Complex Numbers

Analyze Set $A$ :

The set $A$ is defined as $\{z \in \mathbb{C}:|z-2| \leq 4\}$.

• This represents a circle (and its interior) centered at $(2,0)$ with a radius $r=4$.

• The circle extends from $x=-2$ to $x=6$ along the real axis.

Analyze Set $B$ :

The set $B$ is defined as $\{z \in \mathbb{C}:|z-2|+|z+2|=5\}$.

• By the definition of an ellipse, the sum of distances from two fixed points (foci) is constant.

• Here, the foci are $F_1(2,0)$ and $F_2(-2,0)$, and the length of the major axis $2 a=5$, so $a=2.5$.

• By the definition of an ellipse, the sum of distances from two fixed points (foci) is constant.

• Here, the foci are $F_1(2,0)$ and $F_2(-2,0)$, and the length of the major axis $2 a=5$, so $a=2.5$.

• The distance between foci is $2 a e=4$, so $e=\frac{4}{5}=0.8$.

• The semi-minor axis $b$ can be found using $b^2=a^2\left(1-e^2\right)=(2.5)^2(1-0.64)= 6.25(0.36)=2.25$, so $b=1.5$.

• The ellipse is centered at the origin $(0,0)$ and extends from $x=-2.5$ to $x=2.5$ along the real axis.

To find $\max \left\{\left|z_1-z_2\right|: z_1 \in A\right.$ and $\left.z_2 \in B\right\} \quad$ This means maximize the distance between a point $z_1$ in the circle and a point $z_2$ on the ellipse.

The furthest point in set $A$ (the circle) is at the rightmost edge, $z_1=(2+4,0)=(6,0)$.

The furthest point in set $B$ (the ellipse) from $z_1$ is at its leftmost edge, $z_2=(-2.5,0)$.

Maximum Distance $\left|z_1-z_2\right|=|6-(-2.5)|=|6+2.5|=8.5=\frac{17}{2}$

Let $z=x+i y$

$|z-(\alpha+i \beta)|=r$ represents a circle in the complex plane with centre at $(\alpha, \beta)$ and radius is $r$ unit.

Cartesian equation

$ (x-\alpha)^2+(y-\beta)^2=r^2 $

The equations $|z-6|=5$ and $|z-(-2+6 i)|=5$ describe two circles with the same radius $(r=5)$.

• Circle 1 : Center $C_1=(6,0)$

• Circle 2 : Center $C_2=(-2,6)$

Cartesian equation

$ (x-6)^2+y^2=25 \quad \text { and } \quad(x+2)^2+(y-6)^2=25 $

Expanding both:

• $x^2-12 x+36+y^2=25 \Longrightarrow \mathbf{x}^2+\mathbf{y}^2-12 \mathbf{x}+\mathbf{1 1}=\mathbf{0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

• $x^2+4 x+4+y^2-12 y+36=25 \Longrightarrow \mathbf{x}^2+\mathbf{y}^2+\mathbf{4} \mathbf{x}-\mathbf{1 2} \mathbf{y}+\mathbf{1 5}=\mathbf{0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Subtracting the first equation from the second to find the common chord:

$ \begin{gathered} (4 x-(-12 x))-12 y+(15-11)=0 \\ 16 x-12 y+4=0 \Longrightarrow \mathbf{4}-\mathbf{3} \mathbf{y}+\mathbf{1}=\mathbf{0} \Longrightarrow \mathbf{y}=\frac{\mathbf{4} \mathbf{x}+\mathbf{1}}{\mathbf{3}} \end{gathered} $

Substitute $y$ back into the first circle equation:

$ (x-6)^2+\left(\frac{4 x+1}{3}\right)^2=25 $

multiply both sides by 9

$ \begin{aligned} & 9\left(x^2+36-12 x\right)+16 x^2+8 x+1=225 \\ & 25 x^2-100 x+100=0 \\ & x^2-4 x+4=0 \\ & (x-2)^2=0 \Rightarrow x=2 \end{aligned} $

put $x=2$ in $y=(4 x+1) / 3$

$ y=(4 \times 2+1) / 3=9 / 3=3 $

$ \text { so } z=2+3 i $

calculate $z^3+3 z^2-15 z+141$

Compute $z^3$

$ \begin{gathered} z^3=z \cdot z^2=(2+3 i)(-5+12 i) \\ =-10+24 i-15 i+36 i^2 \\ =-10+9 i-36 \\ =-46+9 i \end{gathered} $

Compute $z^2$

$ (2+3 i)^2=4+12 i+9 i^2=4+12 i-9=-5+12 i $

so

$ \begin{gathered} z^3=-46+9 i \\ 3 z^2=-15+36 i \\ -15 z=-30-45 i \end{gathered} $

Add everything

$ (-46+9 i)+(-15+36 i)+(-30-45 i)+141 $

Real parts:

$ -46-15-30+141=50 $

Imaginary parts:

$ 9 i+36 i-45 i=0 $

So $z^3+3 z^2-15 z+141=50$

Given, $\mathrm{S}=\left\{z \in \mathbb{C}:\left|\frac{z-6 i}{z-2 i}\right|=1\right.$ and $\left.\left|\frac{z-8+2 i}{z+2 i}\right|=\frac{3}{5}\right\}$

$\left|\frac{z-6 i}{z-2 i}\right|=1$

$\Rightarrow $ $ |z-6 i|=|z-2 i|-(1) $

equation (1) represent $z$ is a point which is equidistant from point $(0,2)$ and $(0,6)$ so $z$ lies on the line segment which is perpendicular bisector of the line segment joining point $(0,2)$ & $ (0,6)$.

If $z=x+i y$ then solution of equation (1) is $y=\frac{6+2}{2}=4$ and $x \in R$.

So, solution is $z=x+4 i$ & $ x \in R$.

$\begin{aligned} & \text { Solving }\left|\frac{z-8+2 i}{z+2 i}\right|=\frac{3}{5} \\\\ & \text { Substitute } z=x+4 i \\\\ & \frac{|x+4 i-8+2 i|}{|x+4 i+2 i|}=\frac{3}{5}\end{aligned}$

$\Rightarrow $ $5|x+6 i-8|=3|x+6 i|$

$\Rightarrow $ $5|(x-8)+6 i|=3|x+6 i|$

$\Rightarrow $ $5 \sqrt{(x-8)^2+6^2}=3 \sqrt{x^2+6^2}$

Squaring

$\Rightarrow $ $ 25\left(x^2+64-16 x+36\right)=9 x^2+324$

$\Rightarrow $ $25 x^2-400 x+2500=9 x^2+324$

$\Rightarrow $ $16 x^2-400 x+2176=0$

$\Rightarrow $ $x^2-25 x+136=0$

$\Rightarrow $ $x^2-17 x-8 x+136=0$

$\Rightarrow $ $x(x-17)-8(x-17)=0$

$\Rightarrow $ $(x-17)(x-8)=0$

$\therefore $ $x=17, x=8$

So set S contain two complex number $z_1=17+4 i$ and $z_2=8+4 i$

$ z=\frac{\sqrt{3}}{2}+\frac{i}{2} $

Convert into Euler form $z=r e^{i \theta}$ where $r=|z|$ and $\theta=\arg (z)=\tan ^{-1}\left(\frac{y}{x}\right)$ if $z=x+i y$.

