Geometry of complex numbers Questions in English

(B) To find the quadrant,we first simplify the complex number $z = \frac{1 + 2i}{1 - i}$.
Multiply the numerator and denominator by the conjugate of the denominator,which is $1 + i$:
$z = \frac{1 + 2i}{1 - i} \times \frac{1 + i}{1 + i}$
$z = \frac{1 + i + 2i + 2i^2}{1^2 - i^2}$
Since $i^2 = -1$,we have:
$z = \frac{1 + 3i - 2}{1 - (-1)} = \frac{-1 + 3i}{2} = -\frac{1}{2} + i\frac{3}{2}$
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 $II$ quadrant.

(C) Let $z = x + iy$,where $x, y \in \mathbb{R}$. Then its conjugate is $\bar{z} = x - iy$.
Given the equation $z^2 = (\bar{z})^2$.
Substituting the values: $(x + iy)^2 = (x - iy)^2$.
Expanding both sides: $x^2 - y^2 + 2ixy = x^2 - y^2 - 2ixy$.
Subtracting $x^2 - y^2$ from both sides: $2ixy = -2ixy$.
Adding $2ixy$ to both sides: $4ixy = 0$.
This implies $xy = 0$.
Therefore,either $x = 0$ (which means $z$ is purely imaginary) or $y = 0$ (which means $z$ is purely real).
Thus,$z$ is either purely real or purely imaginary.

(D) Given inequality: $|z - 4| < |z - 2|$
Squaring both sides: $|z - 4|^2 < |z - 2|^2$
Let $z = x + iy$. Then $|(x - 4) + iy|^2 < |(x - 2) + iy|^2$
$(x - 4)^2 + y^2 < (x - 2)^2 + y^2$
$x^2 - 8x + 16 + y^2 < x^2 - 4x + 4 + y^2$
$-8x + 16 < -4x + 4$
$12 < 4x$
$x > 3$
Since $x = \text{Re}(z)$,the region is $\text{Re}(z) > 3$.
Thus,the correct option is $D$.

(B) Let $w = \frac{z - 1}{z + 1}$. Since $w$ is purely imaginary,$w + \overline{w} = 0$.
Substituting $w = \frac{z - 1}{z + 1}$,we get $\frac{z - 1}{z + 1} + \overline{\left(\frac{z - 1}{z + 1}\right)} = 0$.
$\frac{z - 1}{z + 1} + \frac{\overline{z} - 1}{\overline{z} + 1} = 0$.
$(z - 1)(\overline{z} + 1) + (\overline{z} - 1)(z + 1) = 0$.
$(z\overline{z} + z - \overline{z} - 1) + (z\overline{z} + \overline{z} - z - 1) = 0$.
$2z\overline{z} - 2 = 0$.
$z\overline{z} = 1$.
Since $|z|^2 = z\overline{z}$,we have $|z|^2 = 1$,which implies $|z| = 1$.

(B) Given $\left| z + \frac{2}{z} \right| = 2$.
Using the triangle inequality,we know that $\left| z + \frac{2}{z} \right| \ge ||z| - |\frac{2}{z}||$.
Also,by the reverse triangle inequality,$|z + \frac{2}{z}| \le |z| + |\frac{2}{z}|$.
From the given condition,we have $|z| - \frac{2}{|z|} \le |z + \frac{2}{z}| = 2$.
Let $|z| = r$. Then $r - \frac{2}{r} \le 2$,which implies $r^2 - 2r - 2 \le 0$.
Solving the quadratic equation $r^2 - 2r - 2 = 0$,we get $r = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$.
Since $r = |z| > 0$,we have $r \le 1 + \sqrt{3}$.
Thus,the maximum value of $|z|$ is $1 + \sqrt{3}$.

(C) Given $\left| \frac{z_1 + z_2}{z_1 - z_2} \right| = 1$.
Squaring both sides,we get $\left| \frac{z_1 + z_2}{z_1 - z_2} \right|^2 = 1$.
This implies $(z_1 + z_2)(\overline{z_1} + \overline{z_2}) = (z_1 - z_2)(\overline{z_1} - \overline{z_2})$.
Expanding both sides: $z_1\overline{z_1} + z_1\overline{z_2} + z_2\overline{z_1} + z_2\overline{z_2} = z_1\overline{z_1} - z_1\overline{z_2} - z_2\overline{z_1} + z_2\overline{z_2}$.
Simplifying,we get $2(z_1\overline{z_2} + z_2\overline{z_1}) = 0$,which means $z_1\overline{z_2} + \overline{z_1\overline{z_2}} = 0$.
This indicates that $2 \text{Re}(z_1\overline{z_2}) = 0$,so the real part of $z_1\overline{z_2}$ is $0$.
Let $\frac{z_1}{z_2} = x + iy$. Then $z_1 = z_2(x + iy)$.
Substituting this into the condition $\text{Re}(z_1\overline{z_2}) = 0$,we get $\text{Re}(z_2(x + iy)\overline{z_2}) = 0$,which implies $|z_2|^2 x = 0$.
Since $z_2 \neq 0$,we must have $x = 0$.
Thus,$\frac{z_1}{z_2}$ is purely imaginary. If $z_1 = 0$,then $\frac{z_1}{z_2} = 0$.
Therefore,$\frac{z_1}{z_2}$ is zero or purely imaginary.

