Complex Numbers

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 $=-8 i$

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

$ \begin{aligned} z & =\frac{(2-i)\left(1+i^3+3 i^2+3 i\right)}{1+i^2-2 i} \\\\ & =\frac{(2-i)(1-i-3+3 i)}{-2 i} \\\\ & =\frac{(2-i)(-2+2 i)}{-2 i}=\frac{(2-i)(1-i)}{i} \end{aligned} $

$ \begin{aligned} & =\frac{2-i-2 i+i^2}{i} \\\\ & =\frac{1-3 i}{i}=-i-3 \\\\ z & =-3-i \\\\ \arg z= & \tan ^{-1}\left(\frac{1}{3}\right)-\pi \end{aligned} $

\begin{array}{l} \operatorname{amp}\left(\frac{2 z-i}{z+2 i}\right)=\frac{\pi}{4}\quad \text { (given) } \\\\ \Rightarrow \operatorname{amp}(2 z-i)-\operatorname{amp}(z+2 i)=\frac{\pi}{4} \\\\ \Rightarrow \operatorname{amp}[2 x+i(2 y-1)]-\operatorname{amp} {[x+i(y+2)]=\frac{\pi}{4}} \\\\ \Rightarrow \tan ^{-1}\left(\frac{2 y-1}{2 x}\right)-\tan ^{-1}\left(\frac{y+2}{x}\right)=\tan ^{-1} 1 \\\\ \Rightarrow \tan ^{-1}\left(\frac{2 y-1}{2 x}\right)=\tan ^{-1} 1+\tan ^{-1}\left(\frac{y+2}{x}\right) \\\\ \Rightarrow \tan ^{-1}\left(\frac{2 y-1}{2 x}\right)=\tan ^{-1} \frac{1+\frac{y+2}{x}}{1-\frac{y+2}{x}} \\\\ \Rightarrow \quad \frac{2 y-1}{2 x}=\frac{x+y+2}{x-y-2} \\\\ \Rightarrow(2 y-1)(x-y-2)=2 x(x+y+2) \\\\ \Rightarrow 2 x y-2 y^2-4 y-x+y+2 = 2 x^2+2 x y+4 x \\\\ \Rightarrow 2 x^2+2 y^2+5 x+3 y-2=0, (x, y) \neq(0,-2) . \end{array}

$x^2+2 x+4=0$

$ \begin{aligned} & x=\frac{-2 \pm \sqrt{4-16}}{2}=\frac{-2 \pm 2 i \sqrt{3}}{2} \\\\ & \alpha=2 e^{i \frac{2 \pi}{3}} \\\\ & x=-1 \pm i \sqrt{3}=2\left(\frac{\cos 2 \pi}{3}+i \sin \frac{2 \pi}{3}\right) \\\\ & \text { and } \quad \beta=2 e^{-i \frac{2 \pi}{3}} \\\\ & \alpha^{2024}-\beta^{2024}=\left[\frac{\alpha^{2025}}{\alpha}-\frac{\beta^{2025}}{\beta}\right] \\\\ & =\frac{2^{2025} e^{i \frac{2 \pi}{3} \times 2025}}{\alpha}-\frac{2^{2025} e^{-i \frac{2 \pi}{3} \times 2025}}{\beta} \\\\ & =2^{2025}\left[\frac{e^{i 1350 \pi}}{\alpha}-\frac{e^{-i 1350 \pi}}{\beta}\right] \\\\ & =2^{2025}\left[\frac{1}{\alpha}-\frac{1}{\beta}\right]=2^{2025}\left(\frac{1}{2} e^{-i \frac{2 \pi}{3}}-\frac{1}{2} e^{-i \frac{i \pi}{3}}\right) \end{aligned} $