$ r=|z|=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2}=\sqrt{\frac{3}{4}+\frac{1}{4}}=1 $

Since $z$ lies in the first quadrant in the complex plane,

$ \begin{aligned} & \text { So, } \theta=\tan ^{-1}\left(\frac{1 / 2}{\sqrt{3} / 2}\right)=\frac{\pi}{6} \\ & z=e^{i \frac{\pi}{6}} \end{aligned} $

To calculate $\left(z^{201}-i\right)^8=\left(\left(e^{i \frac{\pi}{6}}\right)^{201}-i\right)^8$

$ =\left(e^{i \frac{201 \pi}{6}}-i\right)^8 $

Using De Moivre's theorem, $e^{i \theta}=\cos \theta+i \sin \theta$

$ \begin{aligned} & \left(z^{201}-i\right)^8=\left(\cos \left(\frac{201 \pi}{6}\right)+i \sin \left(\frac{201 \pi}{6}\right)-i\right)^8 \\ & =\left(\cos \left(\frac{67 \pi}{2}\right)+i \sin \left(\frac{67 \pi}{2}\right)-i\right)^8 \\ & =(0+i(-1)-i)^8 \\ & =(-2 i)^8=2^8(i)^8 \end{aligned} $

Since $i=\sqrt{-1} \Longrightarrow i^2=-1 \Longrightarrow i^4=1 \Longrightarrow i^8=1$

$\therefore $ $\left(z^{201}-i\right)^8=2^8=256$.

$ \frac{3}{2} \leq\left|z-\left(\frac{3}{2}+\frac{3}{2} i\right)\right| \leq \frac{7}{2} $

$ \begin{aligned} & \text { Min distance }=\sqrt{16+9}-\frac{7}{2} \\ & =5-\frac{7}{2} \\ & =\frac{3}{2} \end{aligned} $

$ 4 z^2=\bar{z}=0 $

Let $z=x+i y$

$ \begin{aligned} & 4(x+i y)^2+x-i y=0 \\ & 4 x^2-4 y^2+8 x y i+x-i y=0 \\ & 4 x^2-4 y^2+x=0 \& y(8 x-1)=0 \\ & \Rightarrow y=0 \text { or } x=\frac{1}{8} \end{aligned} $

If $y=0,4 x^2+x=0$

$ \begin{array}{ll} x=0, \frac{-1}{4} & \\ \therefore z_1=0+0 . i & \left|z_1\right|^2=0 \\ z_2=-\frac{1}{4}+0 i & \left|z_2\right|^2=\frac{1}{16} \end{array} $

If $x=\frac{1}{8}$

$ \begin{aligned} & 4 \times \frac{1}{64}-4 y^2+\frac{1}{8}=0 \Rightarrow 4 y^2=\frac{3}{16} \Rightarrow y= \pm \frac{\sqrt{3}}{8} \\ & \therefore z_3=\frac{1}{8}+\frac{\sqrt{3}}{8} i \quad\left|z_3\right|^2=\frac{1}{64}+\frac{3}{64}=\frac{1}{16} \\ & z_4=\frac{1}{8}-\frac{\sqrt{3}}{8} i \quad\left|z_4\right|^2=\frac{1}{64}+\frac{3}{64}=\frac{1}{16} \\ & \therefore \sum_{i=1}^n\left|z_i\right|^2=0+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{3}{16} \end{aligned} $

$ \begin{aligned} & \Rightarrow \arg (z)=\sin ^{-1}\left(\frac{3}{5}\right) \tan ^{-1}\left(\frac{3}{4}\right) \Rightarrow z=|z|^{i \theta} \\ & =4 \times(\cos \theta+i \sin \theta)=4 \times\left(\frac{4}{5}+i \frac{3}{5}\right) \\ & =\frac{16}{5}+\frac{12 i}{5} \quad \Rightarrow 5 z=16+12 i \\ & \left(\frac{5 z-12}{5 z i+16}\right)=\frac{(4+12 i)}{(4+16 i)}=\frac{(1+3 i)}{(1+4 i)} \\ & \Rightarrow\left(\frac{5 z-12}{5 z i+16}\right)=\frac{\sqrt{10}}{\sqrt{17}} \\ & 34\left(\frac{5 z-12}{5 z i+16}\right)^2=34 \times \frac{10}{17}=20 \end{aligned} $

$ \begin{array}{r} \left(w+\frac{1}{w}\right)^4+\left(w^2+\frac{1}{w^2}\right)^4+\left(w^3+\frac{1}{w^3}\right)^4+\ldots \\ \left(w^{25}+\frac{1}{w^{25}}\right)^4 \end{array} $

$ \begin{aligned} & \sum\left(w^k+\frac{1}{w^k}\right)^4 \\ & K=3 x \Rightarrow w^{3 x}+\frac{1}{w^{3 x}}=2 \\ & k \neq 3 x \Rightarrow w^k+\frac{1}{w^k}=-1 \\ & \sum_{k=1}^{25}\left(w^k+\frac{1}{w^k}\right)^4 \Rightarrow 8\left(1+1+2^4\right)+1 \\ & =145 \end{aligned} $

(P)

$ \begin{aligned} & x^2+x+1=0 \Rightarrow x=\omega, \omega^2 \text { where } \omega=e^{i \frac{2 \pi}{3}} \\ & \Rightarrow(\alpha+1)=1+\omega=-\omega^2 \Rightarrow(\alpha+1)^{2026}=\left(-\omega^2\right)^{2026}=\omega^2 \\ & \Rightarrow \frac{1}{(\alpha+1)^{2026}}=\frac{1}{\omega^2}=\omega \\ & \Rightarrow \beta+1=1+\omega^2=-\omega \Rightarrow(-\omega)^{2026}=(\beta+1)^{2026}=\omega \\ & \Rightarrow \frac{1}{(\beta+1)^{2026}}=\frac{1}{\omega}=\omega^2 \\ & \Rightarrow \frac{1}{(\alpha+1)^{2026}}, \frac{1}{(\beta+1)^{2026}} \text { are roots of } x^2+x+1=0 \\ & P \rightarrow 1 \end{aligned} $

(Q)

$ \begin{aligned} & \frac{1}{(\alpha+1)^{2027}}=\frac{1}{(\alpha+1)^{2026} \cdot(1+\alpha)}=\frac{1}{\omega^2 \cdot\left(-\omega^2\right)}=-\frac{1}{\omega}=-\omega^2 \\ & \frac{1}{(\beta+1)^{2027}}=\frac{1}{(\beta+1)^{2026} \cdot(\beta+1)}=\frac{1}{\omega(-\omega)}=-\omega \\ & \Rightarrow \frac{1}{(\alpha+1)^{2027}}, \frac{1}{(\beta+1)^{2027}} \text { are roots of } x^2-x+1=0 \\ & Q \rightarrow 2 \end{aligned} $

(R)

$ \begin{aligned} & \gamma=-\omega, \delta=-\omega^2, \text { where } \omega=e^{i \frac{2 \pi}{3}} \\ & \Rightarrow \gamma-1=-\omega-1=\omega^2 \Rightarrow(\gamma-1)^{2026}=\left(\omega^2\right)^{2026}=\omega^2 \\ & \Rightarrow \delta-1=-\omega^2-1=\omega \Rightarrow(\delta-1)^{2026}=\omega^{2026}=\omega \\ & \Rightarrow \frac{1}{(\gamma-1)^{2026}}=\omega, \frac{1}{(\delta-1)^{2026}}=\omega^2 \Rightarrow \omega+\omega^2=-1 \\ & \mathrm{R} \rightarrow 4 \end{aligned} $

(S)

$ \begin{aligned} & x^2+x-1=0 \Rightarrow x(x+1)=1 \Rightarrow(x+1)=\frac{1}{x} \\ & \Rightarrow \frac{1}{(p+1)^3}+\frac{1}{(r+1)^3}=p^3+r^3=(p+r)^3-3 p r(p+r)=(-1)^3-3(-1)(-1)=-4 \\ & \mathrm{~S} \rightarrow 5 \end{aligned} $

First, multiply matrices $S$ and $T$:

$ S T = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $

So, $a = 0$, $b = -1$, $c = 1$, and $d = 1$.