(A) Let ${z_1} = a + ib$ and ${z_2} = c - id$,where $a > 0$ and $d > 0$.
Given $|{z_1}| = |{z_2}|$,we have ${a^2} + {b^2} = {c^2} + {d^2}$.
Let $w = \frac{{z_1 + z_2}}{{z_1 - z_2}}$.
Then $\bar{w} = \frac{{\bar{z_1} + \bar{z_2}}}{{\bar{z_1} - \bar{z_2}}}$.
Since $|{z_1}| = |{z_2}| = r$,we have $\bar{z_1} = \frac{r^2}{z_1}$ and $\bar{z_2} = \frac{r^2}{z_2}$.
Substituting these,$\bar{w} = \frac{{\frac{r^2}{z_1} + \frac{r^2}{z_2}}}{{\frac{r^2}{z_1} - \frac{r^2}{z_2}}} = \frac{{z_2 + z_1}}{{z_2 - z_1}} = -\frac{{z_1 + z_2}}{{z_1 - z_2}} = -w$.
Since $\bar{w} = -w$,the complex number $w$ is purely imaginary.
For example,let ${z_1} = 2 + i$ and ${z_2} = 1 - 2i$.
Then $\frac{{z_1 + z_2}}{{z_1 - z_2}} = \frac{{3 - i}}{{1 + 3i}} = \frac{{(3 - i)(1 - 3i)}}{{1^2 + 3^2}} = \frac{{3 - 9i - i - 3}}{{10}} = \frac{{-10i}}{{10}} = -i$,which is purely imaginary.

(A) Let $z = x + iy$,where $x, y \in \mathbb{R}$.
Given $|z + i| = |z - i|$.
Substituting $z = x + iy$:
$|x + iy + i| = |x + iy - i|$
$|x + i(y + 1)| = |x + i(y - 1)|$
Squaring both sides:
$x^2 + (y + 1)^2 = x^2 + (y - 1)^2$
$x^2 + y^2 + 2y + 1 = x^2 + y^2 - 2y + 1$
$2y = -2y$
$4y = 0 \implies y = 0$.
Since $y = 0$,$z = x + i(0) = x$,which represents any real number.

(C) Given $\left| \frac{{z_1} - {z_2}}{{z_1} + {z_2}} \right| = 1$,let $\frac{{z_1} - {z_2}}{{z_1} + {z_2}} = e^{i\alpha} = \cos \alpha + i\sin \alpha$,where $\alpha$ is the angle between ${z_1} - {z_2}$ and ${z_1} + {z_2}$.
Applying componendo and dividendo:
$\frac{({z_1} - {z_2}) + ({z_1} + {z_2})}{({z_1} - {z_2}) - ({z_1} + {z_2})} = \frac{\cos \alpha + i\sin \alpha + 1}{\cos \alpha + i\sin \alpha - 1}$
$\frac{2{z_1}}{-2{z_2}} = \frac{2\cos^2(\alpha/2) + 2i\sin(\alpha/2)\cos(\alpha/2)}{-2\sin^2(\alpha/2) + 2i\sin(\alpha/2)\cos(\alpha/2)}$
$-\frac{{z_1}}{{z_2}} = \frac{2\cos(\alpha/2) [\cos(\alpha/2) + i\sin(\alpha/2)]}{2i\sin(\alpha/2) [\cos(\alpha/2) + i\sin(\alpha/2)]} = \frac{1}{i}\cot(\alpha/2) = -i\cot(\alpha/2)$
So,$\frac{{z_1}}{{z_2}} = i\cot(\alpha/2)$,which implies $i{z_1} = -\cot(\alpha/2) {z_2}$.
Given $i{z_1} = k{z_2}$,we have $k = -\cot(\alpha/2)$,or $\cot(\alpha/2) = -k$.
Using the identity $\tan \alpha = \frac{2\tan(\alpha/2)}{1 - \tan^2(\alpha/2)}$,we have $\tan(\alpha/2) = -1/k$.
$\tan \alpha = \frac{2(-1/k)}{1 - (-1/k)^2} = \frac{-2/k}{1 - 1/k^2} = \frac{-2/k}{(k^2 - 1)/k^2} = \frac{-2k}{k^2 - 1} = \frac{2k}{1 - k^2}$.
Thus,$\alpha = \tan^{-1}\left(\frac{2k}{1 - k^2}\right)$.
Using the property $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$,for $x=k, y=k$,we get $2\tan^{-1}k = \tan^{-1}\left(\frac{2k}{1-k^2}\right)$.
However,checking the sign convention for the argument,the correct result is $-2\tan^{-1}k$.

(B) Given $|z| = 1$,so $|z|^2 = x^2 + y^2 = 1$ .....$(i)$
Now,consider the expression $\frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}$.
Multiply the numerator and denominator by the conjugate of the denominator,$(x + 1) - iy$:
$\frac{z - 1}{z + 1} = \frac{((x - 1) + iy)((x + 1) - iy)}{(x + 1)^2 + y^2}$
$= \frac{(x^2 - 1) + y^2 + i(y(x + 1) - y(x - 1))}{(x + 1)^2 + y^2}$
$= \frac{(x^2 + y^2 - 1) + i(xy + y - xy + y)}{(x + 1)^2 + y^2}$
$= \frac{(1 - 1) + 2iy}{(x + 1)^2 + y^2}$ [using equation $(i)$]
$= \frac{2iy}{(x + 1)^2 + y^2}$
Since the real part is $0$,the expression is purely imaginary.