$ \begin{aligned} & =-2^{2024}\left(e^{i \frac{2 \pi}{3}}-e^{-i \frac{2 \pi}{3}}\right) \\\\ & =-2^{2024}\left(i 2 \sin \frac{2 \pi}{3}\right) \\\\ & =-2^{2024} \cdot 2 \times \frac{\sqrt{3}}{2} i=i K \\\\ & \therefore \quad K=-2^{2024} \sqrt{3} \end{aligned} $

To solve the given equation $ z^{2} + az + a^{2} = 0 $, where $ z = x + iy $ and $ a \in \mathbb{R} $, follow these steps:

First, multiply both sides of the equation by $ (z-a) $:

$ (z-a)(z^{2} + az + a^{2}) = 0 $

Apply the identity for the difference of cubes:

$ z^{3} - a^{3} = 0 $

This implies:

$ z^{3} = a^{3} $

Taking the modulus (or absolute value) of both sides:

$ |z|^{3} = |a|^{3} $

Thus, we conclude:

$ |z| = |a| $

Given the complex numbers $ z_1, z_2, z_3 $ with unit modulus, we know:

$ z_1 \bar{z}_1 = z_2 \bar{z}_2 = z_3 \bar{z}_3 = 1 $

This implies that each complex number has a magnitude of 1. We are also given the condition:

$ |z_1 - z_2|^2 + |z_1 - z_3|^2 = 4 $

We can expand the squared magnitudes as follows:

$ \begin{aligned} |z_1 - z_2|^2 &= (z_1 - z_2)(\bar{z}_1 - \bar{z}_2) = z_1 \bar{z}_1 - z_1 \bar{z}_2 - \bar{z}_1 z_2 + z_2 \bar{z}_2, \\ |z_1 - z_3|^2 &= (z_1 - z_3)(\bar{z}_1 - \bar{z}_3) = z_1 \bar{z}_1 - z_1 \bar{z}_3 - \bar{z}_1 z_3 + z_3 \bar{z}_3. \end{aligned} $

Substitute the expansions back into the equation:

$ \begin{aligned} & (z_1 \bar{z}_1 - z_1 \bar{z}_2 - \bar{z}_1 z_2 + z_2 \bar{z}_2) \\ & + (z_1 \bar{z}_1 - z_1 \bar{z}_3 - \bar{z}_1 z_3 + z_3 \bar{z}_3) = 4. \end{aligned} $

Since $ z_1 \bar{z}_1 = z_2 \bar{z}_2 = z_3 \bar{z}_3 = 1 $, we simplify:

$ (1 - z_1 \bar{z}_2 - \bar{z}_1 z_2 + 1) + (1 - z_1 \bar{z}_3 - \bar{z}_1 z_3 + 1) = 4 $

Simplifying further gives:

$ 4 - (z_1 \bar{z}_2 + \bar{z}_1 z_2 + z_1 \bar{z}_3 + \bar{z}_1 z_3) = 4 $

Thus:

$ z_1 \bar{z}_2 + \bar{z}_1 z_2 + z_1 \bar{z}_3 + \bar{z}_1 z_3 = 0 $

This is the required result.

Given the equation:

$ \left(\frac{a+b \omega+c \omega^{2}}{c+a \omega+b \omega^{2}}\right)^{k}+\left(\frac{a+b \omega+c \omega^{2}}{b+a \omega^{2}+c \omega}\right)^{l}=2 $

where $\omega$ is the complex cube root of unity, meaning $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.

To solve this, we first manipulate the terms by using properties of complex numbers. Consider multiplying and dividing:

Multiply and divide the first term by $\omega^2$:

$ \left(\frac{\omega^2(a+b \omega+c \omega^2)}{\omega^2(c+a \omega+b \omega^2)}\right)^{k} $

Multiply and divide the second term by $\omega$:

$ \left(\frac{\omega(a+b \omega+c \omega^2)}{\omega(b+a \omega^2+c \omega)}\right)^{l} $

This simplifies to:

$ \left(\omega^{2k}\right) + \left(\omega^l\right) = 2 $

For these powers to sum to 2, and knowing the properties of cube roots of unity, we have specific implications for the exponents:

Since $\omega^n$ can only take values 1, $\omega$, or $\omega^2$, for the sum $\omega^{2k} + \omega^l = 2$ to be a real number (i.e., 2), both $\omega^{2k}$ and $\omega^l$ must equal 1.