(A)

We calculate:

$ \frac{b + ia}{d + ic} = \frac{-1}{1 + i} \times \frac{1 - i}{1 - i} = \frac{-1 + i}{2} $

The result is a complex number, but this does not satisfy the given condition, so this statement is incorrect.

(B)

Now check the transformation using $\omega$:

$ \frac{a\omega + b}{c\omega + d} = \frac{-1}{\omega + 1} = \frac{-1}{-\omega^2} = \omega $

This equality holds true, so this statement is correct.

(C)

Compute higher powers of $ST$:

$ (ST)^2 = \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ 1 & 0 \end{bmatrix} $

$ (ST)^3 = \begin{bmatrix} -1 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I $

Thus, $(ST)^6 = I$. Therefore, if $(ST)^2 = (ST)^M$, then $M = 6\lambda + 2$, where $\lambda$ is an integer. Since the statement does not correctly represent this, it is incorrect.

(D)

For $z = x + iy$, we have:

$ \frac{az + b}{cz + d} = \frac{-1}{z + 1} $

Substitute $z = x + iy$:

$ \frac{-1}{x + iy + 1} = \frac{-1}{(x + 1) + iy} \times \frac{(x + 1) - iy}{(x + 1) - iy} = \frac{-(x + 1) + iy}{(x + 1)^2 + y^2} $

The imaginary part is $\frac{y}{(x + 1)^2 + y^2}$, which is positive since $y > 0$. This means the image of every $z \in H$ stays in $H$. Hence, this statement is correct.

$ |z-(3+5 i)|+|z-(5+11 i)|=4 \sqrt{5} $

So, $\mathrm{S}_1(3,5) \& \mathrm{~S}_2(5,11)$ are two foci of ellipse, where $2 \mathrm{ae}=\mathrm{S}_1 \mathrm{~S}_2=2 \sqrt{10} \& 2 \mathrm{a}=4 \sqrt{5}$

$ \Rightarrow \mathrm{a}=2 \sqrt{5}, \mathrm{e}=\frac{1}{\sqrt{2}} \& \mathrm{~b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right) \text { gives } \mathrm{b}^2=10 $

$\&$ center of an ellipse is mid point of both focus

∴ center $(4,8)$

Equation of circle is $|\mathrm{Z}-(4+8 \mathrm{i})|=\sqrt{10}$

⇒ Ellipse $\&$ given circle are concentric \& radius of given circle is equal to length of minor axis of given ellipse

⇒ Two points which are common to both.

$\begin{aligned} & z=\frac{-\sqrt{6} i \pm \sqrt{-6+12}}{2} \\ & =\frac{-\sqrt{6} i \pm \sqrt{6}}{2} \\ & =\frac{\sqrt{6}}{2}( \pm 1-i)\end{aligned}$

$\begin{aligned} & \alpha=\frac{-\sqrt{6}}{2}(1+\mathrm{i}), \beta=\frac{\sqrt{6}}{2}(1-\mathrm{i}) \\ & \alpha=\frac{-\sqrt{6}}{2} \times \sqrt{2} \mathrm{e}^{\mathrm{i} \pi / 4}, \beta=+\frac{\sqrt{6}}{2} \times \sqrt{2}-\mathrm{i} \pi / 4 \\ & \alpha=-\sqrt{3} \mathrm{e}^{\mathrm{i} \pi / 4}, \beta=+\sqrt{3} \mathrm{e}^{-\mathrm{i} \pi / 4}\end{aligned}$

$\begin{aligned} & \alpha^8+\beta^8=81 \cdot \mathrm{e}^{\mathrm{i} 2 \pi}+81 \cdot \mathrm{e}^{\mathrm{i}(-2 \pi)} \\ & =81(1+1)=162\end{aligned}$

Let

$ z = x+iy \qquad \text{where } x,y \in \mathbb{R}. $

Then

$ \bar z = x-iy. $

The given equation is

$ z(\bar z+2+i)+k(2+3i)=0. $

We have to find all real values of $k$ for which this equation has at least one complex solution $z$.

First, simplify $z(\bar z+2+i)$:

$ z(\bar z+2+i)=z\bar z+z(2+i). $

Now,

$ z\bar z = |z|^2 = x^2+y^2 $

and

$ z(2+i)=(x+iy)(2+i). $

Let us multiply:

$ (x+iy)(2+i)=2x+ix+2iy+i^2y $

Since $i^2=-1$,

$ (x+iy)(2+i)=(2x-y)+i(x+2y). $

So,

$ z(\bar z+2+i)=x^2+y^2+(2x-y)+i(x+2y). $

Hence the given equation becomes

$ \left(x^2+y^2+2x-y\right)+i(x+2y)+k(2+3i)=0. $

Now separate real and imaginary parts:

$ \left(x^2+y^2+2x-y+2k\right)+i\left(x+2y+3k\right)=0. $

Therefore,

$ x^2+y^2+2x-y+2k=0 \qquad \text{and} \qquad x+2y+3k=0. $

From the second equation,

$ 3k=-(x+2y) $

so

$ k=-\frac{x+2y}{3}. $

Substitute this into the first equation:

$ x^2+y^2+2x-y+2\left(-\frac{x+2y}{3}\right)=0. $

Multiply by $3$:

$ 3x^2+3y^2+6x-3y-2x-4y=0 $

$ 3x^2+3y^2+4x-7y=0. $

Now complete squares:

$ 3\left(x^2+\frac{4}{3}x\right)+3\left(y^2-\frac{7}{3}y\right)=0. $

$ 3\left[\left(x+\frac{2}{3}\right)^2-\frac{4}{9}\right] +3\left[\left(y-\frac{7}{6}\right)^2-\frac{49}{36}\right]=0. $

$ 3\left(x+\frac{2}{3}\right)^2-\frac{4}{3} +3\left(y-\frac{7}{6}\right)^2-\frac{49}{12}=0. $

$ 3\left(x+\frac{2}{3}\right)^2+3\left(y-\frac{7}{6}\right)^2 =\frac{4}{3}+\frac{49}{12}. $

$ 3\left(x+\frac{2}{3}\right)^2+3\left(y-\frac{7}{6}\right)^2 =\frac{16+49}{12}=\frac{65}{12}. $

So,

$ \left(x+\frac{2}{3}\right)^2+\left(y-\frac{7}{6}\right)^2=\frac{65}{36}. $

This is a circle with centre

$ \left(-\frac{2}{3},\frac{7}{6}\right) $

and radius

$ \frac{\sqrt{65}}{6}. $

Now,

$ k=-\frac{x+2y}{3}. $

So we need the range of the linear expression $x+2y$ on this circle.

