350$.
Since $x, y$ are integers,we look for factors of $350$.
We have $x^2+y^2=25$ and $x^2-y^2=14$ (not integer solutions) or $x^2+y^2=50$ and $x^2-y^2=7$ (not integer solutions).
Wait,let us re-evaluate: $x^2+y^2=25$ and $x^2-y^2=7$ gives $2x^2=32$ $\Rightarrow x^2=16$ $\Rightarrow x=\pm 4$ and $2y^2=18$ $\Rightarrow y^2=9$ $\Rightarrow y=\pm 3$.
The vertices are $(\pm 4, \pm 3)$.
The rectangle has length $2|x| = 8$ and width $2|y| = 6$.
Area $= 8 \times 6 = 48$.
Thus,option $C$ is correct.

(B) Given $z_1 = 12 + 5i$ and $|z_2| = 9$.
First,calculate the modulus of $z_1$:
$|z_1| = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$.
By the triangle inequality,the range of $|z_1 + z_2|$ is given by $||z_1| - |z_2|| \leq |z_1 + z_2| \leq |z_1| + |z_2|$.
The greatest value $b = |z_1| + |z_2| = 13 + 9 = 22$.
The least value $a = ||z_1| - |z_2|| = |13 - 9| = 4$.
Therefore,$a^2 + b^2 = 4^2 + 22^2 = 16 + 484 = 500$.

(D) We have,$\frac{z-1}{z+i} = \frac{x+iy-1}{x+i(y+1)}$.
Multiplying the numerator and denominator by the conjugate of the denominator,$x-i(y+1)$,we get:
$\frac{(x-1)+iy}{x+i(y+1)} \times \frac{x-i(y+1)}{x-i(y+1)} = \frac{x(x-1) - i(x-1)(y+1) + ixy + y(y+1)}{x^2+(y+1)^2}$.
The real part is $\frac{x(x-1)+y(y+1)}{x^2+(y+1)^2}$.
Given that the real part is $1$,we have:
$\frac{x^2-x+y^2+y}{x^2+(y+1)^2} = 1$.
$x^2-x+y^2+y = x^2+y^2+2y+1$.
$-x+y = 2y+1$.
$x+y+1 = 0$.
Checking the options,for $(2016, -2017)$,we have $2016 + (-2017) + 1 = 2016 - 2017 + 1 = 0$.
Thus,the point $(2016, -2017)$ lies on the locus.

(B) Given $z = x + iy$,then $\bar{z} = x - iy$.
Given the equation $z^2 = (i \bar{z})^2$.
Expanding both sides: $z^2 = i^2 (\bar{z})^2$.
Since $i^2 = -1$,we have $z^2 = -(\bar{z})^2$,which implies $z^2 + (\bar{z})^2 = 0$.
Substitute $z = x + iy$ and $\bar{z} = x - iy$:
$(x + iy)^2 + (x - iy)^2 = 0$.
$(x^2 - y^2 + 2ixy) + (x^2 - y^2 - 2ixy) = 0$.
$2(x^2 - y^2) = 0$.
$x^2 - y^2 = 0$.
$x^2 = y^2$,which means $y = \pm x$.

(C) Given that $z_1$ and $z_2$ are roots of $Z^2+pZ+q=0$,we have $z_1+z_2 = -p$ and $z_1z_2 = q$.
Since $OA=OB$,we have $|z_1| = |z_2|$.
Let $z_1 = re^{i\theta_1}$ and $z_2 = re^{i\theta_2}$.
Given $\angle AOB = \alpha$,we have $|\theta_1 - \theta_2| = \alpha$.
Then $z_1/z_2 = e^{i(\theta_1-\theta_2)} = e^{\pm i\alpha}$.
From $z_1+z_2 = -p$,we have $p^2 = (z_1+z_2)^2 = z_1^2 + z_2^2 + 2z_1z_2$.
Also $p^2 - 4q = (z_1+z_2)^2 - 4z_1z_2 = (z_1-z_2)^2$.
Thus $p^2 = 4q + (z_1-z_2)^2 = 4q + z_2^2(z_1/z_2 - 1)^2$.
Using $z_1/z_2 = e^{i\alpha}$,we get $p^2 = 4q + z_2^2(e^{i\alpha}-1)^2 = 4q + z_2^2 e^{i\alpha}(e^{i\alpha/2} - e^{-i\alpha/2})^2$.
Since $z_1z_2 = q$,$z_2^2 e^{i\alpha} = z_2^2 (z_1/z_2) = z_1z_2 = q$.
So $p^2 = 4q + q(2i \sin(\alpha/2))^2 = 4q - 4q \sin^2(\alpha/2) = 4q(1-\sin^2(\alpha/2)) = 4q \cos^2(\alpha/2)$.

(A) The equation of the line in the Argand plane is given by $a \bar{z} + \bar{a} z + b = 0$.
Let $z = x + iy$ and $a = \alpha + i\beta$.
The equation can be rewritten as $a \bar{z} + \overline{a \bar{z}} + b = 0$,which is $2 \text{Re}(a \bar{z}) + b = 0$.
This represents a line in the complex plane.
The perpendicular distance $d$ from a point $z_0$ to the line $a \bar{z} + \bar{a} z + b = 0$ is given by the formula $d = \frac{|a \bar{z_0} + \bar{a} z_0 + b|}{2|a|}$.
Substituting $z_0 = c$,we get the distance $d = \frac{|a \bar{c} + \bar{a} c + b|}{2|a|}$.