Thus, this requires both 2k and l to be multiples of 3 (since $\omega^3 = 1$).

Hence, the expression $2k + l$ is also a multiple of 3, indicating that $2k + l$ is always divisible by 3.

Given, $ z_{1}=\sqrt{3}+i \sqrt{3} $

$ \begin{aligned} z_{2} & =\sqrt{3}+i \\ \frac{z_{1}}{z_{2}} & =\frac{\sqrt{3}(1+i)}{\sqrt{3}+i}=\frac{\sqrt{3}(1+i)}{\sqrt{3}+i} \frac{(\sqrt{3}-i)}{(\sqrt{3}-i)} \\ & =\frac{\sqrt{3}\left(\sqrt{3}-i+\sqrt{3} i-i^{2}\right)}{3-i^{2}} \\ & =\frac{\sqrt{3}}{4}[\sqrt{3}+1+i(\sqrt{3}-1)] \end{aligned} $

multiply and divide by $ 2 \sqrt{2} $.

$ \begin{aligned} \Rightarrow \quad \frac{z_{1}}{z_{2}} & =\frac{2 \sqrt{2} \sqrt{3}}{4} \quad\left[\cos 15^{\circ}+i \sin 15^{\circ}\right] \\ \frac{z_{1}}{z_{2}} & =\frac{\sqrt{6}}{2}\left[e^{i 15^{\circ}}\right] \\ \Rightarrow\left(\frac{z_{1}}{z_{2}}\right)^{50} & =\left(\frac{\sqrt{6}}{2}\right)^{50} e^{i 750^{\circ}} \\ \theta & =750^{\circ}-720^{\circ}=30^{\circ} \end{aligned} $

$ \because $ First quadrant

$ x^{3}-3 x^{2}+3 x+7=0 $

$ \begin{array}{l} \Rightarrow \quad(x+1)\left(x^{2}-4 x+7\right)=0 \\ \Rightarrow \quad x=-1 \text { or } \frac{4 \pm \sqrt{16-28}}{2(1)} \\ \Rightarrow \quad x=-1 \text { or } 2 \pm \sqrt{3} i \\ \therefore \quad \alpha=-1, \beta=2+\sqrt{3} i \text { and } \\ \gamma=2-\sqrt{3} i \end{array} $

New root are $ -1-h_{,}(2-h)+\sqrt{3} i $ and $ (2-h)-\sqrt{3} i $

Equation which have these roots is $ (x+1+h)(x+h-2-\sqrt{3} i) $

$ \begin{array}{l} \quad(x+h-2+\sqrt{3} i)=0 \\ \Rightarrow(x+1+h)\left[(x+h-2)^{2}-(\sqrt{3} i)^{2}\right]=0 \\ \Rightarrow(x+1+h)\left[x^{2}+(h-2)^{2}\right. \\ +2 x(h-2)+3]=0 \end{array} $

In this equation terms containing $ x^{2} $ and $ x $ should be missing.