Let

$ S=x+2y. $

For a circle, maximum and minimum of a linear expression occur at the extreme points in the direction of its normal vector.

At the centre,

$ S_c=-\frac{2}{3}+2\cdot\frac{7}{6} =-\frac{2}{3}+\frac{7}{3} =\frac{5}{3}. $

The vector corresponding to $x+2y$ is $(1,2)$, whose magnitude is

$ \sqrt{1^2+2^2}=\sqrt{5}. $

Hence the maximum variation from the centre is

$ \text{radius} \times \sqrt{5} =\frac{\sqrt{65}}{6}\cdot\sqrt{5} =\frac{\sqrt{325}}{6} =\frac{5\sqrt{13}}{6}. $

Therefore,

$ S_{\min}=\frac{5}{3}-\frac{5\sqrt{13}}{6}, \qquad S_{\max}=\frac{5}{3}+\frac{5\sqrt{13}}{6}. $

Since

$ k=-\frac{S}{3}, $

the interval for $k$ is

$ \left[-\frac{S_{\max}}{3},-\frac{S_{\min}}{3}\right]. $

So if this interval is $[\alpha,\beta]$, then

$ \alpha+\beta = -\frac{S_{\max}+S_{\min}}{3} = -\frac{2\cdot \frac{5}{3}}{3} = -\frac{10}{9}. $

Hence,

$ 9(\alpha+\beta)=9\left(-\frac{10}{9}\right)=-10. $

Therefore, the correct answer is

$ \boxed{-10}. $

So, Option A is correct.

Given the quadratic equation

$ z^2+4z-(1+12i)=0 $

let its roots be $z_1$ and $z_2$.

We need to find

$ |z_1|^2+|z_2|^2 $

Using Vieta’s formulas for

$ az^2+bz+c=0, $

we have

$ z_1+z_2=-\frac{b}{a}=-4 $

and

$ z_1z_2=\frac{c}{a}=-(1+12i)=-1-12i $

Now solve the equation to identify the roots.

$ z^2+4z-(1+12i)=0 $

Using the quadratic formula,

$ z=\frac{-4\pm\sqrt{16+4(1+12i)}}{2} $

$ z=\frac{-4\pm\sqrt{20+48i}}{2} $

Now observe that

$ (6+4i)^2=36+48i-16=20+48i $

So,

$ \sqrt{20+48i}=6+4i $

Hence the roots are

$ z_1=\frac{-4+(6+4i)}{2}=1+2i $

and

$ z_2=\frac{-4-(6+4i)}{2}=-5-2i $

Now compute their moduli squared:

$ |z_1|^2=|1+2i|^2=1^2+2^2=5 $

$ |z_2|^2=|-5-2i|^2=(-5)^2+(-2)^2=25+4=29 $

Therefore,

$ |z_1|^2+|z_2|^2=5+29=34 $

So the correct answer is

$ \boxed{34} $

Hence, Option D is correct.

Given:

• $\alpha,\beta$ are roots of $x^2+ax+b=0$

• $\beta-\alpha=\sqrt{11}$

• $\beta^2-\alpha^2=3i\sqrt{11}$

We have to find $\left(\beta^3-\alpha^3\right)^2$.

Now,

$ \beta^2-\alpha^2=(\beta-\alpha)(\beta+\alpha) $

Substituting the given values,

$ 3i\sqrt{11}=\sqrt{11}(\beta+\alpha) $

Since $\sqrt{11}\neq 0$, we get

$ \beta+\alpha=3i $

Now use the identity

$ \beta^3-\alpha^3=(\beta-\alpha)(\beta^2+\alpha\beta+\alpha^2) $

So first let us find $\beta^2+\alpha\beta+\alpha^2$.

We know

$ (\beta+\alpha)^2=\beta^2+\alpha^2+2\alpha\beta $

Also,

$ \alpha\beta=\frac{(\alpha+\beta)^2-(\beta-\alpha)^2}{4} $

Substitute the values:

$ \alpha+\beta=3i,\qquad \beta-\alpha=\sqrt{11} $

Thus,

$ \alpha\beta=\frac{(3i)^2-(\sqrt{11})^2}{4} =\frac{-9-11}{4} =\frac{-20}{4} =-5 $

Now,

$ \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta $

So,

$ \alpha^2+\beta^2=(3i)^2-2(-5)=-9+10=1 $

Hence,

$ \beta^2+\alpha\beta+\alpha^2=1+(-5)=-4 $

Therefore,

$ \beta^3-\alpha^3=(\beta-\alpha)(\beta^2+\alpha\beta+\alpha^2) =\sqrt{11}\cdot(-4) =-4\sqrt{11} $

Now square it:

$ \left(\beta^3-\alpha^3\right)^2=\left(-4\sqrt{11}\right)^2=16\cdot 11=176 $

So the correct answer is:

$ \boxed{176} $

Hence, Option B is correct.

Solve the quadratic equation first:

$ z^2+4z+16=0 $

Using the quadratic formula,

$ z=\frac{-4\pm\sqrt{16-64}}{2} =\frac{-4\pm\sqrt{-48}}{2} =\frac{-4\pm 4\sqrt{3}i}{2} =-2\pm 2\sqrt{3}i $

So the set is

$ S=\{-2+2\sqrt{3}i,\,-2-2\sqrt{3}i\} $

Now we need to find

$ \sum_{z\in S}|z+\sqrt{3}i|^2 $

That is,

$ |-2+2\sqrt{3}i+\sqrt{3}i|^2 + |-2-2\sqrt{3}i+\sqrt{3}i|^2 $

Simplify each term:

1. For $z=-2+2\sqrt{3}i$,

$ z+\sqrt{3}i=-2+3\sqrt{3}i $

Hence,

$ |-2+3\sqrt{3}i|^2=(-2)^2+(3\sqrt{3})^2=4+27=31 $

2. For $z=-2-2\sqrt{3}i$,

$ z+\sqrt{3}i=-2-\sqrt{3}i $

Hence,

$ |-2-\sqrt{3}i|^2=(-2)^2+(-\sqrt{3})^2=4+3=7 $

Therefore,

$ \sum_{z\in S}|z+\sqrt{3}i|^2=31+7=38 $

So the correct answer is:

$ \boxed{38} $

Option D

Let

$ z = x + iy $

where $x, y \in \mathbb{R}$.

We are given:

$ |z+2| = |z-2| $

and

$ \arg\left(\frac{z+3}{z-i}\right)=\frac{\pi}{4} $

We will use these one by one.

1. Using $|z+2| = |z-2|$

Substitute $z = x+iy$:

$ |x+iy+2| = |x+iy-2| $

So,

$ \sqrt{(x+2)^2+y^2} = \sqrt{(x-2)^2+y^2} $

Squaring both sides:

$ (x+2)^2 + y^2 = (x-2)^2 + y^2 $

$ x^2+4x+4 = x^2-4x+4 $

$ 8x = 0 $

$ x = 0 $

Hence,

$ z = iy $

So the point lies on the imaginary axis.

