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} $

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

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

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 $(0,2)$.

$ \begin{aligned} & \text { }(1-i \sqrt{3})^{2025} \\ & \begin{aligned} \Rightarrow 1-i \sqrt{3} & =2\left(\cos \left(-\frac{\pi}{3}\right)+i \sin \left(-\frac{\pi}{3}\right)\right) \\ = & 2 e^{-i \pi / 3} \end{aligned} \\ & \Rightarrow(1-i \sqrt{3})^{2025}=2^{2025} e^{-i 675 \pi} \\ & \\ & =2^{2025}(\cos 675 \pi-i \sin 675 \pi \\ & \end{aligned} $

$ =-2^{2025} $

$ \begin{aligned} &\text { }(x+1)^4+3^4=0\\ &\left(\frac{x+1}{3}\right)^4=-1\\ &\begin{array}{ll} \text { Let } & z=\frac{x+1}{3} \\ \Rightarrow & z=(-1)^{1 / 4} . \\ \Rightarrow & z=\cos \left(\frac{2 k \pi+\pi}{4}\right)+i \sin \left(\frac{2 k \pi+\pi}{4}\right) \\ z_1 & =\cos \frac{\pi}{4}+i \sin \frac{\pi}{4} \\ z_2 & =\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4} \\ & z_3=\cos \frac{5 \pi}{4}+i \sin \frac{5 \pi}{4}=-\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}} \\ & z_4=\cos \frac{7 \pi}{4}+i \sin \frac{7 \pi}{4} \\ \Rightarrow & \frac{x+1}{3}=\frac{-1-i}{\sqrt{2}} \\ \Rightarrow & x=-\left(\frac{1+i}{\sqrt{2}}\right) 3-\frac{\sqrt{2}}{\sqrt{2}} \\ & x=-\frac{(3+\sqrt{2}+3 i)}{\sqrt{2}} \end{array} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & =\frac{(\sqrt{3}+i)(1-\sqrt{3} i)}{(-1+i)(-1-i)} \\ & =\frac{\sqrt{3}-3 i+i+\sqrt{3}}{1+i-i+1}=\frac{2 \sqrt{3}-2 i}{2} \\ & =\sqrt{3}-i \\ \operatorname{Amp}(z) & =-\tan ^{-1}\left|\frac{1}{\sqrt{3}}\right|=\frac{-\pi}{6} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \operatorname{Im}\left(\frac{z-3}{z+3 i}\right)=0, z+3 i \neq 0 \\ & \therefore x \neq 0, y \neq-3 \\ & \Rightarrow \quad \frac{z-3}{z+3 i}-\left(\frac{\overline{z-3}}{z+3 i}\right)=0 \\ & \Rightarrow \quad \frac{z-3}{z+3 i}-\frac{\bar{z}-3}{\bar{z}-3 i}=0 \\ & \Rightarrow \quad(z-3)(\bar{z}-3 i)-(z+3 i)(\bar{z}-3)=0 \\ & \Rightarrow \quad z \bar{z}-3 i z-3 \bar{z}+9 i -z \bar{z}+3 z-3 i \bar{z}+9 i=0 \\ & \Rightarrow \quad-3 i(z+\bar{z})+3(z-\bar{z})+18 i=0 \\ & \because \quad z=x+i y, \bar{z}=x-i y \\ & \Rightarrow \quad-3 i(2 x)+3(2 i y)+18 i=0 \\ & \Rightarrow \quad x-y-3=0, x, y \neq(0,-3) \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & z=(\sqrt{3}+i)^{10}+(\sqrt{3}-i)^{10} \\ &=\left(2 e^{\frac{\pi}{6}}\right)^{10}+\left(2 e^{\frac{-\pi i}{6}}\right)^{10} \\ &=2^{10}\left(e^{\frac{10 \pi i}{6}}+e^{\frac{-i 10 \pi}{6}}\right) \\ &=2^{10}\left(\cos \frac{10 \pi}{6}+i \sin \frac{10 \pi}{6}+\cos \frac{10 \pi}{6}-i \sin \frac{10 \pi}{6}\right) \\ &=2^{10} \cdot 2 \cos \frac{5 \pi}{3}=2^{11} \cos \left(2 \pi-\frac{\pi}{3}\right) \\ &=2^{11} \times \frac{1}{2}=2^{10}=1024 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \text { } \quad \begin{aligned}Let \,\, &(-1-\sqrt{3} i)^{3 / 4} \\ & z=-1-\sqrt{3} i \end{aligned} \\ & \quad \begin{aligned} z & =2 e^{i\left(\frac{-2 \pi}{3}\right)} \\ z^{3 / 4} & =e^{i\left(\frac{-2 \pi}{3}\right)} \\ & =\left(2 e^{-i \frac{2 \pi}{3}}\right)^{3 / 4} \\ & =2^{3 / 4}\left(e^{-i \pi}\right)^{1 / 2} \\ \therefore \text { At } \quad k & =0, z_1=2^{3 / 4} e^0=2^{3 / 4} \\ \text { At } \quad k & =1, z_2=2^{3 / 4} e^{-i \pi}=-2^{3 / 4} \end{aligned} \\ & \text { Number of real value }=2 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} z & =\sqrt{24-70 i}+\sqrt{-24+70 i} \\ z^2 & =24-70 i+(-24+70 i) \\ & \quad+2 \sqrt{24-70 i} \sqrt{-24+70 i} \\ & =2 i(24-70 i)=2(24 i+70) \\ & =4(12 i+35) \\ z & =\sqrt{4} \sqrt{35+12 i} \\ & =2\left( \pm\left(\frac{\operatorname{Re} z+|z|}{2}\right)^{1 / 2}\right. \\ & \left.\quad+i\left(\frac{-\operatorname{Re} z+|z|}{2}\right)^{1 / 2}\right) \\ & = \pm 2\left(\left(\frac{35+37}{2}\right)^{1 / 2}+i\left(\frac{-35+37}{2}\right)^{1 / 2}\right) \\ & = \pm 2(6+i)= \pm(12 \pm 2 i) \end{aligned} \end{aligned} $