$ \therefore(1+h)+2(h-2)=0 $ and

$ 2(1+h)(h-2)+3+(h-2)^{2}=0 $

Now, $ \frac{\alpha-h}{\beta-h}+\frac{\beta-h}{\gamma-h}+\frac{\gamma-h}{\alpha-h} $

$ =\frac{-1-1}{2+\sqrt{3} i-1}+\frac{2+\sqrt{3} i-1}{2-\sqrt{3} i-1}+\frac{2-\sqrt{3} i-1}{-1-1} $

$ =\frac{-2}{1+\sqrt{3} i}+\frac{1+\sqrt{3} i}{1-\sqrt{3} i}+\frac{1-\sqrt{3} i}{-2} $

$ \begin{array}{l} =\frac{-2(1-\sqrt{3} i)}{4}+\frac{(1+\sqrt{3} i)^{2}}{4}+\frac{(-2)(1-\sqrt{3} i)}{4} \\ =\frac{1}{4}[-2+2 \sqrt{3} i+1-3+2 \sqrt{3} i-2+2 \sqrt{3} i] \\ =\frac{1}{4}[-6+6 \sqrt{3} i] \\ =3\left[-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right] \end{array} $

Given, $ \sqrt{-5+12 i}=\sqrt{\frac{13-5}{2}}+i \sqrt{\frac{13+5}{2}} $

$ \begin{aligned} & =2+3 i \\\\ \therefore x+i y & =\frac{13(2+3 i)}{(2-3 i)(3+2 i)} \\\\ & =\frac{13(2+3 i)}{6-9 i+4 i-6 i^{2}}=\frac{13(2+3 i)}{12-5 i} \\\\ & =\frac{13(2+3 i)(12+5 i)}{(12-5 i)(12+5 i)} \\\\ & =\frac{13\left(24+36 i+10 i+15 i^{2}\right)}{169} \\\\ & =\frac{13(9+46 i)}{169}=\frac{9}{13}+\frac{46 i}{13} \\\\ \Rightarrow \quad x & =\frac{9}{13} \text { and } y=\frac{46}{13} \\\\ 13 y-26 \dot{x} & =46-18=28 \end{aligned} $

Given, $ z=x+i y $

$ \begin{array}{l} |x+i y-1|+|x+i y+i|=2 \\\\ \sqrt{(x-1)^{2}+y^{2}}+\sqrt{x^{2}+(y+1)^{2}}=2 \\\\ \sqrt{(x-1)^{2}+y^{2}}=2-\sqrt{x^{2}+(y+1)^{2}} \\\\ x^{2}+y^{2}-2 x+1=4+x^{2}+y^{2}+2 y+1 -4 \sqrt{x^{2}+(y+1)^{2}} \\\\ x+y+2=2 \sqrt{x^{2}+y^{2}+2 y+1} \end{array} $

$ \begin{array}{r} x^{2}+y^{2}+4+2 x y+4 x+4 y \\\\ =4 x^{2}+4 y^{2}+8 y+4 \\\\ 3 x^{2}+3 y^{2}-2 x y-4 x+4 y=0 \end{array} $

To find one of the values of $(-64i)^{5/6}$, let's proceed with the computation:

Let $ Z = (-64i)^{5/6} $.

We can express $ Z $ as:

$ Z = \left((-i)^5\right)^{1/6} \times (64)^{5/6} $

Calculate $(64)^{5/6}$:

$ 64^{5/6} = 32 $

Handle the imaginary unit:

Since $(-i)^5 = -i$, we represent this in polar coordinates:

$ -i = \cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right) $

Find $(-i)^{1/6}$:

Using De Moivre’s theorem, $ (-i)^{1/6} $ can be expressed as:

$ (-i)^{1/6} = \left[\cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right)\right]^{1/6} $

$ = \cos\left(\frac{3\pi}{12}\right) + i\sin\left(\frac{3\pi}{12}\right) $

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

Combine the results:

Therefore, $ Z $ becomes:

$ Z = 32 \times \frac{1}{\sqrt{2}}(1+i) = 16\sqrt{2}(1+i) $

The calculation yields one of the values of $(-64i)^{5/6}$ as $ 16\sqrt{2}(1+i) $.