(C) Let $f(z) = |2z - 1| + |3z - 2| = 2|z - 1/2| + 3|z - 2/3|$.
By the triangle inequality,$|a| + |b| \ge |a - b|$.
We can rewrite the expression as $f(z) = |2z - 1| + |2/3 - z| \times 3$ is not quite right,let us use the property of the sum of distances.
The expression is $2|z - 1/2| + 3|z - 2/3|$.
This represents the sum of distances from $z$ to $1/2$ and $2/3$ weighted by $2$ and $3$ respectively.
Consider the function $g(x) = |2x - 1| + |3x - 2|$ for $x \in \mathbb{R}$.
If $x < 1/2$,$g(x) = -(2x - 1) - (3x - 2) = -5x + 3$,which is decreasing.
If $1/2 \le x \le 2/3$,$g(x) = (2x - 1) - (3x - 2) = -x + 1$,which is decreasing.
If $x > 2/3$,$g(x) = (2x - 1) + (3x - 2) = 5x - 3$,which is increasing.
The minimum occurs at $x = 2/3$.
At $x = 2/3$,$g(2/3) = |2(2/3) - 1| + |3(2/3) - 2| = |4/3 - 1| + 0 = 1/3$.

(C) Let $z = r(\cos \theta + i \sin \theta) = re^{i\theta}$.
Then $-iz = -i(re^{i\theta}) = e^{-i\pi/2} (re^{i\theta}) = re^{i(\theta - \pi/2)}$.
The argument of $z$ is $\theta$ and the argument of $-iz$ is $\theta - \frac{\pi}{2}$.
The angle between the vectors is the difference of their arguments:
$\text{Angle} = (\theta - \frac{\pi}{2}) - \theta = -\frac{\pi}{2}$.
Thus,the angle is $-\frac{\pi}{2}$.

(A) Given $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2$.
Using the property $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 + 2|z_1||z_2| \cos(\theta_1 - \theta_2)$,where $\theta_1 = \arg(z_1)$ and $\theta_2 = \arg(z_2)$.
Substituting this into the given equation:
$|z_1|^2 + |z_2|^2 + 2|z_1||z_2| \cos(\theta_1 - \theta_2) = |z_1|^2 + |z_2|^2$.
This implies $2|z_1||z_2| \cos(\theta_1 - \theta_2) = 0$.
Assuming $z_1, z_2 \neq 0$,we get $\cos(\theta_1 - \theta_2) = 0$.
This means $\theta_1 - \theta_2 = \pm \frac{\pi}{2}$.
Therefore,$\arg\left( \frac{z_1}{z_2} \right) = \arg(z_1) - \arg(z_2) = \pm \frac{\pi}{2}$.
Since the argument of $\frac{z_1}{z_2}$ is $\pm \frac{\pi}{2}$,the complex number $\frac{z_1}{z_2}$ is purely imaginary.
Thus,$\text{Re} \left( \frac{z_1}{z_2} \right) = 0$.

(C) Given $|z_1 + z_2| = |z_1 - z_2|$.
Squaring both sides,we get $|z_1 + z_2|^2 = |z_1 - z_2|^2$.
$(z_1 + z_2)(\overline{z_1} + \overline{z_2}) = (z_1 - z_2)(\overline{z_1} - \overline{z_2})$.
$|z_1|^2 + z_1\overline{z_2} + \overline{z_1}z_2 + |z_2|^2 = |z_1|^2 - z_1\overline{z_2} - \overline{z_1}z_2 + |z_2|^2$.
$2(z_1\overline{z_2} + \overline{z_1}z_2) = 0$,which implies $z_1\overline{z_2} + \overline{z_1}z_2 = 0$.
This means $2Re(z_1\overline{z_2}) = 0$,so $z_1\overline{z_2}$ is purely imaginary.
Let $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$.
Then $z_1\overline{z_2} = r_1 r_2 e^{i(\theta_1 - \theta_2)} = r_1 r_2 (\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2))$.
Since $z_1\overline{z_2}$ is purely imaginary,its real part must be zero: $\cos(\theta_1 - \theta_2) = 0$.
Therefore,$\theta_1 - \theta_2 = \pm \frac{\pi}{2}$.

(A) Given that $\arg\left( \frac{z_1}{z_2} \right) = \pi$.
This implies $\arg(z_1) - \arg(z_2) = \pi$,or $\arg(z_1) = \arg(z_2) + \pi$.
Let $\arg(z_2) = \theta$. Then $\arg(z_1) = \theta + \pi$.
Since $|z_1| = |z_2|$,let $|z_1| = |z_2| = r$.
We can write $z_2 = r(\cos \theta + i \sin \theta)$.
Then $z_1 = r(\cos(\theta + \pi) + i \sin(\theta + \pi)) = r(-\cos \theta - i \sin \theta) = -r(\cos \theta + i \sin \theta) = -z_2$.
Therefore,$z_1 + z_2 = -z_2 + z_2 = 0$.

