Complex Numbers

$ \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.

We have, $(-8-8 \sqrt{3} i)^{1 / 4}$

let, $z=-8-8 \sqrt{3} i$

$ |z|=\sqrt{(-8)^2+(-8 \sqrt{3})^2}=16 $

and $\theta=\tan ^{-1}\left(\frac{-8 \sqrt{3}}{-8}\right)=\pi / 3$

so, but the complex number lies in 3rd quadrant

So, $\theta=\pi+\pi / 3=4 \pi / 3$

then, $z=16 e^{i 4 \pi / 3}$

So, $z^{1 / 4}=(16)^{1 / 4} \cdot e^{i\left(\frac{4 \pi}{3}+2 \pi k\right) / 4}$

$ k=0,1,2,3, \ldots $

for $k=0$,

$ z^{1 / 4}=2 e^{i\left(\frac{4 \pi}{3}\right) / 4}=2 e^{i \frac{\pi}{3}}=1+i \sqrt{3} $

for $k=1$,

$ \begin{aligned} z^{1 / 4} & =2 e^{i\left(\frac{4 \pi}{3}+2 \pi\right) / 4}=2 e^{i\left(\frac{\pi}{3}+\frac{\pi}{2}\right)} \\ & =-\sqrt{3}+i \end{aligned} $

for $k=2$,

$ \begin{aligned} z^{1 / 4}=2 e^{i\left(\frac{4 \pi}{3}+4 \pi\right) / 4} & =2 e^{i\left(\pi+\frac{\pi}{3}\right)} \\ & =-1-i \sqrt{3} \end{aligned} $

for $k=3$,

$ \begin{aligned} z^{1 / 4} & =2 e^{i\left(\frac{4 \pi}{3}+6 \pi\right) / 4}=2 e^{i(\pi / 3+3 \pi / 2)} \\ & =\sqrt{3}-i \end{aligned} $

Hence, two values are $\sqrt{3}-i,-1-\sqrt{3} i$

$ \begin{aligned} &\\ &\begin{gathered} \text { Given, }|z w|=1 \Rightarrow z \bar{z} w \bar{w}=1 \\ \arg (z)-\arg (w)=\frac{\pi}{2} \\ \arg \left(\frac{z}{w}\right)=\frac{\pi}{2} \\ \frac{z}{w}=e^{i \frac{\pi}{2}} \\ z=w e^{i \frac{\pi}{2}} \\ z=i w \Rightarrow|z|=|w| \end{gathered} \end{aligned} $

$ \begin{aligned} &\text { Now, } \bar{z} w\\ &\begin{aligned} & =\bar{z} \cdot \frac{z}{i} \\ & =\frac{z \bar{z}}{i}(\because|z w|=1,|z||w|=1,|z|=1) \\ & =\frac{1}{i}=-i \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }|z|=1\\ &\begin{aligned} & \begin{array}{c} z=1-\bar{z} \text { and } \operatorname{Im}(z)>0 \\ \text { Let } z=x+i y \\ |z|=1 \Rightarrow x^2+y^2=1 \end{array} \\ & \text { } \begin{aligned} and\,\,z+\bar{z} & =1 \\ 2 x & =1 \\ x & =\frac{1}{2} \\ \therefore \quad y & = \pm \frac{\sqrt{3}}{2} \\ \therefore \quad z & =\frac{1}{2}+\frac{\sqrt{3}}{2} i \\ & \begin{aligned} & \arg (z)=\tan ^{-1}\left(\frac{\sqrt{3}}{\frac{2}{1}}\right) \\ & \quad \end{aligned} \\ & =\tan ^{-1} \sqrt{3}=\frac{\pi}{3} \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Given, }\left|a w_1+b w_2\right|=\left|a w_1-b w_2\right| \\ & \Rightarrow\left|a w_1+b w_2\right|^2=\left|a w_1-b w_2\right|^2 \\ & \Rightarrow\left(a w_1+b w_2\right)\left(a \bar{w}_1+b \bar{w}_2\right)=\left(a w_1-b w_2\right) \left(a \bar{w}_1-b \bar{w}_2\right) \\ & \Rightarrow a^2 w_1 \bar{w}_1+a b w_1 \bar{w}_2+a b \bar{w}_1 w_2+b^2 w_2 \bar{w}_2 \\ & =a^2 w_1 \bar{w}_1-a b w_1 \bar{w}_2-a b \bar{w}_1 w_2+b^2 w_2 \bar{w}_2 \\ & 2 a b\left(w_1 \bar{w}_2+\bar{w}_1 \bar{w}_2\right)=0 \Rightarrow w_1 \bar{w}_2=-\bar{w}_1 w_2 \\ & \Rightarrow \frac{w_1}{w_2}=-\left(\frac{\bar{w}_1}{w_2}\right) \\ & \Rightarrow \frac{w_1}{w_2}+\left(\frac{\bar{w}_1}{w_2}\right)=0 \end{aligned} $

$\therefore \frac{w_1}{w_2}$ is pure imaginary number.