We have,

$ \begin{aligned} & \frac{(2-i) x+(1+i)}{2+i}+\frac{(1-2 i) y+(1-i)}{1+2 i}=1-2i \\\\ & \Rightarrow \frac{(2 x+1)+(1-x) i}{2+i}+\frac{(y+1)+(-2 y-1) i}{1+2 i}=1-2i \\\\ & \Rightarrow \frac{[(2 x+1)+(1-x) i](2-i)}{4-i^2}+\frac{[(y+1)+(-2 y-1) i](1-2 i)}{1-4 i^2}=1-2i \\\\ & \Rightarrow {[2(2 x+1)+(1-x)+(y+1)+2(-2 y-1)] } \\\\ &+i[-2 x-1+2(1-x)-2(y+1)+(-2 y-1)] \\\\ &=5(1-2 i) \\\\ & \Rightarrow(3 x-3 y+2)+i(-4 x-4 y-2)=5-10 i \end{aligned} $

We have,

$ z=1-\sqrt{3} i=2 e^{-i \frac{\pi}{3}} $

Now, $z^3-3 z^2+3 z$

$ \begin{aligned} & \begin{array}{r} =\left(2 e^{-\frac{\pi}{3} i}\right)^3-3\left(2 e^{-\frac{\pi}{3} i}\right)^2+3\left(2 e^{-\frac{\pi}{3} i}\right) \\\\ \begin{aligned} &= 8 e^{-\pi i}-12 e^{-\frac{2 \pi}{3} i}+6 e^{-\frac{\pi}{3} i} \\\\ &=8(\cos (-\pi)+i \sin (-\pi)) \end{aligned} \\\\ \quad-12\left(\cos \left(-\frac{2 \pi}{3}\right)+i \sin \left(-\frac{2 \pi}{3}\right)\right) \\\\ \quad+6\left(\cos \left(-\frac{\pi}{3}\right)+i \sin \left(-\frac{\pi}{3}\right)\right) \\\\ =8(-1+i(0))-12\left(-\frac{1}{2}+i\left(-\frac{\sqrt{3}}{2}\right)\right) \\\\ +6\left(\frac{1}{2}+i\left(-\frac{\sqrt{3}}{2}\right)\right)\end{array} \\\\ & =(-8+6+3)+i(0+6 \sqrt{3}-3 \sqrt{3}) \\\\ & =1+3 \sqrt{3 i} \end{aligned} $

Let, $(\sqrt{3}-i)^{2 / 5}=x$

$ \begin{array}{ll} \Rightarrow & (\sqrt{3}-i)^2=x^5 \\\\ \Rightarrow & x^5=(\sqrt{3})^2+(i)^2-2 \sqrt{3} i \\\\ \Rightarrow & x^5=3+(-1)-2 \sqrt{3} i \\\\ \Rightarrow & x^5=2-2 \sqrt{3} i=2(1-\sqrt{3} i) \end{array} $

Clearly, the product of all the values of $(\sqrt{3}-i)^{2 / 5}$ is $2(1-\sqrt{3} i)$

12 th roots of unity are $l, e^{i \frac{\pi}{6}}, e^{i \frac{2 \pi}{6}}$, $e^{i \frac{3 \pi}{6}}, e^{i \frac{4 \pi}{6}}, \ldots, e^{i \frac{11 \pi}{6}}$.

30th roots of unity are $1, e^{i \frac{\pi}{15}}, e^{i \frac{2 \pi}{15}}$, $e^{i \frac{3 \pi}{15}}, e^{i \frac{4 \pi}{15}}, \ldots, e^{i \frac{29 \pi}{15}}$.

Clearly, common roots are $1, e^{i \frac{\pi}{3}}, e^{i \frac{2 \pi}{3}}$, $e^{i \pi}, e^{i \frac{4 \pi}{3}}$ and $e^{i \frac{5 \pi}{3}}$

Hence, the number of common roots is 6.