(C) The condition $|{z_1} + {z_2}| = |{z_1}| + |{z_2}|$ represents the triangle inequality equality case.
This occurs when the complex numbers ${z_1}$ and ${z_2}$ are in the same direction from the origin in the complex plane.
Mathematically,this implies that their arguments are equal,i.e.,$arg({z_1}) = arg({z_2})$ or one of them is a non-negative real multiple of the other.
Thus,the correct option is $C$.

(A) We know that ${e^{i\theta } = \cos \theta + i\sin \theta }$.
Therefore,${e^{e^{i\theta }} = e^{\cos \theta + i\sin \theta } = e^{\cos \theta } \cdot e^{i\sin \theta }}$.
Using Euler's formula,${e^{i\sin \theta } = \cos(\sin \theta ) + i\sin(\sin \theta )}$.
Thus,${e^{e^{i\theta }} = e^{\cos \theta } [\cos(\sin \theta ) + i\sin(\sin \theta )] = e^{\cos \theta } \cos(\sin \theta ) + i e^{\cos \theta } \sin(\sin \theta )}$.
The real part is ${e^{\cos \theta } \cos(\sin \theta )}$.

(C) Let the two complex numbers be $z_1 = -1 - i$ and $z_2 = 2 + 3i$.
The distance between two points $z_1$ and $z_2$ in the complex plane is given by the modulus of their difference,$|z_1 - z_2|$.
$|z_1 - z_2| = |(-1 - i) - (2 + 3i)|$
$= |-1 - i - 2 - 3i|$
$= |-3 - 4i|$
$= \sqrt{(-3)^2 + (-4)^2}$
$= \sqrt{9 + 16}$
$= \sqrt{25}$
$= 5$.
Thus,the length of the line segment is $5$.

(B) In a parallelogram $ABCD$,the diagonals $AC$ and $BD$ bisect each other at the same midpoint.
The midpoint of diagonal $AC$ is given by $\frac{z_1 + z_3}{2}$.
The midpoint of diagonal $BD$ is given by $\frac{z_2 + z_4}{2}$.
Since the diagonals bisect each other,we have $\frac{z_1 + z_3}{2} = \frac{z_2 + z_4}{2}$.
Therefore,$z_1 + z_3 = z_2 + z_4$.

(B) Given the equation $z\overline{z} + a\overline{z} + \overline{a}z + b = 0$.
Adding $|a|^2$ to both sides,we get:
$z\overline{z} + a\overline{z} + \overline{a}z + |a|^2 = |a|^2 - b$
This can be factored as:
$(z + a)(\overline{z} + \overline{a}) = |a|^2 - b$
Since $(z + a)(\overline{z} + \overline{a}) = |z + a|^2$,the equation becomes:
$|z + a|^2 = |a|^2 - b$
This represents a circle with center $-a$ and radius $\sqrt{|a|^2 - b}$ if the radius squared is positive,i.e.,$|a|^2 - b > 0$.
Therefore,$|a|^2 > b$.

(C) Let $r$ be the circumradius of the equilateral triangle and $\omega$ be the cube root of unity. Let $ABC$ be the equilateral triangle with $z_1, z_2$ and $z_3$ as its vertices $A, B$ and $C$ respectively,with circumcentre $O'(z_0)$.
The vectors $O'A, O'B, O'C$ are equal in magnitude and separated by an angle of $\frac{2\pi}{3}$.
Then,we can write:
$z_1 - z_0 = r e^{i\theta}$
$z_2 - z_0 = r e^{i(\theta + \frac{2\pi}{3})} = r \omega e^{i\theta}$
$z_3 - z_0 = r e^{i(\theta + \frac{4\pi}{3})} = r \omega^2 e^{i\theta}$
Thus,$z_1 = z_0 + r e^{i\theta}$,$z_2 = z_0 + r \omega e^{i\theta}$,and $z_3 = z_0 + r \omega^2 e^{i\theta}$.
Squaring and adding these gives:
$z_1^2 + z_2^2 + z_3^2 = (z_0 + r e^{i\theta})^2 + (z_0 + r \omega e^{i\theta})^2 + (z_0 + r \omega^2 e^{i\theta})^2$
$= 3z_0^2 + 2 z_0 r e^{i\theta} (1 + \omega + \omega^2) + r^2 e^{i2\theta} (1 + \omega^2 + \omega^4)$
Since $1 + \omega + \omega^2 = 0$ and $1 + \omega^2 + \omega^4 = 1 + \omega^2 + \omega = 0$,we get:
$z_1^2 + z_2^2 + z_3^2 = 3z_0^2$.

(B) The given equation is $\overline{b}z + b\overline{z} = c$.
Let $z = x + iy$ and $b = b_1 + ib_2$,where $x, y, b_1, b_2 \in \mathbb{R}$.
Substituting these into the equation,we get:
$(b_1 - ib_2)(x + iy) + (b_1 + ib_2)(x - iy) = c$
$(b_1x + ib_1y - ib_2x + b_2y) + (b_1x - ib_1y + ib_2x + b_2y) = c$
$2b_1x + 2b_2y = c$.
This is a linear equation in $x$ and $y$ of the form $Ax + By + C = 0$,which represents a straight line.

(B) Let $z_1, z_2, z_3$ be three complex numbers in $A.P.$
Then $2z_2 = z_1 + z_3$.
This implies $z_2 = \frac{z_1 + z_3}{2}$.
Thus,the complex number $z_2$ is the midpoint of the line segment joining the points $z_1$ and $z_3$.
Therefore,the three points $z_1, z_2,$ and $z_3$ are collinear and lie on a straight line in the complex plane.