$ \begin{aligned} &\\ &\begin{aligned} & \text { Given, } \sinh ^{-1} 2+\sinh ^{-1} 3=\alpha \\ & \Rightarrow \sin \left(\sinh ^{-1} 2+\sinh ^{-1} 3\right)=\sinh \alpha \\ & \Rightarrow \sin \left(\sinh ^{-1} 2\right) \cos \left(\sinh ^{-1} 3\right)+\cos \left(\sinh ^{-1} 2\right) \\ & \Rightarrow \sin \left(\sinh ^{-1} 3\right)=\sinh \alpha \\ & \Rightarrow \sinh \alpha=2 \cdot \sqrt{1+9}+\sqrt{1+4} \cdot 3 \\ & =2 \sqrt{10}+3 \sqrt{5} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Given, } x=3-2 \sqrt{3} i \\ & \Rightarrow \quad(x-3)=-2 \sqrt{3} i \end{aligned} $

Squaring both sides, we get

$ \begin{aligned} (x-3)^2 & =(-2 \sqrt{3} i)^2 \\ \Rightarrow x^2-6 x+9 & =-12 \Rightarrow x^2-6 x=-21 \end{aligned} $

Again, squaring both sides, we get

$ \left(x^2-6 x\right)^2=441 $

$ \begin{aligned} & \Rightarrow x^4-12 x^3+36 x^2=441 \\ & \text { Now, } x^4-12 x^3+54 x^2-108 x-54 \\ & \quad=x^4-12 x^3+36 x^2+18 x^2-108 x-54 \\ & \quad=441+18\left(x^2-6 x-3\right) \\ & \quad=441+18(-21-3) \\ & \quad=441-18(24) \\ & \quad=441-432=9 \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \qquad\left|z_1-z_2\right|=\sqrt{25-12 \sqrt{3}}=c \\ & \text { and } \frac{\left|z_1-z_3\right|}{\left|z_2-z_3\right|}=\frac{3}{4}=\frac{b}{a} \angle A C B=30^{\circ} \\ & \text { Let } b=3 k, a=4 k \\ & \because \cos C=\frac{b^2+a^2-c^2}{2 a b} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \frac{\sqrt{3}}{2}=\frac{9 k^2+16 k^2-25+12 \sqrt{3}}{24 k} \\ & \Rightarrow \quad 12 \sqrt{3} k=25 k^2-25+12 \sqrt{3} \\ & \Rightarrow \quad 25 k^2-12 \sqrt{3} k-25+12 \sqrt{3}=0 \\ & \Rightarrow \quad 25\left(k^2-1\right)-12 \sqrt{3}(k-1)=0 \\ & \Rightarrow \quad(k-1)(25(k+1)-12 \sqrt{3})=0 \\ & \Rightarrow \quad k=1 \text { or } 25(k+1)=12 \sqrt{3} \\ & \Rightarrow \quad k=\frac{12 \sqrt{3}-25}{25}<0 \end{aligned}\\ &\because k \text { cannot be negative. }\\ &\begin{aligned} & \therefore \quad k=1 \Rightarrow b=3, a=4 \\ & \therefore \text { Area of } \triangle A B C=\frac{1}{2} a b \sin C \\ & =\frac{1}{2} \times 3 \times 4 \times \sin 30^{\circ}=3 \end{aligned} \end{aligned} $

Given, $z=(1+i)^{3 / 4}$

$ \begin{aligned} & =(\sqrt{2})^{\frac{3}{4}}\left(e^{i \frac{\pi}{4}}\right)^{\frac{3}{4}}=2^{\frac{3}{8}}\left(e^{i \frac{3 \pi}{4}}\right)^{\frac{1}{4}} \\ & =2^{3 / 8} e^{\frac{i}{4}}\left(2 n \pi+\frac{3 \pi}{4}\right) ; n=0,1,2,3 \end{aligned} $