We have,

$ \begin{aligned} & \sqrt{5}-i \sqrt{15}=r(\cos \theta+i \sin \theta) \\\\ & \begin{aligned} \Rightarrow \quad r & =\sqrt{(\sqrt{5})^2+(-\sqrt{15})^2} \\\\ & =\sqrt{5+15}=2 \sqrt{5} \end{aligned} \\\\ & \text { and } \theta=\tan ^{-1}\left(\frac{\text { Imaginary part }}{\text { Real part }}\right) \\\\ & =\tan ^{-1}\left(\frac{-\sqrt{15}}{\sqrt{5}}\right)=\tan ^{-1}(-\sqrt{3}) \end{aligned} $

$\begin{aligned} & \Rightarrow \quad \theta=-\frac{\pi}{3} \\\\ & \text { So, } r^2\left(\sec \theta+3 \operatorname{cosec}^2 \theta\right) \\\\ & =(2 \sqrt{5})^2\left[\sec \left(-\frac{\pi}{3}\right)+3 \operatorname{cosec}^2\left(-\frac{\pi}{3}\right)\right] \\\\ & =20\left[2+3 \times \frac{4}{3}\right]=20 \times 6=120\end{aligned}$

We have, $z=x+i y$

$ \begin{aligned} \therefore \frac{2 z-i}{z-2}= & \frac{2(x+i y)-i}{x+i y-2} \\\\ & =\frac{2 x+(2 y-1) i}{x-2+i y} \\\\ & =\frac{2 x+(2 y-1) i}{(x-2)+y i} \times\left[\frac{(x-2)-y i}{(x-2)-y i}\right] \\\\ & 2 x(x-2)+y(2 y-1) \\\\ & =\frac{+[(x-2)(2 y-1)-2 x y] i}{(x-2)^2+y^2} \\\\ = & \frac{2 x^2-4 x+2 y^2-y}{(x-2)^2+y^2} \\\\ & \quad+\frac{[(x-2)(2 y-1)-2 x y] i}{(x-2)^2+y^2} \end{aligned} $

$\because \frac{2 z-i}{z-2}$ is purely real number

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

Given $x$ and $y$ are two complex number with $|x|=|y|=1$ and $x=e^{i 2 \alpha}$

$ \begin{aligned} y & =e^{i .3 \beta}, \text { where } \alpha+\beta=\frac{\pi}{36} \\\\ \Rightarrow \quad x^6 & =\left(e^{i 2 \alpha}\right)^6=e^{i .12 \alpha} \\\\ y^4 & =\left(e^{i .3 \beta}\right)^4=e^{i .12 \beta} \end{aligned} $

So, $x^6 \cdot y^4=e^{i .12 \alpha} \cdot e^{i, 12 \beta}=e^{i .12(\alpha+\beta)}$

Since, $\alpha+\beta=\frac{\pi}{36}$

$ \Rightarrow \quad x^6 \cdot y^4=e^{i .12 \frac{\pi}{36}=e^{i \frac{\pi}{3}}} $

and $\frac{1}{x^6 \cdot y^4}=e^{-i \cdot \frac{\pi}{3}}$

$ \begin{aligned} \Rightarrow x^6 \cdot y^4+\frac{1}{x^6 \cdot y^4} & =e^{i \cdot \frac{\pi}{3}}+e^{-i \cdot \frac{\pi}{3}} \\\\ & =2 \cos \frac{\pi}{3}=1 \end{aligned} $