(B) Let $z = x + iy$. Then $|z| = \sqrt{x^2 + y^2} = 4$,so $x^2 + y^2 = 16$.
The vertices of the triangle are $A = z$,$B = iz$,and $C = z + iz$.
Note that $C - A = iz$ and $C - B = z$.
The vectors representing the sides $AC$ and $BC$ are $iz$ and $z$ respectively.
The area of the triangle formed by complex numbers $z_1, z_2, z_3$ is given by $\frac{1}{2} |\text{Im}(\bar{z_1}z_2 + \bar{z_2}z_3 + \bar{z_3}z_1)|$.
Alternatively,since the triangle is formed by $z, iz$,and $z+iz$,it is a right-angled triangle with vertices at the origin $O(0,0)$,$A(z)$,and $B(iz)$ if we consider the triangle $OAB$ where $O, z, z+iz$ form a triangle with base $|z|$ and height $|iz|$.
Actually,the vertices are $z, iz, z+iz$. This is a right-angled triangle because the vector $z$ and $iz$ are perpendicular.
The lengths of the two legs are $|z|$ and $|iz|$.
Since $|iz| = |i||z| = |z| = 4$,the area is $\frac{1}{2} \times |z| \times |iz| = \frac{1}{2} \times 4 \times 4 = 8$ sq. units.

(B) Given,$|z^2-1|=|z|^2+1$. \\ Let $z=x+iy$. \\ Then,$|(x+iy)^2-1| = |x+iy|^2+1$. \\ $|x^2-y^2+2ixy-1| = x^2+y^2+1$. \\ $|(x^2-y^2-1)+i(2xy)| = x^2+y^2+1$. \\ Squaring both sides,we get: \\ $(x^2-y^2-1)^2 + (2xy)^2 = (x^2+y^2+1)^2$. \\ $(x^2-y^2)^2 + 1 - 2(x^2-y^2) + 4x^2y^2 = (x^2+y^2)^2 + 1 + 2(x^2+y^2)$. \\ $x^4+y^4-2x^2y^2 + 1 - 2x^2+2y^2 + 4x^2y^2 = x^4+y^4+2x^2y^2 + 1 + 2x^2+2y^2$. \\ $x^4+y^4+2x^2y^2 + 1 - 2x^2+2y^2 = x^4+y^4+2x^2y^2 + 1 + 2x^2+2y^2$. \\ Simplifying the equation: \\ $-2x^2 = 2x^2$. \\ $4x^2 = 0 \implies x=0$. \\ Since $x=0$,the real part of $z$ is $0$,which means $z$ lies on the imaginary axis.

(A) We are given $|z_1+z_2|^2 = |z_1|^2 + |z_2|^2$.
Using the property $|z|^2 = z \bar{z}$, we can write:
$(z_1+z_2)(\bar{z_1}+\bar{z_2}) = z_1 \bar{z_1} + z_2 \bar{z_2}$.
Expanding the left side:
$z_1 \bar{z_1} + z_1 \bar{z_2} + z_2 \bar{z_1} + z_2 \bar{z_2} = z_1 \bar{z_1} + z_2 \bar{z_2}$.
Subtracting $|z_1|^2$ and $|z_2|^2$ from both sides, we get:
$z_1 \bar{z_2} + z_2 \bar{z_1} = 0$.
This can be written as $z_1 \bar{z_2} + \overline{z_1 \bar{z_2}} = 0$.
Since $z + \bar{z} = 2 \operatorname{Re}(z)$, we have $2 \operatorname{Re}(z_1 \bar{z_2}) = 0$, which implies $\operatorname{Re}(z_1 \bar{z_2}) = 0$.
Dividing by $|z_2|^2$, we get $\operatorname{Re}\left(\frac{z_1 \bar{z_2}}{z_2 \bar{z_2}}\right) = 0$, which simplifies to $\operatorname{Re}\left(\frac{z_1}{z_2}\right) = 0$.

(D) Given that $|z \omega| = 1$,we have $|z| |\omega| = 1$.
Also,$\operatorname{Arg}(z) - \operatorname{Arg}(\omega) = \frac{\pi}{2}$,which implies $\operatorname{Arg}(\frac{z}{\omega}) = \frac{\pi}{2}$.
Let $z = r_1 e^{i \theta_1}$ and $\omega = r_2 e^{i \theta_2}$.
Then $|z| = r_1$ and $|\omega| = r_2$,so $r_1 r_2 = 1$.
$\operatorname{Arg}(z) = \theta_1$ and $\operatorname{Arg}(\omega) = \theta_2$,so $\theta_1 - \theta_2 = \frac{\pi}{2}$.
We need to find $\bar{z} \omega$.
$\bar{z} = r_1 e^{-i \theta_1}$.
$\bar{z} \omega = (r_1 e^{-i \theta_1}) (r_2 e^{i \theta_2}) = (r_1 r_2) e^{i(\theta_2 - \theta_1)}$.
Since $r_1 r_2 = 1$ and $\theta_2 - \theta_1 = -\frac{\pi}{2}$,we have:
$\bar{z} \omega = 1 \cdot e^{-i \frac{\pi}{2}} = \cos(-\frac{\pi}{2}) + i \sin(-\frac{\pi}{2}) = 0 - i = -i$.