∴ Product of four values

$ \begin{aligned} & =\left(2^{3 / 8}\right)^4 e^{\frac{i}{4}}\left(\frac{3 \pi}{4}+2 \pi+\frac{3 \pi}{4}+4 \pi+\frac{3 \pi}{4}+6 \pi+\frac{3 \pi}{4}\right) \\ & \quad=2^{3 / 2}\left(e^i \frac{15 \pi}{4}\right)=2^{3 / 2}\left(e^i 4 \pi-\frac{\pi}{4}\right) \end{aligned} $

$ \begin{aligned} & =2^{3 / 2}\left(e^{i 4 \pi}\right) \cdot e^{i\left(-\frac{\pi}{4}\right)} \\ & =2^{3 / 2}(1)\left(\cos \left(\frac{-\pi}{4}\right)+i \sin \left(\frac{-\pi}{4}\right)\right) \\ & =2 \sqrt{2}\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)=2(1-i) \end{aligned} $

We note that the quotient $\frac{z-(2-i)}{z+(1+2 i)}$ is not defined if $z=-(1+2 i) \Leftrightarrow x=-1$ and $y=-2$

Since $x=x+i y$, then

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

Since, $\frac{z-(2-i)}{z+(1+2 i)}$ is purely imaginary,

So real part $=0$

$ \begin{aligned} & \therefore \frac{(x-2)(x+1)+(y+1)(y+2)}{(x+1)^2+(y+2)}=0 \\ & \Leftrightarrow x^2+x-2 x-2+y^2+2 y+y+2=0 \\ & x^2+y^2-x+3 y=0 \end{aligned} $

And $x \neq-1$ and $y \neq-2$

This is the locus of point $P$. This is a circle.

$ \begin{aligned} &\text { Given, }(\sqrt{3}-i)^n, n \in N\\ &\text { Modulus } \begin{aligned} |\sqrt{3}-1| & =\sqrt{(\sqrt{3})^2+(-1)^2} \\ & =\sqrt{3+1}=\sqrt{4}=2 \end{aligned} \end{aligned} $

$ \theta=\arg (\sqrt{3}-i)=\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-\pi}{6} $

Argument :

$ \theta=\arg (\sqrt{3}-i)=\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-\pi}{6} $

So, in polar form,

$ \begin{aligned} \sqrt{3}-i & =2 \cdot(\cos \theta+i \sin \theta) \\ & =2\left[\cos \left(\frac{-\pi}{6}\right)+\sin \left(\frac{-\pi}{6}\right)\right] \\ & =2\left[\cos \frac{\pi}{6}-i \sin \frac{\pi}{6}\right] \end{aligned} $

$ \begin{aligned} & \Rightarrow(\sqrt{3}-i)^n=\left\{2 \cdot\left[\cos \left(\frac{\pi}{6}\right)-i \sin \left(\frac{\pi}{6}\right)\right]\right\}^n \\ & =2^n \cdot\left[\cos \left(\frac{n \pi}{6}\right)-i \sin \left(\frac{n \pi}{6}\right)\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

But given that $(\sqrt{3}-1)^n=2^n\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii), we get

$ \begin{array}{ll} & 2^n\left[\cos \left(\frac{n \pi}{6}\right)-i \sin \left(\frac{n \pi}{6}\right)\right]=2^n \\ \Rightarrow & \cos \left(\frac{n \pi}{6}\right)-i \sin \left(\frac{n \pi}{6}\right)=1 \\ \Rightarrow & \cos \left(\frac{n \pi}{6}\right)=1 \text { and } \sin \left(-\frac{n \pi}{6}\right)=1 \\ \Rightarrow & -\frac{n \pi}{6}=2 k \pi \\ \Rightarrow & n=\frac{-12 k \pi}{\pi}=-12 k \end{array} $

Since, $n \in N$, for the smallest positive $n$, let $k=-1$

$ \Rightarrow \quad n=-12(-1)=12 $

$\therefore$ The least positive value of $n$ is 12 .