We have,

$ \begin{aligned} & x^{14}+x^9-x^5-1=0 \\\\ & \Rightarrow \quad\left(x^9-1\right)\left(x^5+1\right)=0 \\\\ & x^9=1 \\\\ & x=(1)^{1 / 9} \\\\ & x=\left(\cos 0^{\circ}\right. \\\\ & \left.+i \sin 0^{\circ}\right)^{1 / 9} \\\\ & =(\cos 2 K \pi \\\\ & +i \sin 2 K \pi)^{1 / 9} \\\\ & =\cos \frac{2 K \pi}{9}+i \sin \frac{2 K \pi}{9} \\\\ & \text { using Demoivre's theorm here, } K=0 \text {, } 1, \ldots ., 8 \\\\ & \begin{array}{l} x^5=-1 \\\\ x=(\cos \pi+i \sin \pi)^{1 / 5} \\\\ =[\cos (2 K \pi+\pi) \\\\ +i \sin (2 K \pi+\pi)]^{1 / 5} \\\\ =[\cos (2 K+1) \pi \\\\ +i \sin (2 K+1) \pi]^{1 / 5} \\\\ =\cos \frac{(2 K+1) \pi}{5} \\\\ +i \sin \frac{(2 K+1) \pi}{5} \\\\ \text { where } K=0,1, \ldots ., 5 \end{array} \end{aligned} $

So, one of root of the given equation is

$ \begin{aligned} & \text { So, one of root } \frac{2 \pi}{9}+i \sin \frac{2 \pi}{9} \text { for } K=1 \\\\ & =\frac{\sqrt{5}+1}{4}+i \frac{\sqrt{10-2 \sqrt{5}}}{4} \end{aligned} $

Let $\alpha=\omega$ and $\beta=\omega^2$

We know that

$ (a+b)\left(a \omega+b \omega^2\right)\left(a \omega^2+b \omega\right)=a^3+b^3 $

Now, check

$ \begin{aligned} & x+y+z=a(1+\alpha+\beta)+b(1+\alpha+\beta) \\ & \Rightarrow \quad x+y+z=0 \quad\left[\because 1+\omega+\omega^2=0\right] \\ & \Rightarrow \quad x^3+y^3+z^3=3 x y z=3\left(a^3+b^3\right) \end{aligned} $

$ \begin{aligned} & \text { Given, } z=\frac{3+2 i \cos \theta}{1-2 i \sin \theta} \times \frac{1+2 i \sin \theta}{1+2 i \sin \theta} \\ & =\frac{3-4 \sin \theta \cos \theta+6 i \sin \theta+2 i \cos \theta}{1+4 \sin ^2 \theta} \\ & \because \text { Purely imaginary } \Rightarrow \operatorname{Re}(z)=0 \\ & \frac{3-4 \sin \theta \cos \theta}{1+4 \sin ^2 \theta}=0 \Rightarrow 2 \sin 2 \theta=3 \\ & \Rightarrow \sin 2 \theta=\frac{3}{2}, \text { which is not possible. } \end{aligned} $

$ \begin{aligned} &\text { Given, } z=x+i y\\ &\begin{aligned} & \because \quad z \bar{z}^3+\bar{z} z^3=350 \\ & \Rightarrow \quad z \bar{z}\left(z^2+\bar{z}^2\right)=350 \\ & \Rightarrow\left(x^2+y^2\right)\left(x^2-y^2+2 i x y+x^2-y^2\right. \\ & -2 i x y)=350 \\ & \Rightarrow\left(x^2+y^2\right)\left(x^2-y^2\right)=175=25 \times 7 \\ & \Rightarrow\left(x^2+y^2\right)\left(x^2-y^2\right)=(16+9)(16-9) \\ & x^2=16, y^2=9 \\ & \text { Now, }|z|=\sqrt{x^2+y^2} \\ & =\sqrt{16+9}=5 \end{aligned} \end{aligned} $