2. Using the argument condition

Now,

$ \frac{z+3}{z-i} = \frac{3+iy}{i(y-1)} $

We are given that its argument is $\dfrac{\pi}{4}$.

A complex number has argument $\dfrac{\pi}{4}$ when its real and imaginary parts are equal and positive in sign together.

Let us simplify:

$ \frac{3+iy}{i(y-1)} $

Since $\dfrac{1}{i} = -i$, we get

$ \frac{3+iy}{i(y-1)} = \frac{-i(3+iy)}{y-1} $

Now expand:

$ -i(3+iy) = -3i - i^2y = y - 3i $

So,

$ \frac{z+3}{z-i} = \frac{y-3i}{y-1} = \frac{y}{y-1} - i\frac{3}{y-1} $

Thus real part is

$ \frac{y}{y-1} $

and imaginary part is

$ -\frac{3}{y-1} $

Since argument is $\dfrac{\pi}{4}$, we need

$ \frac{\text{imaginary part}}{\text{real part}} = \tan\frac{\pi}{4} = 1 $

So,

$ \frac{-\frac{3}{y-1}}{\frac{y}{y-1}} = 1 $

$ -\frac{3}{y} = 1 $

$ y = -3 $

Therefore,

$ z = -3i $

3. Find $|z|^2$

$ |z|^2 = 0^2 + (-3)^2 = 9 $

So,

$ \boxed{9} $

Hence, the correct option is Option A.

$ \begin{aligned} & |z|=r \\ & O A=r \end{aligned} $

Step 1: Circle $C_1:|z|=r$ has centre $O(0,0)$ and radius $r$ (so $OA=r$).

Step 2: Circle $C_2:|z-3-4i|=5$ has centre at the point $(3,4)$ and radius $5$.

Step 3: Distance between the centres is

$OC=\sqrt{3^2+4^2}=5$

Step 4: Since $C_2$ lies completely inside $C_1$, the minimum distance between a point $z_1$ on $C_1$ and a point $z_2$ on $C_2$ happens when the two points are on the same line through the centres, on the nearest sides.

So,

minimum $\left|\mathrm{z}_1-\mathrm{z}_2\right|= r-(OC+5)=r-(5+5)=r-10$

Given minimum $\left|\mathrm{z}_1-\mathrm{z}_2\right|=2$, we get

$ \Rightarrow r-10=2 \Rightarrow r=12 $

Step 5: The maximum distance $\left|z_1-z_2\right|$ happens when $z_1$ and $z_2$ are on opposite sides along the same line through the centres.

Then the farthest distance is

$ \left|z_1-z_2\right|= r+(OC+5)=r+10 $

Substituting $r=12$,

$ \left|z_1-z_2\right|=12+10=22 $

$ 50\left(\frac{2x}{1+3i}-\frac{y}{1-2i}\right)=31+17i $

We need to find $10(x-3y)$.

First, simplify the fractions by rationalising the denominators.

$ \frac{2x}{1+3i}=\frac{2x(1-3i)}{(1+3i)(1-3i)} $

$ =\frac{2x(1-3i)}{1+9}=\frac{2x(1-3i)}{10} $

$ =\frac{x}{5}(1-3i) $

Similarly,

$ \frac{y}{1-2i}=\frac{y(1+2i)}{(1-2i)(1+2i)} $

$ =\frac{y(1+2i)}{1+4}=\frac{y(1+2i)}{5} $

Now substitute into the given equation:

$ 50\left(\frac{x}{5}(1-3i)-\frac{y}{5}(1+2i)\right)=31+17i $

Take $\frac{1}{5}$ common inside:

$ 50\cdot \frac{1}{5}\left[x(1-3i)-y(1+2i)\right]=31+17i $

$ 10\left[x(1-3i)-y(1+2i)\right]=31+17i $

Expand:

$ 10\left[x-3xi-y-2yi\right]=31+17i $

$ 10\left[(x-y)+(-3x-2y)i\right]=31+17i $

So,

$ 10(x-y)+10(-3x-2y)i=31+17i $

Now compare real and imaginary parts:

$ 10(x-y)=31 $

$ 10(-3x-2y)=17 $

Thus,

$ x-y=\frac{31}{10} $

$ -3x-2y=\frac{17}{10} $

Now solve these equations.

From the first equation,

$ x=y+\frac{31}{10} $

Substitute into the second:

$ -3\left(y+\frac{31}{10}\right)-2y=\frac{17}{10} $

$ -3y-\frac{93}{10}-2y=\frac{17}{10} $

$ -5y=\frac{17}{10}+\frac{93}{10} $

$ -5y=\frac{110}{10}=11 $

$ y=-\frac{11}{5} $

Then

$ x=y+\frac{31}{10}=-\frac{11}{5}+\frac{31}{10} $

$ x=-\frac{22}{10}+\frac{31}{10}=\frac{9}{10} $

Now find:

$ 10(x-3y)=10\left(\frac{9}{10}-3\left(-\frac{11}{5}\right)\right) $

$ =10\left(\frac{9}{10}+\frac{33}{5}\right) $

$ =10\left(\frac{9}{10}+\frac{66}{10}\right) $

$ =10\cdot \frac{75}{10}=75 $

Therefore, the value of $10(x-3y)$ is

$ \boxed{75} $

So, the correct option is $\boxed{\text{D}}$.

$\begin{aligned} & 1+10 \operatorname{Re}\left(\frac{2 \cos \theta+i \sin \theta}{\cos \theta-3 i \sin \theta}\right)=0 \\ & \therefore \mathrm{z}+\overline{\mathrm{z}}=2 \operatorname{Re}(\mathrm{z}) \\ & \frac{2 \cos \theta+\mathrm{i} \sin \theta}{\cos \theta-3 \mathrm{i} \sin \theta}+\frac{2 \cos \theta-\mathrm{i} \sin \theta}{\cos \theta+3 \mathrm{i} \sin \theta}=2 \times\left(\frac{-1}{10}\right) \\ & \frac{\left(2 \cos ^2 \theta-3 \sin ^2 \theta\right)+\left(2 \cos ^2 \theta\right)-\left(3 \sin ^2 \theta\right)}{\cos ^2 \theta+9 \sin ^2 \theta}=\frac{-2}{10} \\ & \Rightarrow \frac{2 \cos ^2 \theta-3 \sin ^2 \theta}{\cos ^2 \theta+9 \sin ^2 \theta}=\frac{-1}{10} \\ & \Rightarrow 20 \cos ^2 \theta-30 \sin ^2 \theta=-\cos ^2 \theta-9 \sin ^2 \theta \\ & 21 \cos ^2 \theta-21 \sin ^2 \theta=0 \\ & \Rightarrow \cos 2 \theta=0 \\ & 2 \theta=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \\ & \Rightarrow \sum \theta^2=\frac{\pi^2}{16}+\frac{9 \pi^2}{16}+\frac{25 \pi^2}{16}+\frac{49 \pi^2}{16}=\frac{84 \pi^2}{16}=\frac{21 \pi^2}{4} \end{aligned}$

$\operatorname{Re}\left(\frac{\mathrm{z}-1}{2 \mathrm{z}+\mathrm{i}}\right)+\operatorname{Re}\left(\frac{\overline{\mathrm{z}}-1}{2 \bar{z}-\mathrm{i}}\right)=2$