Let $z=1+\sqrt{5}+i \sqrt{10-2 \sqrt{5}}$

$ \begin{aligned} & \text { } \begin{aligned}So, |z| & =\sqrt{(1+\sqrt{5})^2+(\sqrt{10-2 \sqrt{5}})^2} \\ & =\sqrt{1+5+2 \sqrt{5}+10-2 \sqrt{5}} \\ & =\sqrt{16+2 \sqrt{5}-2 \sqrt{5}}=\sqrt{16}=4 \\ \text { And } \theta & =\arg (z)=\tan ^{-1}\left(\frac{\sqrt{10-2 \sqrt{5}}}{1+\sqrt{5}}\right) \\ & =\tan ^{-1}\left(\tan \left(\frac{\pi}{5}\right)\right)=\frac{\pi}{5} \end{aligned} \end{aligned} $

So, $z=|z| \cdot(\cos \theta+i \sin \theta)$

$ \begin{aligned} & =4\left[\cos \left(\frac{\pi}{5}\right)+i \sin \left(\frac{\pi}{5}\right)\right] \\ \therefore z^5 & =4^5\left[\cos \left(\frac{\pi}{5}\right)+i \sin \left(\frac{\pi}{5}\right)\right]^5 \\ & =1024\left[\cos \left(5 \cdot \frac{\pi}{5}\right)+i \sin \left(5 \cdot \frac{\pi}{5}\right)\right] \\ & =1024[\cos \pi+i \sin \pi] \\ & =1024[-1+0]=-1024 \end{aligned} $

$ \text { } \begin{aligned} \frac{z-1}{z-i} & =\frac{(x+i y)-1}{(x+i y)-i} \\ & =\frac{(x-1)+i y}{x+i(y-1)} \\ & =\frac{(x-1)+i y}{x+i(y-1)} \times \frac{(x-i(y-1))}{(x-i(y-1))} \end{aligned} $

$ =\frac{[x(x-1)+y(y-1)]+i[x y-(x-1)(y-1)]}{x^2+(y-1)^2} $

$\because$ The fraction is purely imaginary.

So, $x^2-x+y^2-y=0$ {real part should be zero)

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

Center of circle $(\alpha, \beta)=\left(\frac{1}{2}, \frac{1}{2}\right)$

and Radius $(r)=\frac{1}{\sqrt{2}}$

Thus, $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{(1 / 2)}{(1 / 2)}+\frac{(1 / 2)}{(1 / 2)}=2$

and $r=\frac{1}{\sqrt{2}}$

$ \Rightarrow r^2=\frac{1}{2} $

So, $4 r^2=4 \times \frac{1}{2}=2$

Hence, $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=4 r^2$

$\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^n=-1$

$\because$ Modulus of numerator $=\sqrt{(\sqrt{3})^2+1^2}$

$ =2 \theta=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6} $

and modulus of denominator $=2$

$ \begin{aligned} & \theta=\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{6} \\ & \begin{aligned} \therefore\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^n & =\left(\frac{2 \operatorname{cis}\left(\frac{\pi}{6}\right)}{2 \operatorname{cis}\left(-\frac{\pi}{6}\right)}\right)^n \\ & =\left[\operatorname{cis}\left(\frac{\pi}{6}-\left(-\frac{\pi}{6}\right)\right)\right]^n \\ & =\left[\operatorname{cis}\left(\frac{\pi}{3}\right)\right]^n=\operatorname{cis}\left(\frac{n \pi}{3}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { and } \operatorname{cis}\left(\frac{n \pi}{3}\right)=-1 \\ & =\operatorname{cis}(\pi) \\ & \therefore \quad \frac{n \pi}{3}=\pi \\ & \Rightarrow \quad n=3 \end{aligned} $

thus, $p=3$

Similarly, second expression

$ \begin{aligned} &\begin{aligned} & \left(\frac{1-\sqrt{3} i}{1+\sqrt{3} i}\right)^m=\operatorname{cis}\left(\frac{2 \pi}{3}\right) \\ \Rightarrow & {\left[\frac{2\left(\operatorname{cis}\left(-\frac{\pi}{3}\right)\right)}{2\left(\operatorname{cis}\left(\frac{\pi}{3}\right)\right)}\right]^m=\operatorname{cis}\left(\frac{2 \pi}{3}\right) } \\ \Rightarrow & {\left[\operatorname{cis}\left(-\frac{\pi}{3}-\frac{\pi}{3}\right)\right]^m=\operatorname{cis}\left(\frac{2 \pi}{3}\right) } \\ \Rightarrow & \operatorname{cis}\left(-\frac{2 \pi}{3} \cdot m\right)=\operatorname{cis}\left(\frac{2 \pi}{3}\right) \end{aligned}\\ &\text { Therefore, }-\frac{2 \pi}{3} \cdot m=\frac{2 \pi}{3} \end{aligned} $