(B) Since the triangle with vertices $z_1 = a + i$,$z_2 = 1 + bi$,and $z_3 = 0$ is equilateral,we have the condition $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.
Substituting the values,we get $(a + i)^2 + (1 + bi)^2 + 0 = (a + i)(1 + bi)$.
Expanding both sides: $(a^2 - 1 + 2ai) + (1 - b^2 + 2bi) = a + abi + i - b$.
Simplifying: $(a^2 - b^2) + 2i(a + b) = (a - b) + i(1 + ab)$.
Equating real and imaginary parts:
$a^2 - b^2 = a - b$ ... $(i)$
$2(a + b) = 1 + ab$ ... $(ii)$
From $(i)$,$(a - b)(a + b) = (a - b)$,which implies $(a - b)(a + b - 1) = 0$.
So,either $a = b$ or $a + b = 1$.
Case $1$: If $a = b$,then from $(ii)$,$2(2a) = 1 + a^2$,so $a^2 - 4a + 1 = 0$.
Solving for $a$,$a = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}$.
Since $0 < a < 1$,we must have $a = b = 2 - \sqrt{3}$.
Case $2$: If $a + b = 1$,then $b = 1 - a$. Substituting into $(ii)$,$2(1) = 1 + a(1 - a)$,which gives $a^2 - a + 1 = 0$. This equation has no real roots.
Thus,the only solution is $a = b = 2 - \sqrt{3}$.

(A) Let $\omega = -1 + 5z$. Then $\omega + 1 = 5z$.
Taking the modulus on both sides,we get $|\omega + 1| = |5z| = 5|z|$.
Given that $|z| = 2$,we have $|\omega + 1| = 5 \times 2 = 10$.
The equation $|\omega - (-1)| = 10$ represents a circle with center at $-1$ and radius $10$.
Thus,the points representing the complex numbers lie on a circle.

(C) The vertices are represented as points in the complex plane: $A(1, 2),$ $B(-3, 1),$ $C(-2, -3),$ and $D(2, -2).$
Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(-3 - 1)^2 + (1 - 2)^2} = \sqrt{(-4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}$
$BC = \sqrt{(-2 - (-3))^2 + (-3 - 1)^2} = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17}$
$CD = \sqrt{(2 - (-2))^2 + (-2 - (-3))^2} = \sqrt{4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17}$
$DA = \sqrt{(1 - 2)^2 + (2 - (-2))^2} = \sqrt{(-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}$
Since all sides are equal,it is a rhombus. Now,check the diagonals:
$AC = \sqrt{(-2 - 1)^2 + (-3 - 2)^2} = \sqrt{(-3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$
$BD = \sqrt{(2 - (-3))^2 + (-2 - 1)^2} = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$
Since all sides are equal and the diagonals are equal,the quadrilateral is a square.

(D) Let the initial complex number be $z = 4 - 3i$.
Rotating a complex number by $180^o$ in the clockwise sense is equivalent to multiplying it by $e^{-i\pi} = -1$.
So,the rotated vector is $z' = -z = -(4 - 3i) = -4 + 3i$.
Stretching the vector three times means multiplying its magnitude by $3$,which is equivalent to multiplying the complex number by $3$.
Thus,the new complex number is $z_{new} = 3 \times z' = 3(-4 + 3i) = -12 + 9i$.

(B) Let the given complex number be $z = 3 - 4i$.
Rotating a complex number $z$ anticlockwise by an angle $\theta$ is equivalent to multiplying it by $e^{i\theta}$.
For $\theta = 180^{\circ} = \pi$ radians,the rotation factor is $e^{i\pi} = \cos(\pi) + i\sin(\pi) = -1$.
After rotation,the number becomes $z' = z \times (-1) = -(3 - 4i) = -3 + 4i$.
Stretching the vector by $2.5$ times is equivalent to multiplying the complex number by the scalar $2.5 = \frac{5}{2}$.
Thus,the final complex number is $z'' = 2.5 \times (-3 + 4i) = \frac{5}{2}(-3 + 4i) = \frac{-15}{2} + 10i$.

(D) Given that $POQ$ is a straight line passing through the origin $O$,the points $P$,$O$,and $Q$ are collinear.
Since $OP = OQ$ and they lie on a straight line through the origin,$P$ and $Q$ are reflections of each other with respect to the origin.
Thus,the complex number $z_2 = -z_1$.
$c + id = -(a + ib) = -a - ib$.
Comparing real and imaginary parts,we get $c = -a$ and $d = -b$,which implies $a + c = 0$ and $b + d = 0$.
Also,the magnitude condition $OP = OQ$ implies $|z_1| = |z_2|$,which means $|a + ib| = |c + id|$.
Therefore,both conditions $A$ and $B$ are correct.

(C) We have $z_k = 1 + a + a^2 + \dots + a^{k-1} = \frac{1 - a^k}{1 - a}$.
Subtracting $\frac{1}{1 - a}$ from both sides,we get:
$z_k - \frac{1}{1 - a} = \frac{1 - a^k}{1 - a} - \frac{1}{1 - a} = \frac{-a^k}{1 - a}$.
Taking the modulus on both sides:
$\left| z_k - \frac{1}{1 - a} \right| = \left| \frac{-a^k}{1 - a} \right| = \frac{|a|^k}{|1 - a|}$.
Since $|a| < 1$,it follows that $|a|^k < 1$ for all $k \ge 1$.
Therefore,$\left| z_k - \frac{1}{1 - a} \right| < \frac{1}{|1 - a|}$.
This implies that the vertices $z_k$ lie within the circle defined by $\left| z - \frac{1}{1 - a} \right| = \frac{1}{|1 - a|}$.