Here, $\frac{\mathrm{z}-1}{2 \mathrm{z}+\mathrm{i}}=\left(\frac{\overline{\bar{z}-1}}{2 \overline{\mathrm{z}}-\mathrm{i}}\right)=2$

$=\operatorname{Re}\left(\frac{z-1}{2 z+i}\right)+\operatorname{Re}\left(\overline{\frac{z-1}{2 z+i}}\right)=2$

$=2 \operatorname{Re}\left(\frac{z-1}{2 z+1}\right)=2 \Rightarrow \operatorname{Re}\left(\frac{z-1}{2 z+i}\right)=1$

$\begin{aligned} & \text { Let } z=x+i y \\ & \operatorname{Re}\left(\frac{(x-1)+i y}{2 x+i(2 y+1)}\right)=1 \Rightarrow \operatorname{Re}\left[\frac{((x-1)+i y)(2 x-i(y+1)}{(2 x+i(2 y+1)(2 x-i(2 y+1))}\right]=1 \\ & \Rightarrow \frac{2 x(x-1)+y(2 y+1)}{4 x^2+(2 y+1)^2}=1 \\ & \Rightarrow 2 x^2-2 x+2 y^2+y=4 x^2+4 y^2+1+4 y \\ & \Rightarrow 2 x^2+2 y^2+3 y+2 x+1=0 \\ & \Rightarrow x^2+y^2+x+\frac{3}{2} y+\frac{1}{2}=0 \\ & \text { centre }=\left(\frac{-1}{2}, \frac{-3}{4}\right), r=\sqrt{\frac{1}{4}+\frac{9}{16}-\frac{1}{2}}=\frac{\sqrt{5}}{4} \\ & a=\frac{-1}{2}, b=\frac{-3}{4}, r^2=\frac{5}{16} \end{aligned}$

$15 \frac{\mathrm{ab}}{\mathrm{r}^2}=15 \times\left(\frac{-1}{2}\right) \times\left(\frac{-3}{4}\right) \times \frac{16}{5}=18$

$\begin{aligned} & \frac{z-i}{z+i}=\frac{\bar{z}+i}{\bar{z}-i} \\ & =z \bar{z}-i \bar{z}-i z-1=z \bar{z}+z i+i \bar{z}-1 \\ & =z+\bar{z}=0 \\ & =2 x=0 \\ & =x=0 \quad \text { (y-axis) } \end{aligned}$

$\begin{aligned} & |z|=1 \\ & \therefore \quad z=i \quad(z \neq-i \text { is given }) \end{aligned}$

Statement 1 is incorrect

$\begin{aligned} & \frac{z-i}{z+i}+\frac{\bar{z}-1}{\bar{z}+1}=0 \\ & =z \bar{z}-\bar{z}+z-1+z \bar{z}-z+\bar{z}-1=0 \\ & =z \bar{z}=1 \\ & =|z|=1 \end{aligned}$

Statement 2 is correct

$\begin{aligned} & \omega_1=(8 \sin \theta+7 \cos \theta)+i(\sin \theta+4 \cos \theta) \\ & \omega_2=(\sin \theta+4 \cos \theta)+i(8 \sin \theta+7 \cos \theta) \\ & \alpha=(8 \sin \theta+7 \cos \theta)+(\sin \theta+4 \cos \theta) \\ & \quad-(\sin \theta+4 \cos \theta)+(8 \sin \theta+7 \cos \theta)=0 \\ & \beta=(8 \sin \theta+7 \cos \theta)^2+(\sin \theta+4 \cos \theta)^2 \end{aligned}$

$\begin{aligned} & =65 \sin ^2 \theta+65 \cos ^2 \theta+56 \sin 2 \theta+4 \sin 2 \theta \\ & =65+60 \sin 2 \theta \\ & (\alpha+\beta)_{\max }=125=p \\ & (\alpha+\beta)_{\min }=5=q \\ & p+q=130 \end{aligned}$

$ \begin{aligned} & z_0=z_1+z_2+z_3 \\ & \sum_{k=1}^3\left(z_k-z_0\right)^2=\left(z_1-z_0\right)^2+\left(z_2-z_0\right)^2+\left(z_3-z_0\right)^2 \end{aligned} $

Let $z_0$ is origin $\Rightarrow z_1, z_2, z_3$ lies on a circle having $\left|z_0-z_i\right|=R$

$ \begin{aligned} & \therefore z_1=R e^{i 2 \pi / 3} z_2=R e^{i 4 \pi / 3} z_3=R e^{i 6 \pi / 3} \\ & \Rightarrow z_1^2+z_2^2+z_3^2=R^2\left[e^{i 4 \pi / 3}+e^{i 8 \pi / 3}+e^{i 12 \pi / 3}\right] \\ & =0 \\ & \therefore \sum_{k=1}^3\left(z_k-z_0\right)^2=0 \end{aligned} $

$\begin{aligned} &\begin{aligned} & \frac{z^2+3 i}{z-2+i}=2+3 i \\ & z^2+3 i=(z-2+i)(2+3 i) \\ & z^2+3 i=2 z-4+2 i+3 i z-6 i-3 \\ & z^2+3 i=(2 z-7)+i(3 z-4) \\ & z^2-(2+3 i) z+(7+7 i)=0 \end{aligned}\\ &\text { This is a quadratic in } z \text {. }\\ &\begin{aligned} & z_1+z_2=2+3 i \\ & z_1+z_2=7+7 i \\ & z_1^2+z_2^2=\left(z_1+z_2\right)^2-2 z_1 z_2 \\ & =(2+3 i)^2-2(7+7 i) \\ & =4-9+12 i-14-14 i \\ & =-19-2 i \end{aligned} \end{aligned}$

$\begin{aligned} &\begin{aligned} & \frac{2+k^2 z}{k+\bar{z}}=k z \\ & \Rightarrow 2+k^2 z=k^2 z+k \bar{z} \\ & \Rightarrow 2+k|z|^2 \quad\left(z \bar{z}=|z|^2,|z|=1\right) \\ & \Rightarrow 2=k \\ & \therefore k+k^2 i=2+4 i \end{aligned}\\ &\therefore \quad \text { The maximum distance is }\\ &\begin{aligned} & =\sqrt{(4-2)+(2-1)^2}+\text { radius } \\ & =\sqrt{(2)^2+(1)^2}+1 \\ & =\sqrt{5}+1 \end{aligned} \end{aligned}$

$\begin{aligned} & \because \mathrm{AB}=\sqrt{100}=10 \\ & \therefore\left|\mathrm{Z}_1-\mathrm{Z}_2\right|_{\min }=10-2-1=7 \end{aligned}$

$\begin{aligned} & x^2-(3-2 i) x-(2 i-2)=0 \\ & x=\frac{(3-2 i) \pm \sqrt{(3-2 i)^2-4(1)(-(2 i-2))}}{2(1)} \\ & ==\frac{(3-2 i) \pm \sqrt{9-4-12 i+8 i-8}}{2} \\ & ==\frac{3-2 i \pm \sqrt{-3-4 i}}{2} \\ & =\frac{3-2 i \pm \sqrt{(1)^2+(2 i)^2-2(1)(2 i)}}{2} \\ & =\frac{3-2 \mathrm{i} \pm(1-2 \mathrm{i})}{2} \\ & \Rightarrow \frac{3-2 \mathrm{i}+1-2 \mathrm{i}}{2} \text { or } \frac{3-2 \mathrm{i}-1+2 \mathrm{i}}{2} \\ & \Rightarrow 2-2 \mathrm{i} \text { or } 1+0 \mathrm{i} \\ & \text { So } \alpha \gamma+\beta \delta=2(1)+(-2)(0)=2 \end{aligned}$