$ \begin{array}{cc} \Rightarrow & -2 \pi m \equiv 2 \pi \cdot \bmod (6 \pi) \\ \Rightarrow & -2 m \equiv 2 \bmod (6) \\ \Rightarrow & 2 m \equiv-2 \equiv 4 \bmod 6 \\ \Rightarrow & m=2 \bmod 3 \end{array} $

$\therefore$ Least positive integer $=m=2$ thus, $q=2$

Hence, $\sqrt{q^2+p^2}=\sqrt{3^2+2^2}=\sqrt{13}$

$ \begin{aligned} & \text { } z^8-20 z^4+64=0 \\ & \Rightarrow \quad\left(z^4\right)^2-16 z^4-4 z^4+64=0 \\ & \Rightarrow \quad\left(z^4-16\right)\left(z^4-4\right)=0 \\ & \quad z^4=16 \text { or } z^4=4 \\ & \Rightarrow \quad z= \pm 2 \pm 2 i, \pm \sqrt{2}, \pm \sqrt{2} i \end{aligned} $

imaginary roots are $2 i,-2 i, \sqrt{2} i,-\sqrt{2} i$ sum of square the imaginary roots

$ \begin{aligned} & =(2 i)^2+(-2 i)^2+(\sqrt{2} i)^2+(-\sqrt{2} i)^2 \\ & =-4-4-2-2=-12 \end{aligned} $

$ \begin{aligned} & \text { Given }\left|z_1+z_2\right|^2=\left|z_1\right|^2+\left|z_2\right|^2 \\ & \Rightarrow\left|z_1\right|^2+\left|z_2\right|^2+z_1 \bar{z}_2+\bar{z}_1 z_2=\left|z_1\right|^2+\left|z_2\right|^2 \\ & \Rightarrow z_1 \bar{z}_2+\bar{z}_1 z_2=0 \Rightarrow \frac{z_1}{z_2}+\frac{\bar{z}_1}{\bar{z}_2}=0 \\ & \text { (on dividing by } z_2 \cdot \bar{z}_2 )\\ & \Rightarrow \frac{z_1}{z_2}+\left(\frac{\bar{z}_1}{z_2}\right)=0 \end{aligned} $

$\therefore \frac{z_1}{z_2}$ is purely imaginary i.e. its real pant

$ =0 $

Given, $1, \omega, \omega^2$ are cube roots of unity.

Also, we can write the general term $\left(t_n\right)$ of given expression

$ \begin{aligned} t_n & =(n-1)\left(n+\frac{1}{\omega}\right)\left(n+\frac{1}{\omega^2}\right) \\ & =(n-1)\left(n+\omega^2\right)(n+\omega) \\ & =(n-1)\left(n^2+n\left(\omega+\omega^2\right)+\omega^3\right) \\ & =(n-1)\left(n^2-n+1\right)\left[\because 1+\omega+\omega^2=0\right] \\ & =\left(n^3-2 n^2+2 n-1\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

Now, $\sum_{r=1}^n r=\frac{n(n+1)}{2}$,

$ \begin{aligned} & \sum_{r=1}^n r^2=\frac{n(n+1)(2 n+1)}{6} \\ & \sum_{r=1}^n r^3=\left(\frac{n(n+1)}{2}\right)^2 \end{aligned} $

So, we have;

$ \begin{aligned} & \begin{aligned} \sum_{n=2}^{11} n & =\left(\sum_{n=2}^{11} n-1\right) \\ & =\frac{11(12)}{2}-1=66-1=65 \end{aligned} \\ & \sum_{n=2}^{11} n^2-\left(\sum_{n=1}^{11} n^2\right)-1 \end{aligned} $

$ \begin{aligned} & \left.=\frac{11(12)(23)}{6}-1=506-1=505\right. \\ & \begin{aligned} \sum_{n=2}^{11} n^3 & =\left(\sum_{n=1}^{11} n^3\right)-1 \\ & =\left(\frac{11(12)}{2}\right)^2-1=(66)^2-1 \\ & =4356-=4355 \\ and\,\,\sum_{n=2}^{11} 1 & =10 \end{aligned} \end{aligned} $

So, final answer

$ \begin{aligned} & =4355-2(505)+2(65)-10 \\ & =4355-1010+130-10=3465 \end{aligned} $