(A) Let $O$ be the origin $(z=0)$ and $A$ be the vertex with affix $z_1$. Since it is a regular polygon of $n$ sides,the angle subtended by any side at the centre is $\frac{2\pi}{n}$.
$z_2$ can be obtained by rotating $z_1$ about the origin through an angle of $\pm \frac{2\pi}{n}$.
Using the rotation formula,$z_2 = z_1 e^{\pm i \frac{2\pi}{n}}$.
Applying Euler's formula,$z_2 = z_1 \left( \cos \frac{2\pi}{n} \pm i \sin \frac{2\pi}{n} \right)$.

(B) Let the vertices be $A, B, C, D$ in order. The diagonals $AC$ and $BD$ bisect each other at $E$.
The midpoint $E$ of $BD$ is $\frac{(1 - 2i) + (4 + 2i)}{2} = \frac{5}{2}$.
Since $AC = 2BD$,the length $AE = EC = BD = |(4 + 2i) - (1 - 2i)| = |3 + 4i| = \sqrt{3^2 + 4^2} = 5$.
Since the diagonals are at right angles,the vector $\vec{EA}$ is obtained by rotating $\vec{ED}$ by $90^\circ$ (or $-90^\circ$).
$\vec{ED} = (4 + 2i) - \frac{5}{2} = \frac{3}{2} + 2i$.
Rotating by $90^\circ$ (multiplying by $i$): $\vec{EA} = i(\frac{3}{2} + 2i) = -2 + \frac{3}{2}i$.
Thus,$A = E + \vec{EA} = \frac{5}{2} + (-2 + \frac{3}{2}i) = \frac{1}{2} + \frac{3}{2}i$.
Rotating by $-90^\circ$ (multiplying by $-i$): $\vec{EA} = -i(\frac{3}{2} + 2i) = 2 - \frac{3}{2}i$.
Thus,$A = E + \vec{EA} = \frac{5}{2} + (2 - \frac{3}{2}i) = \frac{9}{2} - \frac{3}{2}i$.
Checking the options,option $(b)$ is $3i - \frac{3}{2} = -1.5 + 3i$. Re-evaluating the geometry,if $A$ is $z$,then $z - E = \pm i(D - E)$.
$z = \frac{5}{2} \pm i(\frac{3}{2} + 2i) = \frac{5}{2} \pm (\frac{3}{2}i - 2) = (\frac{5}{2} \mp 2) \pm \frac{3}{2}i$.
This gives $A = \frac{1}{2} + \frac{3}{2}i$ or $A = \frac{9}{2} - \frac{3}{2}i$.
Given the provided options,there might be a typo in the question or options. Based on standard interpretation,none match exactly,but we provide the derivation.

(A) Given that $|z - z_1| = |z - z_2| = |z - z_3| = |z - z_4| = r$ (say).
This implies that the points represented by ${z_1}, {z_2}, {z_3}, {z_4}$ are at a constant distance $r$ from the point represented by $z$.
By definition,a set of points equidistant from a fixed point (the center $z$) lie on a circle with radius $r$.
Therefore,the points ${z_1}, {z_2}, {z_3}, {z_4}$ are concyclic.

(A) In a rhombus,diagonals bisect each other at right angles. Thus,$M$ is the midpoint of $AC$ and $BD$,and $AC \perp BD$.
Given $BD = 2AC$,we have $2DM = 2(2AM)$,which simplifies to $DM = 2AM$.
The complex number for $D$ is $1 + i$ (point $(1, 1)$) and for $M$ is $2 - i$ (point $(2, -1)$).
The vector $\vec{MD} = (1 - 2, 1 - (-1)) = (-1, 2)$.
Since $AC \perp BD$,the vector $\vec{MA}$ must be perpendicular to $\vec{MD}$.
Rotating $\vec{MD}$ by $90^\circ$ gives vectors $(\pm 2, \pm 1)$.
Since $DM = 2AM$,the vector $\vec{MA} = \pm \frac{1}{2} \vec{MD}_{rotated} = \pm \frac{1}{2} (2, 1) = \pm (1, \frac{1}{2})$.
Thus,$A = M \pm (1, \frac{1}{2}) = (2 \pm 1, -1 \pm \frac{1}{2})$.
This gives $A = (3, -1/2)$ or $A = (1, -3/2)$.
In complex form,$A = 3 - \frac{1}{2}i$ or $1 - \frac{3}{2}i$.