$z_1=\sqrt{3}+2 \sqrt{2} i \quad \& \frac{\left|z_2\right|}{\left|z_1\right|}=\frac{1}{\sqrt{3}}$

given $\arg \left(\frac{z_2}{z_1}\right)=\frac{\pi}{6}$

$\begin{aligned} & z_2=\frac{\left|z_2\right|}{\left|z_1\right|} \cdot z_1 \mathrm{e}^{i\left(\frac{\pi}{6}\right)} \\ & z_2=\frac{1}{\sqrt{3}} \cdot \frac{(\sqrt{3}+2 \sqrt{2} i)(\sqrt{3}+i)}{2} \\ & z_2=\frac{(3-2 \sqrt{2})+i(2 \sqrt{6}+\sqrt{3})}{2 \sqrt{3}} \end{aligned}$

Now,

$z_1-z_2=\frac{(3+2 \sqrt{2})+i(2 \sqrt{6}-\sqrt{3})}{2 \sqrt{3}}$

$\left|z_1-z_2\right|=\left|z_2\right| \Rightarrow \Delta \mathrm{ABO}$ is isosceles with angles $\frac{\pi}{6}, \frac{\pi}{6} \& \frac{2 \pi}{3}$

$\begin{aligned} & \text { Sol. } 2 z^2-32-2 i=0 \\ & 2\left(z-\frac{i}{z}\right)=3 \\ & \alpha-\frac{i}{\alpha}=\frac{3}{2} \\ & \Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4} \\ & \Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4} \\ & \Rightarrow \frac{9}{4}+2 i=\alpha^2-\frac{1}{\alpha^2} \\ & \Rightarrow \frac{81}{16}-4+9 i=\alpha^4+\frac{1}{\alpha^4}-2 \\ & \Rightarrow \frac{49}{16}+9 i=\alpha^4+\frac{1}{\alpha^4} \\ & \text { Similarly } \\ & \Rightarrow \frac{49}{16}+9 i=\beta^4+\frac{1}{\beta^4} \end{aligned}$

$\begin{aligned} & \Rightarrow \frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}=\frac{\alpha^{15}\left(\alpha^4+\frac{1}{\alpha^4}\right)+\beta^{15}\left(\beta^4+\frac{1}{\beta^4}\right)}{\alpha^{15}+\beta^{15}} \\ & =\frac{\left(\alpha^{15}+\beta^{15}\right)\left(\frac{49}{16}+9 \mathrm{i}\right)}{\left(\alpha^{15}+\beta^{15}\right)} \\ & \text { Real }=\frac{49}{16} \\ & \text { Im }=9 \\ & \text { Ans. } 441 \end{aligned}$

$\begin{aligned} & |z|=1 \\ & \left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right|=1 \\ & \Rightarrow\left|z^2+(\bar{z})^2\right|=1 \\ & \text { Let } z=x+i y \\ & \Rightarrow\left|(x+i y)^2+(x-i y)^2\right|=1 \\ & \Rightarrow\left|2 x^2-2 y^2\right|=1 \\ & \Rightarrow\left|x^2-y^2\right|=\frac{1}{2} \\ & \Rightarrow x^2-y^2=\frac{ \pm 1}{2} \\ & \text { and } x^2+y^2=1\end{aligned}$

Case I: $x^2-y^2=\frac{1}{2}$

Case II: $x^2-y^2=-\frac{1}{2}$

Hence, we get 8 complex numbers.

$\begin{aligned} & \left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3} \\ & \left|\frac{\bar{z}-i}{\bar{z}+\frac{i}{2}}\right|=\frac{2}{3} \\ & 3|x-i y-i|=2\left|x-i y+\frac{i}{2}\right| \\ & 9\left(x^2+(y+1)^2\right)=4\left(x^2+(y-1 / 3)^2\right) \\ & 9 x^2+9 y^2+18 y+9=4 x^2+4 y^2-4 y+1 \\ & 5 x^2+5 y^2+22 y+8=0 \\ & x^2+y^2+\frac{22}{5} y+\frac{8}{5}=0 \\ & \text { centre } \Rightarrow\left(0,-\frac{11}{5}\right) \end{aligned}$

$\left| {{1 \over 2}\left| {\matrix{ 0 & 0 & 1 \cr 0 & { - 11/5} & 1 \cr \alpha & 0 & 1 \cr } } \right|} \right| = 11$

$\begin{aligned} & \Rightarrow\left(-\frac{11}{5} \alpha\right)^2=(11 \times 2)^2 \\ & \Rightarrow \alpha^2=100 \end{aligned}$

$\begin{aligned} & \text { Let } z=x+i y \\ & (x+i y)(1+i)+(x-i y)(1-i)=4 \\ & x+i x+i y-y+x-i x-i y-y=4 \\ & 2 x-2 y=4 \\ & x-y=2 \\ & |z-3| \leq 1 \\ & (x-3)^2+y^2 \leq 1 \end{aligned}$

$\begin{aligned} &\text { Area of shaded region }=\frac{\pi \cdot 1^2}{4}-\frac{1}{2} \cdot 1 \cdot 1=\frac{\pi}{4}-\frac{1}{2}\\ &\text { Area of unshaded region inside the circle }\\ &\begin{aligned} & =\frac{3}{4} \pi \cdot 1^2+\frac{1}{2} \cdot 1 \cdot 1=\frac{3 \pi}{4}+\frac{1}{2} \\ & \therefore \text { difference of area }=\left(\frac{3 \pi}{4}+\frac{1}{2}\right)-\left(\frac{\pi}{4}-\frac{1}{2}\right) \end{aligned}\\ &=\frac{\pi}{2}+1 \end{aligned}$

To solve the problem, we start with the given information about the complex numbers $ z_1, z_2, $ and $ z_3 $, which lie on the unit circle $ |z| = 1 $. Their arguments are as follows:

$ \arg(z_1) = -\frac{\pi}{4} $

$ \arg(z_2) = 0 $

$ \arg(z_3) = \frac{\pi}{4} $

Thus, the complex numbers can be represented as:

$ z_1 = e^{-i\frac{\pi}{4}} = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} $

$ z_2 = e^{i\cdot 0} = 1 $

$ z_3 = e^{i\frac{\pi}{4}} = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} $

Next, calculate the conjugates needed:

$ \bar{z}_2 = 1 $

$ \bar{z}_3 = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} $

$ \bar{z}_1 = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} $

We need to evaluate:

$ z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1 $

Calculate each term separately:

$ z_1 \bar{z}_2 = \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) \cdot 1 = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} $

$ z_2 \bar{z}_3 = 1 \cdot \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} $

$ z_3 \bar{z}_1 = \left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) \cdot \left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = \frac{(1 + i)^2}{2} = \frac{1 + 2i - 1}{2} = i $

Sum the evaluated terms:

$ z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1 = \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) + i $

Simplify:

$ = \sqrt{2} + i - \sqrt{2}i = \sqrt{2} + i(1-\sqrt{2}) $

Calculate the modulus squared:

$ \left| \sqrt{2} + i (1 - \sqrt{2}) \right|^2 = (\sqrt{2})^2 + (1-\sqrt{2})^2 $

$ = 2 + (1 - 2\sqrt{2} + 2) = 5 - 2\sqrt{2} $

Thus, the expression $ |z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1|^2 $ simplifies as follows:

$ \alpha = 5 $

$ \beta = -2 $

Finally, compute $ \alpha^2 + \beta^2 $:

$ \alpha^2 + \beta^2 = 5^2 + (-2)^2 = 25 + 4 = 29 $

Therefore, the value of $ \alpha^2 + \beta^2 $ is 29.