$ \begin{aligned} & \text { }(1+\sqrt{3} i)^6-(\sqrt{3}+i)^6 \\ & =2^6 \cdot\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^6-\left(\frac{\sqrt{3}}{2}+\frac{1}{2} i\right)^6 2^6 \\ & =2^6\left[\left(\cos \frac{\pi}{3}+\sin \frac{\pi}{3} i\right)^6-\left(\cos \frac{\pi}{6}+\sin \frac{\pi}{6} i\right)\right] \end{aligned} $

(Using De-Moivre's theorem)

$ \begin{aligned} & =2^6[(\cos 2 \pi+\sin 2 \pi i)-(\cos \pi+\sin \pi i)] \\ & =64[(1+0)-(-1+0)] \\ & =64(1+1)=128 \end{aligned} $

$ \begin{aligned} &\text { } z=x+i y \text { and } x^2+y^2=1\\ &\begin{aligned} \frac{1+x+i y}{1+x-i y} & =\frac{((1+x)+i y)((1+x)+i y)}{(1+x)^2-i^2 y^2} \\ & =\frac{(1+x)^2+i^2 y^2+2(1+x) y i}{1+x^2+2 x+y^2} \\ & =\frac{1+x^2+2 x-y^2+2(1+x) y i}{2+2 x} \\ & =\frac{2 x^2+2 x+2(1+x) y i}{2(1+x)} \\ & =\frac{2 x(1+x)+2(1+x) y i}{2(1+x)} \\ & =x+i y=z \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \sqrt{3}-i=2\left(\cos \left(-\frac{\pi}{6}\right)+i \sin \left(-\frac{\pi}{6}\right)\right) \\ & =2 e^{-i \pi / 6} \\ & \Rightarrow(\sqrt{3}-i)^5=\left(2 e^{-\frac{i \pi}{6}}\right)^5=32 e^{-\frac{i 5 \pi}{6}} \end{aligned} $

The product of the root of $x^6=c$ is

$ (-1)^{6-1} c=-c $

The product of the root is $-32 e^{-\frac{i 5 \pi}{6}}$

$ \begin{aligned} & =-32\left(\cos \left(\frac{-5 \pi}{6}\right)+i \sin \left(-\frac{5 \pi}{6}\right)\right) \\ & =-32\left(-\frac{\sqrt{3}}{2}-\frac{1}{2} i\right) \\ & =16 \sqrt{3}+16 i \\ & =\frac{2^6}{\sqrt{3}-i} \end{aligned} $

$ \begin{aligned} & \text { }|z-1|+|z-5| \geq|z-1-z+5| \\ & \geq 4\left[\because\left|z_1\right|+\left|z_2\right| \geq\left|z_1-z_2\right|\right] \\ & \text { Minimum value }=4 \end{aligned} $

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

$ \begin{aligned} &\text { Squaring on both sides, we get }\\ &\begin{aligned} & \qquad \begin{aligned} & 4 x^2-16 x+16+4 y^2-24 y+36 \\ & =9 x^2-36 x+36+9 y^2+18 y+9 \\ \Rightarrow \quad & 5 x^2+5 y^2-20 x+42 y-7=0 \end{aligned} \\ & \text { Centre } \equiv\left(2, \frac{-21}{5}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } x^7=1 \\ & \Rightarrow x^7-1=0 \\ & \therefore z^7-1=0 \\ & (z-1)\left(1+z+z^2+z^3+z^4+z^5+z^6\right)=0 \\ & z \neq 1 \\ & 1+z+z^2+z^3+z^4+z^5+z^6=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now,

$ \begin{aligned} &\text { If } \cosh 2 x=199\\ &\begin{aligned} \operatorname{coth} x & =\sqrt{\frac{1+\cosh 2 x}{\cosh 2 x-1}}=\sqrt{\frac{200}{198}} \\ \Rightarrow \quad \operatorname{coth} 2 x & =\frac{10}{3 \sqrt{11}} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }|z|=1\\ &\begin{aligned} & \begin{aligned} & \because \text { Let } z=\cos \theta+i \sin \theta=e^{i \theta} \\ & \qquad \begin{aligned} 1+z^2 & =1+\cos 2 \theta+i \sin 2 \theta \\ & =2 \cos ^2 \theta+2 i \sin \theta \cos \theta \\ & =2 \cos \theta(\cos \theta+i \sin \theta)=2 \cos \theta \cdot e^{i \theta} \end{aligned} \\ & \text { Now, } \frac{1+z^2}{2 i z}=\frac{2 \cos \theta \cdot e^{i \theta}}{2 i e^{i \theta}}=\frac{\cos \theta}{i}=-i \cos \theta \end{aligned} \\ & \text { Now, Im }\left(\frac{1+z^2}{2 i z}\right)=-\cos \theta=-\operatorname{Re}(z) \end{aligned} \end{aligned} $