$z=\frac{1-i \cos \theta}{1+2 i \sin \theta}$ is purely imaginary

$ \begin{aligned} & \therefore z+\bar{z}=0 \\ & \Rightarrow \frac{1-i \cos \theta}{1+2 i \sin \theta}+\frac{1+i \cos \theta}{1-2 i \sin \theta}=0 \\ & 1-2 i \sin \theta-i \cos \theta-2 \sin \theta \cos \theta+1 \\ & \frac{+i \cos \theta+2 i \sin \theta-2 \sin \theta \cos \theta}{(1+2 i \sin \theta)(1-2 i \sin \theta)}=0 \\ & \Rightarrow(2-4 \sin \theta \cos \theta)=0 \\ & \Rightarrow 2 \sin 2 \theta=2 \Rightarrow \sin 2 \theta=1 \\ & \Rightarrow \sin 2 \theta=\sin \frac{\pi}{2} \\ & \therefore \quad 2 \theta=n \pi+(-1)^n \frac{\pi}{2}, n \in z \\ & \therefore \quad \theta=\frac{n \pi}{2}+(-1)^n \frac{\pi}{4}, n \in z . \end{aligned} $

$ \begin{aligned} & x^2-x+1=0 \rightarrow \alpha \\ & \therefore \alpha=-\omega \Rightarrow \alpha^3=-\omega^3=-1 \end{aligned} $

Now, $\left(\alpha+\frac{1}{\alpha}\right)^3+\left(\alpha^2+\frac{1}{\alpha^2}\right)^3+\ldots$ to 12 terms

$ \begin{array}{r} =\left(\frac{\alpha^2+1}{\alpha}\right)^3+\left(\frac{\alpha^4+1}{\alpha^2}\right)^3+(-1-1)^3+\ldots +12 \text { terms } \\ =\left[(1)^3+(1)^3+(-2)^3+(-1)^3+(-1)^3\right. \left.+(2)^3\right] 2=0 \end{array} $