(D) Let $A, B, C$ be the points represented by the complex numbers $z_1, z_2, z_3$ and $P$ be the point represented by $z$.
For the four points $A, B, C, P$ to form a parallelogram,the point $P$ can be placed in three possible positions relative to the triangle $ABC$:
$(i)$ If $A, B, P, C$ form a parallelogram,then $\overrightarrow{AB} = \overrightarrow{CP}$,which implies $z_2 - z_1 = z - z_3$,so $z = z_2 + z_3 - z_1$.
$(ii)$ If $B, C, P, A$ form a parallelogram,then $\overrightarrow{BC} = \overrightarrow{AP}$,which implies $z_3 - z_2 = z - z_1$,so $z = z_3 + z_1 - z_2$.
$(iii)$ If $C, A, P, B$ form a parallelogram,then $\overrightarrow{CA} = \overrightarrow{BP}$,which implies $z_1 - z_3 = z - z_2$,so $z = z_1 + z_2 - z_3$.
Thus,all the given expressions for $z$ are correct.

(B) Let $z = x + iy$,then $\overline{z} = x - iy$.
Substituting these into the given equation:
$(x + iy)(x - iy) + (2 - 3i)(x + iy) + (2 + 3i)(x - iy) + 4 = 0$
$(x^2 + y^2) + (2x + 2iy - 3ix + 3y) + (2x - 2iy + 3ix + 3y) + 4 = 0$
$(x^2 + y^2) + 4x + 6y + 4 = 0$
This is the equation of a circle in the form $x^2 + y^2 + 2gx + 2fy + c = 0$,where $g = 2$,$f = 3$,and $c = 4$.
The radius $r$ is given by $\sqrt{g^2 + f^2 - c}$.
$r = \sqrt{2^2 + 3^2 - 4} = \sqrt{4 + 9 - 4} = \sqrt{9} = 3$.

(D) The vertex of the rectangle in the first quadrant is given by the complex number $z = a + ib\sqrt{3}$,which corresponds to the point $(a, b\sqrt{3})$ in the Cartesian plane.
Since the rectangle is centered at the origin $(0, 0)$ and its sides are parallel to the axes,the vertices of the rectangle are $(a, b\sqrt{3})$,$(-a, b\sqrt{3})$,$(-a, -b\sqrt{3})$,and $(a, -b\sqrt{3})$.
The length of the rectangle along the $X$-axis is $2|a| = 2a$ (assuming $a, b > 0$).
The height of the rectangle along the $Y$-axis is $2|b\sqrt{3}| = 2b\sqrt{3}$.
Therefore,the area of the rectangle is $\text{Length} \times \text{Height} = (2a) \times (2b\sqrt{3}) = 4ab\sqrt{3}$.

(A) The figure shows a parallelogram $OP_1P_3P_2$ where $O$ is the origin $(0,0)$.
By the parallelogram law of vector addition,the position vector of point $P_3$ is the sum of the position vectors of points $P_1$ and $P_2$.
Thus,$\vec{OP_3} = \vec{OP_1} + \vec{OP_2}$.
Since the points $P_1$ and $P_2$ represent the complex numbers $z_1$ and $z_2$ respectively,the point $P_3$ represents the complex number $z_1 + z_2$.

(D) Given,$\frac{|z - 2|}{|z - 3|} = 2$
Squaring both sides,we get $|z - 2|^2 = 4|z - 3|^2$
Let $z = x + iy$,then $(x - 2)^2 + y^2 = 4[(x - 3)^2 + y^2]$
$x^2 - 4x + 4 + y^2 = 4(x^2 - 6x + 9 + y^2)$
$x^2 + y^2 - 4x + 4 = 4x^2 + 4y^2 - 24x + 36$
$3x^2 + 3y^2 - 20x + 32 = 0$
Dividing by $3$,we get $x^2 + y^2 - \frac{20}{3}x + \frac{32}{3} = 0$
Comparing with the standard equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $g = -\frac{10}{3}$,$f = 0$,and $c = \frac{32}{3}$
The radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-\frac{10}{3})^2 + 0^2 - \frac{32}{3}} = \sqrt{\frac{100}{9} - \frac{96}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$

(D) Given that $\triangle ABC$ is an isosceles right-angled triangle with $\angle C = 90^\circ$ and $AC = BC$.
Using the property of rotation in the complex plane,the vector $\vec{CB}$ is obtained by rotating $\vec{CA}$ by $90^\circ$ (or $\pi/2$ radians).
Thus,$(z_2 - z_3) = \pm i(z_1 - z_3)$.
Squaring both sides,we get $(z_2 - z_3)^2 = -(z_1 - z_3)^2$.
$(z_2 - z_3)^2 + (z_1 - z_3)^2 = 0$.
Expanding this: $z_2^2 + z_3^2 - 2z_2z_3 + z_1^2 + z_3^2 - 2z_1z_3 = 0$.
$z_1^2 + z_2^2 + 2z_3^2 - 2z_3(z_1 + z_2) = 0$.
We know $(z_1 - z_2)^2 = z_1^2 + z_2^2 - 2z_1z_2$.
Adding and subtracting $2z_1z_2$ and rearranging terms,we find that $(z_1 - z_2)^2 = 2(z_1 - z_3)(z_3 - z_2)$.

(D) In a regular hexagon,the distance from the center to any vertex is equal to the side length of the hexagon.
Let the center be $O(0,0)$ and one vertex be $z = 1 + 2i$.
The distance from the origin to the vertex $z$ is the modulus $|z|$.
$|z| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$.
Since the hexagon is regular,the distance from the center to each vertex is the same,and the side length $s$ of the regular hexagon is equal to the distance from the center to any vertex.
Therefore,the side length $s = \sqrt{5}$.
The perimeter of a regular hexagon with side length $s$ is $6s$.
Perimeter $= 6 \times \sqrt{5} = 6\sqrt{5}$.