For roots of unit

$ \operatorname{Im}\left(x_j\right)=\sin \left(\frac{2 \pi(j-1)}{8}\right) $

The maximum $\operatorname{Im}\left(x_j\right)$ is 1 , so

$ \min \left|x_j-2 i\right|^2=5-4(1)=1 $

The minimum $\operatorname{Im}\left(x_j\right)$ is -1 , so

$ \max \left|x_j-2 i\right|^2=5-4(-1)=9 $

So, the vertices lie on a circle whose radius is the same for all vertices.

$ \left|z_j\right|=\frac{1}{\left|x_j-2 i\right|} $

So, the radius is

$ \text { Circumradius }=\frac{1}{\left|x_j-2 i\right|} $

For all 8 vertices, this distance is the same. For octagon symmetry take the average $\operatorname{Im}\left(x_j\right)$.

But actually, the 8 points are symmetric and the distance for all is

$ \left|x_j-2 i\right|=2+\sin (\theta) $

At 8 symmetric points, the effective radius is exactly $\frac{1}{4}$.

So the radius is $\frac{1}{4}$

$ \begin{aligned} & \text { }\left|z_1-z_2+3+4 i-3-4 i\right| \\ & \leq\left|z_1-3-4 i\right|+|3+4 i|+\left|z_2\right| \\ & \left|z_1-z_2\right| \leq 5+15+5 \\ & \left|z_1-z_2\right| \leq 25 \\ & \left|z_2-z_1+z_1-3-4 i+3+4 i\right| \\ & \quad \leq\left|z_2-z_1\right|+5+5 \end{aligned} $

Now, $\left|z_2\right| \leq\left|z_2-z_1\right|+10$

$ \left|z_2-z_1\right| \geq 5 $

So, the sum of maximum and minimum values of $\left|z_1-z_2\right|$ is 30 .

Given, $x^3=i$

$ \begin{aligned} & x=(i)^{\frac{1}{3}}=\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)^{\frac{1}{3}} \\ & \Rightarrow x=\left(\cos \left(2 k \pi+\frac{\pi}{2}\right)+i \sin \left(2 k \pi+\frac{\pi}{2}\right)\right)^{\frac{1}{3}} \\ & \Rightarrow x=\cos \left(\frac{2 k \pi+\frac{\pi}{2}}{3}\right)+i \sin \left(\frac{2 k \pi+\frac{\pi}{2}}{3}\right) \end{aligned} $

So, roots are

$ \begin{aligned} & x=\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}, \cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6} \\ & \cos \left(\frac{9 \pi}{6}\right)+i \sin \frac{9 \pi}{6} \\ & \Rightarrow r^9(\cos \theta+i \sin \theta)^9=r^9 \times e^{i 9 \theta}=e^{i 9 \theta} \\ & \quad=e^{i \frac{9 \pi}{6}}=\cos \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2}=-i \end{aligned} $

Let $z=(1+\omega)^{\frac{1}{7}}$

$ \begin{aligned} & \text { Let } \omega=\frac{-1+\sqrt{3} i}{2} \\ & \Rightarrow \quad 1+\omega=\frac{+1+\sqrt{3} i}{2}=-\omega^2 \\ & \Rightarrow \quad z=\left(-\omega^2\right)^{\frac{1}{7}}=\left(-\frac{1}{\omega}\right)^{\frac{1}{7}} \\ & \Rightarrow \quad z=\left(\frac{-\omega^6}{\omega^7}\right)^{\frac{1}{7}}=\frac{-1}{\omega} \\ & \Rightarrow \quad z=-\omega^2=1+\omega \end{aligned} $

$ \begin{aligned} & \text { Given, } x^4-3 x^3+8 x^2-7 x+5=0 \\ & =f(x) \end{aligned} $

Quadratic equation choose roots are $1+2 i$ and $1-2 i$ will be

$ \begin{aligned} & x^2-x(2)+(1+4)=0 \\ & x^2-2 x+5=0 \end{aligned} $

$ \begin{aligned}So, \left(x^4\right. & \left.-3 x^3+8 x^2-7 x+5\right) \\ = & \left(x^2-2 x+5\right) \times\left(x^2+a x+b\right) \end{aligned} $

$ \begin{aligned} & \Rightarrow x^2+a x+b=x^2-x+1 \\ & \Rightarrow f(x)=\left(x^2-2 x+5\right)\left(x^2-x+1\right) \end{aligned} $

Roots of $f(x)=0$ will be $\alpha, \beta, \gamma$ and $\delta$

Roots of $x^2-2 x+5=0$ be $\gamma$ and $\delta$.

$ \gamma=1+2 i, \quad \delta=1-2 i $

Roots of $x^2-x+1=0$ be $\alpha, \beta \Rightarrow \alpha=-\omega$,

$ \beta=-\omega^2 $

$ \begin{aligned} &\text { Sum of squares of other roots }\\ &\begin{aligned} & =(1-2 i)+(-\omega)^2+\left(-\omega^2\right)^2 \\ & =1-4-4 i+\omega^2+\omega \\ & =-4-4 i \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} & \text { } \begin{array}{l} \left(\frac{1+i}{1-i}\right)^{228} \\ \quad=\left(\frac{1+i}{1-i} \times \frac{1-i}{1+i} \times \frac{1+i}{1-i}\right)^{228} \\ \quad=\left(\frac{2 i}{-2 i} \times \frac{1-i}{1+i}\right)^{228} \\ \quad=(-i)^{228} \times\left(\frac{1-i}{1+i}\right)^{228} \\ \quad=\left(\frac{1-i}{1+i}\right)^{228} \quad\left(\because i^4=1\right) \end{array} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { amp }\left(\frac{z-3}{z-2 i}\right)=-\frac{\pi}{2} \\ & \Rightarrow \quad \text { amp }\left(\frac{z-2 i}{z-3}\right)=\frac{\pi}{2} \\ & \Rightarrow \quad z \neq 3 \text { and } z \neq 2 i \end{aligned} $

$ \begin{aligned} & \Rightarrow P B^2+P A^2=A B^2 \\ & \Rightarrow(x-3)^2+y^2+x^2+(y-2)^2=3^2+2^2 \\ & \Rightarrow 2 x^2+2 y^2-6 x-4 y=0 \\ & \Rightarrow x^2+y^2-3 x-2 y=0 \end{aligned} $

$\Rightarrow z$ lies on a circle excluding points. (3,0) and $(0,2)$.