We have, $|z-1| \leq 2$

$ \begin{aligned}And\,\, & \left|\omega z-1-\omega^2\right|=r \\ & |\omega|\left|z-\omega^2-\omega\right|=r \\ & |z+1|=r \\ & \quad\left[\because|\omega|=1,1+\omega+\omega^2=0\right] \\ & |z-1| \leq 2 \Rightarrow|z+1-2| \leq 2 \\ & |z+1|-2 \leq 2 \Rightarrow|z+1| \leq 4 \end{aligned} $

For no solution

$ \therefore \quad|z+1|>4 \Rightarrow r>4 $

$ \begin{aligned} & \text { Given, }|z|=2, z_1=\frac{z}{2} e^{i \alpha} \\ & \therefore \quad z=2 e^{i \theta} \end{aligned} $

( $\theta$ is the argument of $z$ )

And  $ z_1=\frac{2 e^{i \theta}}{2} e^{i \alpha}=e^{i(\theta+\alpha)} $

Now, $\quad z_1^n=e^{i n(\theta+\alpha)}$

$ \Rightarrow \quad z_1^{-n}=e^{-i n(\theta+\alpha)} $

We know that $\frac{e^{i x}-e^{-i x}}{e^{i x}+e^{-i x}}=i \tan x$

$ \quad \begin{aligned}\therefore \frac{z_1^n-z_1^{-n}}{z_1^n+z_1^{-n}} & =\frac{e^{i n(\theta+\alpha)}-e^{-i n(\theta+\alpha)}}{e^{i n(\theta+\alpha)}+e^{-i n(\theta+\alpha)}} \\ & =i \tan (n(\theta+\alpha)) \\ & =i \tan (n \theta+n \alpha) \end{aligned} $

Given, $n, K \in N$ such that $n \neq 3 K$

Let $z=\sqrt{3}+i$, then $\bar{z}=\sqrt{3}-i$

$ \begin{aligned} & |z|=|\bar{z}|=\sqrt{3+1}=2 \\ & \arg (z)=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6} \\ & \arg (z)=\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)=\frac{-\pi}{6} \\ & \therefore \quad z=2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right) \\ & \Rightarrow z^{2 n}=2^{2 n}\left(\cos \frac{2 n \pi}{6}+i \sin \frac{2 n \pi}{6}\right) \\ & \text { And } \bar{z}=2\left(\cos \frac{\pi}{6}-i \sin \frac{\pi}{6}\right) \\ & \Rightarrow \bar{z}^{2 n}=2^{2 n}\left(\cos \frac{2 n \pi}{6}-i \sin \frac{2 n \pi}{6}\right) \\ & \therefore z^{2 n}+\bar{z}^{2 n}=2 \cdot 2^{2 n} \cos \left(\frac{n \pi}{3}\right) \end{aligned} $

$ =2^{2 n+1} \cos \left(\frac{n \pi}{3}\right) $

Since, $n \neq 3 K \Rightarrow n$ is not a multiple of 3 $\cos \left(\frac{n \pi}{3}\right) \neq \cos (k \pi)= \pm 1=(-1)^{n+1} 2^{2 n}$

Let $z=1-i \sqrt{3}$

$ \begin{aligned} & \therefore \quad|z|=\sqrt{1+(\sqrt{3})^2}=2 \\ & \arg (z)=\tan ^{-1}\left(\frac{-\sqrt{3}}{1}\right)=-\tan \frac{\pi}{3} \\ & \therefore \quad z=2\left(\cos \left(\frac{-\pi}{3}\right)+i \sin \left(-\frac{\pi}{3}\right)\right) \\ & \quad=2 e^{-i \pi / 3} \\ & \therefore \quad \sqrt[3]{z}=2^{1 / 3} \cdot e^{4-i \pi / 3+2 k \pi i} \\ & \quad=2^{1 / 3} e^{i\left(-\frac{\pi}{9}+\frac{2 k \pi}{3}\right)} \end{aligned} $