(C) Let the origin $O$ be represented by $0$.
Let the point $P$ be represented by the complex number $z$.
Let the point $Q$ be represented by the complex number $z + iz$.
The vector $\vec{OP}$ corresponds to the complex number $z - 0 = z$.
The vector $\vec{PQ}$ corresponds to the complex number $(z + iz) - z = iz$.
We know that multiplying a complex number by $i$ corresponds to a rotation of $90^\circ$ or $\frac{\pi}{2}$ radians in the counter-clockwise direction.
Since $\vec{PQ} = i \vec{OP}$,the vector $\vec{PQ}$ is perpendicular to $\vec{OP}$.
Therefore,the angle $\angle OPQ$ is $\frac{\pi}{2}$.

(A) The equation of a circle with centre $z_0$ and radius $r$ is given by $|z - z_0| = r$.
Squaring both sides,we get $|z - z_0|^2 = r^2$.
Using the property $|w|^2 = w\bar{w}$,we have $(z - z_0)(\overline{z - z_0}) = r^2$.
$(z - z_0)(\bar{z} - \bar{z_0}) = r^2$.
Expanding the product,we get $z\bar{z} - z\bar{z_0} - \bar{z}z_0 + z_0\bar{z_0} = r^2$.

(D) The vertices ${z_1}, {z_2}, {z_3}$ form an equilateral triangle circumscribing the circle $|z| = \frac{1}{2}$.
Since the circle is the incircle of the equilateral triangle,its center is at the origin $0$.
The distance from the origin to each vertex is $R = \frac{r}{\cos(30^{\circ})} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$.
However,given ${z_1} = \frac{1}{2} + \frac{\sqrt{3}i}{2}$,we note $|z_1| = \sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = 1$.
For an equilateral triangle centered at the origin,the vertices are obtained by rotating ${z_1}$ by $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ radians.
${z_2} = {z_1} e^{i(2\pi/3)} = (\frac{1}{2} + \frac{\sqrt{3}}{2}i) (\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3})$
${z_2} = (\frac{1}{2} + \frac{\sqrt{3}}{2}i) (-\frac{1}{2} + \frac{\sqrt{3}}{2}i)$
${z_2} = -\frac{1}{4} + \frac{\sqrt{3}}{4}i - \frac{\sqrt{3}}{4}i - \frac{3}{4} = -1$.

(B) The given equations represent two circles in the complex plane:
Circle $C_1$: Center $O(0, 0)$,radius $r_1 = 12$.
Circle $C_2$: Center $C(3, 4)$,radius $r_2 = 5$.
The distance between the centers is $d = |(3 + 4i) - 0| = \sqrt{3^2 + 4^2} = 5$.
Since $d + r_2 = 5 + 5 = 10 < r_1 = 12$,the circle $C_2$ lies entirely inside the circle $C_1$.
The minimum distance between a point on $C_1$ and a point on $C_2$ is given by $r_1 - (d + r_2) = 12 - (5 + 5) = 12 - 10 = 2$.

(B) Let the complex numbers be $z_P = 4 + i$,$z_Q = 1 + 6i$,$z_R = -4 + 3i$,and $z_S = -1 - 2i$.
Calculating the side lengths:
$|PQ| = |(1 + 6i) - (4 + i)| = |-3 + 5i| = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$.
$|QR| = |(-4 + 3i) - (1 + 6i)| = |-5 - 3i| = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$.
$|RS| = |(-1 - 2i) - (-4 + 3i)| = |3 - 5i| = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$.
$|SP| = |(4 + i) - (-1 - 2i)| = |5 + 3i| = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34}$.
Since all sides are equal,it is a rhombus.
Now check the diagonals:
$|PR| = |(-4 + 3i) - (4 + i)| = |-8 + 2i| = \sqrt{(-8)^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68}$.
$|QS| = |(-1 - 2i) - (1 + 6i)| = |-2 - 8i| = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68}$.
Since the diagonals are equal and all sides are equal,$PQRS$ is a square.

(A) The value of the determinant $\Delta = \left| \begin{array}{ccc} z_1 & \overline{z_1} & 1 \\ z_2 & \overline{z_2} & 1 \\ z_3 & \overline{z_3} & 1 \end{array} \right|$ represents $2i \times (\text{Area of the triangle formed by } z_1, z_2, z_3)$.
If the points $z_1, z_2, z_3$ are collinear,the area of the triangle is $0$,and the determinant is $0$.
However,in the general case,the expression is purely imaginary. Given the options provided,the question implies the condition where the points are collinear or the determinant evaluates to $0$ in the context of standard multiple-choice problems of this type.

(B) Let $z_1 = 1 + 3i$,$z_2 = 5 + i$,and $z_3 = 3 + 2i$.
These points correspond to the coordinates $(1, 3)$,$(5, 1)$,and $(3, 2)$ in the Cartesian plane.
The area of the triangle formed by these points is given by:
$A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$A = \frac{1}{2} |1(1 - 2) + 5(2 - 3) + 3(3 - 1)|$
$A = \frac{1}{2} |1(-1) + 5(-1) + 3(2)|$
$A = \frac{1}{2} |-1 - 5 + 6| = \frac{1}{2} |0| = 0$.
Since the area is $0$,the points are collinear.