Given, $(3+4 i)^{2025}=5^{2023}(x+i y)$

Taking modulus on both sides, we get

$ \begin{aligned} 5^{2025} & =5^{2023} \sqrt{x^2+y^2} \\ \because \sqrt{x^2+y^2} & =5^2=25 \end{aligned} $

$ \begin{aligned} &\text { We have, }\left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^{2024}+\left(\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta+i \sin \theta}\right)^{2025}\\ &\begin{aligned} & =\left\{\frac{(\cos \theta+i \sin \theta)}{i(\cos \theta-i \sin \theta)}\right\}^{2024}+\left(\frac{2 \cos ^2 \frac{\theta}{2}+2 i \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^2 \frac{\theta}{2}+2 i \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)^{2025} \\ & =\left(\frac{e^{i \theta}}{e^{-i \theta}}\right)^{2024}+\frac{\left(2 \cos \frac{\theta}{2}\right)^{2025}\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)^{2025}}{\left(2 \sin \frac{\theta}{2}\right)^{2025}\left(\sin \frac{\theta}{2}+i \cos \frac{\theta}{2}\right)^{2025}} \\ & =\left(e^{2 i \theta}\right)^{2024}+\left(\cot \frac{\theta}{2}\right)^{2025} \frac{\left(e^{i \frac{\theta}{2}}\right)^{2025}}{i^{2025}\left(e^{-i \frac{\theta}{2}}\right)^{2025}} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\left(e^{2 i \theta}\right)^{2024}+\left(\cot \frac{\theta}{2}\right)^{2025} \frac{1}{i}\left(e^{i \theta}\right)^{2025} \\ & =(\cos 2 \theta+i \sin 2 \theta)^{2024}-\left(\cot \frac{\theta}{2}\right)^{2025} i(\cos \theta+i \sin \theta)^{2025} \end{aligned}\\ &\text { Now, put } \theta=\frac{\pi}{2}\\ &=(-1)^{2024}-i(i)^{2025}=1-i^2=1+1=2 \end{aligned} $

Given,

$ x^4-10 x^3+50 x^2-130 x+169=0 $

Sum of roots $2 a+2 b=10 \Rightarrow a+b=5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Product of roots $\left(a^2+b^2\right)\left(b^2+a^2\right)=169$

$ a^2+b^2=13\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

From, Eq.s (i) and (ii) $a b=6$

$ \frac{a^2+b^2}{a b}=\frac{13}{6} \Rightarrow \frac{a}{b}+\frac{b}{a}=\frac{13}{6} $

$ \begin{aligned} &\text { We have, } i=\sqrt{-1}\\ &\begin{aligned} & \sum_{n=2}^{30} i^n=i^2+i^3+\ldots+i^{30}=0+i^{30}=0-1 \\ & \sum_{n=30}^{65} i^{n+3}=i^{33}+i^{34}+\ldots+i^{68}=0 \\ & \because \sum_{n=2}^{30} i^n+\sum_{n=30}^{65} i^{n+3}=-1+0=-1 \end{aligned} \end{aligned} $

$z_1$ and $z_2$ are both $n$th roots of unity. This means each one can be written in the form $e^{\frac{i 2 \pi r}{n}}$, where $r$ is an integer.

Let $z_1 = e^{\frac{i 2\pi r_1}{n}}$ and $z_2 = e^{\frac{i 2\pi r_2}{n}}$ for some integers $r_1$ and $r_2$.

If the line segment joining $z_1$ and $z_2$ makes a right angle at the origin, this means the points $z_1$, $z_2$, and the origin form a right-angled triangle with the right angle at the origin.

This can only happen if $\frac{z_1}{z_2}$ is a purely imaginary number. A complex number is purely imaginary when its argument (angle) is $\frac{\pi}{2}$ or $\frac{3\pi}{2}$, or in general, $\frac{\pi}{2} + m\pi$ where $m$ is an integer.