For any integer $k \in\{0,1,2\}$

$\omega_0=$ lies in 4th quadrant

$\omega_1=$ lies in 2nd quadrant

$\omega_2=$ lies in 3rd quadrant

Given, $\frac{2+3 i}{i-2}-\frac{4 i-3}{3+4 i}=x+i y$

Now, $\frac{2+3 i}{i-2}=\frac{2+3 i}{-2+i} \times \frac{(-2-i)}{(-2-i)}$

$ \Rightarrow \quad \frac{-4-2 i-6 i-3 i^2}{(-2)^2-i^2}=\frac{-1-8 i}{5} $

And, $\frac{4 i-3}{3+4 i} \times \frac{3-4 i}{3-4 i}$

$ \Rightarrow \quad \frac{-9+12 i+12 i-16 i^2}{3^2-(4 i)^2}=\frac{7+24 i}{25} $

So, $\quad \frac{2+3 i}{i-2}-\frac{4 i-3}{3+4 i}$

$ \begin{aligned} & =\frac{-1}{5}-\frac{8 i}{5}-\frac{7}{25}-\frac{24}{25} i \\ & =\frac{-12}{25}-\frac{64}{25} i \end{aligned} $

On comparing, we get

$ x=\frac{-12}{25}, y=\frac{-64}{25} $

So, $3 x+y=3\left(\frac{-12}{25}\right)-\frac{64}{25}$

$ =-\frac{36}{25}-\frac{64}{25}=\frac{-100}{25}=-4 $

Given, arg $\left(\frac{z-3 i}{z+2 i}\right)=\frac{\pi}{4}$

Let $z=x+i y$

So, $\frac{z-3 i}{z+2 i}=\frac{x+i y-3 i}{x+i y+2 i}$

$\Rightarrow \frac{x+i(y-3)}{x+i(y+2)} \times \frac{x-i(y+2)}{x-i(y+2)}$

$ \begin{aligned} & \Rightarrow \frac{x^2-i x(y+2)+i x(y-3)+(y-3)(y+2)}{x^2+(y+2)^2} \\ & \Rightarrow \frac{x^2+y^2-y-6-i 5 x}{x^2+(y+2)^2} \end{aligned} $

Let the complex numbers be $A+i B$, where

$ A=\frac{x^2+y^2-y-6}{x^2+(y+2)^2} \text { and } B=\frac{-5 x}{x^2+(y+2)^2} $

So, $\arg \left(\frac{z-3 i}{z+2 i}\right)=\left(\frac{\pi}{4}\right)$

$ \begin{aligned} & \Rightarrow \quad \tan \left(\frac{B}{A}\right)=\frac{\pi}{4} \\ & \Rightarrow \quad \frac{-5 x}{x^2+y^2-y-6}=\tan ^{-1}\left(\frac{\pi}{4}\right)=1 \\ & \Rightarrow \quad x^2+y^2-y-6=-5 x \\ & \Rightarrow \quad x^2+5 x+y^2-y-6=0, \text { where } \\ & x \neq 0 \text { and } \quad y \neq-2 \end{aligned} $

Given, $x=\omega^2-\omega+2$ where $w$ is a complex cube root of unity.

We know that $1+\omega+\omega^2=0$

$ \Rightarrow \omega^2=-1-\omega $

So, $x=(-1-\omega)-\omega+2=1-2 \omega$

$ \Rightarrow \omega=\frac{1-x}{2} $

So, $1+\omega+\omega^2=0$

$ \begin{aligned} & \Rightarrow \quad 1+\left(\frac{1-x}{2}\right)+\left(\frac{1-x}{2}\right)^2=0 \\ & \Rightarrow \quad 1+\frac{(1-x)}{2}+\frac{(1-x)^2}{4}=0 \\ & \Rightarrow \quad 4+2(1-x)+1-2 x+x^2=0 \\ & \Rightarrow \quad 4+2-2 x+1-2 x+x^2=0 \\ & \Rightarrow \quad x^2-4 x+7=0 \end{aligned} $

Let $z=\sqrt{3}-i$

$ |z|=\sqrt{(\sqrt{3})^2+(-1)^2}=\sqrt{3+1}=2 $

And $\quad \theta=\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=-\frac{\pi}{6}$ or $\frac{11 \pi}{6}$

So, $\quad z=2\left(\cos \left(-\frac{\pi}{6}\right)+i \sin \left(-\frac{\pi}{6}\right)\right)$