Given, $\alpha$ and $\beta$ are the roots of equation $x^2+x+1=0$

$ \because x=\frac{-1 \pm i \sqrt{3}}{2}=\omega, \omega^2 $

$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} & (\alpha+\beta)^2+\left(\alpha^2+\beta^2\right)^2+\left(\alpha^3+\beta^3\right)^2 \\ & +\ldots .+\left(\alpha^{12}+\beta^{12}\right)^2 \\ & =\left(\omega+\omega^2\right)^2+\left(\omega^2+\omega^4\right)^2+\left(\omega^3+\omega^6\right)^2 \\ & \ldots\left(\omega^{12}+\omega^{24}\right)^2 \\ & =\left(\omega+\omega^2\right)^2+\left(\omega^2+\omega\right)^2 \\ & +(1+1)^2+\left(\omega+\omega^2\right) \ldots(1+1)^2 \\ & =8(-1)^2+4(1+1)^2 \\ & {\left[\because \omega+\omega^2=-1 \text { and } \omega^3=1\right]} \\ & =24 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { }\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^n=1 \\ & \Rightarrow\left(\frac{1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18}}{1+\cos \frac{5 \pi}{18}-i \sin \frac{5 \pi}{18}}\right)^n=1 \\ & {\left[\because \frac{2 \pi}{9}=\frac{\pi}{2}-\frac{5 \pi}{18} \text { and } \sin \left(\frac{\pi}{2}-\theta\right)=\cos \theta\right.} \\ & \left.\cos \left(\frac{\pi}{2}-\theta\right)=\sin \theta\right] \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow\left(\frac{2 \cos ^2 \frac{5 \pi}{36}+i 2 \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}}{2 \cos ^2 \frac{5 \pi}{36}-i 2 \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}}\right)^n=1 \\ & {\left[\because 1+\cos \theta=2 \cos ^2 \frac{\theta}{2}\right.} \text { and sin } \left.\theta=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right] \\ & \Rightarrow\left(\frac{\cos \frac{5 \pi}{36}+i \sin \frac{5 \pi}{36}}{\cos \frac{5 \pi}{36}-i \sin \frac{5 \pi}{36}}\right)^n=1 \\ & \{\text { multiplying with conjugate }\} \\ & \Rightarrow\left\{\frac{\left(\cos \frac{5 \pi}{36}+i \sin \frac{5 \pi}{36}\right)^2}{\cos ^2 \frac{5 \pi}{36}+\sin \frac{5 \pi}{36}}\right\}^n=1 \\ & \Rightarrow\left[\cos \frac{5 n \pi}{18}+i \sin \frac{5 n \pi}{18}\right]^n=1 \end{aligned}\\ &\text { So, least positive integral value of } n=18 \end{aligned} $

$ \begin{aligned} & P(x)=2 x^4+a x^3+b x^2+c x+d \\ & P(1)=1^2+3=4 \\ & P(2)=2^2+3=7 \\ & P(3)=3^2+3=12 \\ & P(4)=4^2+3=19 \end{aligned} $

$ \begin{aligned} & \text { Let } \\ & P(x)=2(x-1)(x-2)(x-3)(x-4)+x^2+3 \\ & \Rightarrow P(5)=2 \times 4 \times 3 \times 2 \times 1+5^2+3=76 \end{aligned} $

$ \begin{aligned} & \text { Given, } x^3+x^2+x+1=0 \\ & \Rightarrow \quad x^2(x+1)+1(x+1)=0 \\ & \Rightarrow \quad\left(x^2+1\right)(x+1)=0 \\ & x=-1, i,-i \end{aligned} $

$ \begin{aligned}(i) \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} & =\frac{1}{-1}+\frac{1}{i}+\frac{1}{-i} \\ & =-1-i+i=-1(\mathrm{a}) \end{aligned} $

$ \begin{array}{r}(ii) \alpha^3+\beta^3+\gamma^3=(-1)^3+(i)^3+(-i)^3 \\ =-1+(-i)+i=-1(\mathrm{a}) \end{array} $

$ \begin{gathered}(iii) \alpha^4+\beta^4+\gamma^4=(-1)^4+(i)^4+(-i)^4 \\ =1+1+1=3(\mathrm{~d}) \end{gathered} $

$ \text { } \begin{aligned} (iv) & (\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2 \\ = & (-1-i)^2+(2 i)^2+(-i+1)^2 \\ = & 2 i-4-2 i=-4 \text { (b) } \end{aligned} $