We find: $ \frac{z_1}{z_2} = e^{\frac{i 2\pi}{n}(r_1-r_2)} $

For this to be purely imaginary, we need: $ \frac{2\pi}{n} (r_1 - r_2) = \frac{\pi}{2} + m\pi $

Divide both sides by $\pi$: $ \frac{2}{n} (r_1 - r_2) = \frac{2m + 1}{2} $

Now solve for $n$: $ n = \frac{4(r_1 - r_2)}{2m + 1} $

Since $n$ must be a positive integer, $2m+1$ must divide evenly into $4(r_1 - r_2)$. The simplest way for this to always work for integer values is when $n$ is a multiple of $4$. Therefore, $n = 4k$ for some positive integer $k$.

Let, $Z=\sqrt{\sqrt{2}+1}+i \sqrt{\sqrt{2}-1}$

$ \begin{aligned} & \Rightarrow \quad|Z|=\sqrt{\sqrt{2}+1+\sqrt{2}-1}=\sqrt{2 \sqrt{2}} \\ & \quad \cos \theta=\frac{\sqrt{\sqrt{2}+1}}{\sqrt{2 \sqrt{2}}}, \sin \theta=\frac{\sqrt{\sqrt{2}-1}}{\sqrt{2 \sqrt{2}}} \\ & \therefore \quad \cos ^2 \theta=\frac{\sqrt{2}+1}{2 \sqrt{2}}, \sin ^2 \theta=\frac{\sqrt{2}-1}{2 \sqrt{2}} \\ & \Rightarrow \frac{1+\cos 2 \theta}{2}=\frac{\sqrt{2}+1}{2 \sqrt{2}} \\ & \Rightarrow \cos 2 \theta=\frac{1}{\sqrt{2}} \end{aligned} $

$\Rightarrow 2 \theta=\frac{\pi}{4}$ (since, real and imaginary part are positive.)

Now,

$ Z^8=(\sqrt{2 \sqrt{2}})^8\left(\cos \left(8 \times \frac{\pi}{8}\right)+i \sin \left(8 \times \frac{\pi}{8}\right)\right) $

$ \begin{aligned} & =(2 \sqrt{2})^4(\cos \pi+i \sin \pi) \\ & =\left(2^{\frac{3}{2}} \times 4\right)(-1+0) \\ & =2^6(-1)=-64 \end{aligned} $

Let $\alpha$ be a purely imaginary root of $a z^2+z+1=0$

Then,   $ a \alpha^2+\alpha+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Taking conjugate on both sides, we get

$ \begin{array}{ll} & \bar{a}(\bar{\alpha})^2+\bar{\alpha}+1=0 \\ \Rightarrow & \bar{a} \alpha^2-\alpha+1=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \,\,\,\,\,\,\,\,\,[\because \bar{\alpha}=-\alpha]\end{array} $

Solving, Eqs. (i) and (ii) by cross-multiplication, we get

$ \begin{array}{rlrl} \frac{\alpha^2}{1+1} & =\frac{\alpha}{\bar{a}-a}=\frac{1}{-a-\bar{a}} \\ \Rightarrow & \alpha^2 =\frac{-2}{a+\bar{a}} \text { and } \alpha=\frac{\bar{a}-a}{-a-\bar{a}} \end{array} $

Now, $\alpha^2=(\alpha)^2$

$ \begin{aligned} & \Rightarrow \frac{-2}{a+\bar{a}}=\left(\frac{\bar{a}-a}{-a-\bar{a}}\right)^2=\frac{(\bar{a}-a)^2}{(a+\bar{a})^2} \\ & \Rightarrow-2(a+\bar{a})=(\bar{a}-a)^2 \\ & \Rightarrow(\bar{a}-a)^2+2(\bar{a}+a)=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Now, $|a|=1 \Rightarrow a=\cos \theta+i \sin \theta$

Put the value of $a$ in Eq. (iii), we get

$ \begin{aligned} & (-2 i \sin \theta)^2+4 \cos \theta=0 \\ \Rightarrow & -4 \sin ^2 \theta+4 \cos \theta=0 \\ \Rightarrow & \cos ^2 \theta+\cos \theta-1=0 \\ \Rightarrow & \cos \theta=\frac{\sqrt{5}-1}{2}, \frac{-\sqrt{5}-1}{2} \quad[\because \cos \theta \in[-1,1]] \\ \Rightarrow & \quad \theta=\cos ^{-1}\left(\frac{\sqrt{5}-1}{2}\right) \end{aligned} $

Hence, $a=\cos \left\{\cos ^{-1}\left(\frac{\sqrt{5}-1}{2}\right)\right\}$$ +i \sin \left\{\cos ^{-1}\left(\frac{\sqrt{5}-1}{2}\right)\right\} $