Now, $z^{\frac{3}{7}}=(\sqrt{3}-i)^{\frac{3}{7}}$

First, $z^3=(\sqrt{3}-i)^3$

$ \begin{aligned} & =2^3\left(\cos \left(\frac{-3 \pi}{6}\right)+i \sin \left(\frac{-3 \pi}{6}\right)\right) \\ & =8\left(\cos \frac{\pi}{2}-i \sin \frac{\pi}{2}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, $z^{\frac{3}{7}}=(\sqrt{3}-i)^{\frac{3}{7}}$

$ =(8)^{\frac{1}{7}}\left(\cos \left(\frac{-\frac{\pi}{2}+2 k \pi}{7}\right)+i \sin \left(\frac{-\frac{\pi}{2}+2 k \pi}{7}\right)\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

for $k=0,1, \ldots, 6$

So, $z^3=8(0-i)=-8 i \quad$ [Using Eq. (i)]

Now, we know that the product of $n t$ h root of a complex number $A$ is given by $(-1)^{n-1} A$, if $A$ is a real number.

Since, $n=7$ is odd, the product of the 7 roots of $A$ is $A$ itself.

$\therefore$ The product of the 7th roots of $A$ is $A$ itself.

$ \therefore \quad(\sqrt{3}-i)^{\frac{3}{7}}=z^3 $

So, product $=-8 i$

$\because \arg \left(\frac{Z-3+2 i}{Z+2-3 i}\right)=\frac{\pi}{4}$ let $A=3-2 i, B=-2+3 i$

this means, the angle between $Z-A$ and $Z-B$ is $\pi / 4$ i.e., the point $Z$ subtends an angle of $\pi / 4$ at the fixed point $Z$ lies on circle that subtends angle $\frac{\pi}{4}$ at the chord joining points $A$ and $B$.

Now, distance between $A$ and $B=|A-1|$

$ \begin{aligned} & =|(3-2 i)-(-2+3 i)|=|5-5 i| \\ & =5 \sqrt{2} \end{aligned} $

therefore,

$ \begin{aligned} \operatorname{Radius}(R) & =\frac{A B}{2 \sin \theta}=\frac{5 \sqrt{2}}{2 \sin \pi / 4} \\ & =\frac{5 \sqrt{2}}{2 \times \frac{1}{\sqrt{2}}}=5 \end{aligned} $

Hence, the locus of $P$ is a circle with radius 5 .

Let $\sqrt{\frac{A}{B}}=\sqrt{\frac{21+12 \sqrt{2} i}{21-12 \sqrt{2} i}}$

Now,

$ \frac{21+12 \sqrt{2} i}{21-12 \sqrt{2} i}=\frac{21+12 \sqrt{2} i}{21-12 \sqrt{2} i} \times \frac{21+12 \sqrt{2i}}{21+12 \sqrt{2i}} $

$ \begin{aligned} & =\frac{(21+12 \sqrt{2} i)^2}{(21)^2-(12 \sqrt{2} i)^2} \\ & =\frac{153+504 \sqrt{2} i}{729} \\ & =\frac{153}{729}+\frac{504 \sqrt{2}}{729} i \\ \text { then, } & \sqrt{\frac{153}{729}+\frac{504 \sqrt{2} i}{729}}=a+i b \end{aligned} $

Squaring both sides, we get

$ \begin{aligned} & \frac{153}{729}+\frac{504 \sqrt{2}}{729} i=a^2-b^2+(2 a b) i \\ & \text { So, } a^2-b^2=\frac{153}{729} \text { and } a b=\frac{252 \sqrt{2}}{729} \\ & \Rightarrow a^2-b^2=\frac{17}{81} \text { and } a b=\frac{28 \sqrt{2}}{81} \end{aligned} $

let $b=a k$

$ \Rightarrow b / a=k $

So, $a^2\left(1-k^2\right)=\frac{17}{81}$ and $a^2 k=\frac{28 \sqrt{2}}{81}$

So, $\frac{k}{1-k^2}=\frac{28 \sqrt{2}}{81} \times \frac{81}{17}=\frac{28 \sqrt{2}}{17}$

From the options, $k=\frac{4 \sqrt{2}}{7}$ satisfy the equation.