Complex Numbers

$ \text { Given, } i=\sqrt{-1} $

$ \begin{aligned} \text { Consider, } z= & \frac{(1+i)^{2025}}{(1-i)^{2022}}=\frac{(1+i)^{2022}}{(1-i)^{2022}}(1+i)^3 \\ & =\frac{\left[(1+i)^2\right]^{1011}(1+i)^3}{\left[(1-i)^2\right]^{1011}} \\ & =\frac{[1-1+2 i]^{1011}}{[1-1-2 i]^{1011}}(1+i)^3 \\ & =\left(\frac{2 i}{-2 i}\right)^{1011}(1+i)^2(1+i) \\ & =(-1)^{1011}[1-1+2 i](1+i) \\ & =-1(2 i)(1+i)=-\left[2 i+2(i)^2\right] \\ & =-2 i-(-2)=2-2 i \end{aligned} $

We have,

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

Squaring on both sides, we get

$ \begin{gathered} |x+(y-1) i|^2=(2|x+(y+1) i|)^2 \\ x^2+(y-1)^2=4\left(x^2+(y+1)^2\right) \\ x^2+y^2+1-2 y=4 x^2+4 y^2+4+8 y \\ 3 x^2+3 y^2+8 y+2 y+4-1=0 \\ 3 x^2+3 y^2+10 y+3=0 \end{gathered} $

Hence, the locus of $z$ is

$ 3 x^2+3 y^2+10 y+3=0 $

$ \begin{aligned} &\text { Given, } x_n=\cos \frac{\pi}{2^n}+i \sin \frac{\pi}{2^n}\\ &\prod_{n=1}^{\infty} x_n=x_1 \cdot x_2 \cdot x_3 \ldots \end{aligned} $

$ \begin{aligned} & =\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) \left(\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}\right) \\ & =\cos \left(\frac{\pi}{2}+\frac{\pi}{4}+\frac{\pi}{8}+\ldots\right)+i \sin \left(\frac{\pi}{2}+\frac{\pi}{4}+\frac{\pi}{8}+\ldots\right) \end{aligned} $

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

Given, the roots of the equation $z^2-i=0$ are $\alpha$ and $\beta$.

Now, $z^2-i=0$

$ \begin{array}{ll} \Rightarrow & z^2=i \\ \Rightarrow & z= \pm \sqrt{i} \\ \Rightarrow & z= \pm \sqrt{\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}} \\ \Rightarrow & z= \pm\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)^{1 / 2} \\ \Rightarrow & z= \pm\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) \end{array} $

$ \begin{aligned} &\text { Thus, } \alpha=\cos \frac{\pi}{4}+i \sin \frac{\pi}{4} \text { and }\\ &\begin{aligned} & \beta=-\cos \frac{\pi}{4}-i \sin \frac{\pi}{4} \\ & \alpha=\cos \frac{\pi}{4}+i \sin \frac{\pi}{4} \text { and } \\ & \beta=\cos \frac{5 \pi}{4}+i \sin \frac{5 \pi}{4} \\ & \therefore|(\arg \beta-\arg \alpha)|=\left|\frac{5 \pi}{4}-\frac{\pi}{4}\right|=\left|\frac{4 \pi}{4}\right|=\pi \end{aligned} \end{aligned} $

Given, $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+2 x^2-x-2=0$

On comparing the given equation with $a x^3+b x^2+c x+d=0$, we get

$ \begin{aligned} & a=1, b=2, c=-1 \text { and } d=-2 \\ & \therefore \alpha+\beta+\gamma=-\frac{b}{a}=-2 \\ & \alpha \beta \gamma=\frac{-d}{a}=2 \text { and } \alpha \beta+\beta \gamma+\alpha \gamma=\frac{c}{a}=-1 \end{aligned} $

Now, $(\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2$

$ \begin{aligned} & \quad(\alpha \beta+\beta \gamma+\alpha \gamma) \\ \Rightarrow & \quad(-2)^2=\alpha^2+\beta^2+\gamma^2+2(-1) \\ \Rightarrow & \alpha^2+\beta^2+\gamma^2=4+2=6 \\ \therefore & \left(\alpha^2+\beta^2+\gamma^2\right)^3=(6)^3 \\ \Rightarrow & \left(\alpha^6+\beta^6+\gamma^6\right)+3\left[\left(\alpha^2+\beta^2+\gamma^2\right)\right. \\ & \left.\quad\left(\alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2\right)-\alpha^2 \beta^2 \gamma^2\right]=216 \\ \Rightarrow & \quad \alpha^6+\beta^6+\gamma^6+3[(6)(9)-4]=216 \\ & \quad\left[\because \alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2=9\right] \\ \Rightarrow & \alpha^6+\beta^6+\gamma^6=216-150 \\ \Rightarrow & \alpha^6+\beta^6+\gamma^6=66 \end{aligned} $

Given,

$ \begin{aligned} & \frac{3 x+2}{(x+1)\left(2 x^2+3\right)}=\frac{A}{x+1}+\frac{B x+C}{2 x^2+3} \\ & \Rightarrow 3 x+2=A\left(2 x^2+3\right) \\ & +(B x+C)(x+1) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

On putting $x=-1$ into the Eq. (i), we get

$ \begin{aligned} 3(-1)+2=A\left[2(-1)^2+3\right] & \\ & +(B(-1)+C)(-1+1) \\ -1=5 A \Rightarrow A= & -\frac{1}{5} \end{aligned} $

On putting $x=0$ and $A=-\frac{1}{5}$ into the

Eq. (i), we get

$ \begin{aligned} & 3(0)+2=-\frac{1}{5}\left(2(0)^2+3\right)+(B(0)+C)(0+1) \\ & \Rightarrow \quad 2=-\frac{3}{5}+C \Rightarrow C=\frac{13}{5} \end{aligned} $

On putting $x=1, A=-\frac{1}{5}$ and $C=\frac{13}{5}$ into the Eq. (i), we get

$ \begin{aligned} & 3(1)+2=-\frac{1}{5}\left(2(1)^2+3\right)+\left(B(1)+\frac{13}{5}\right)(1+1) \\ & \Rightarrow \quad 5=-1+2 B+\frac{26}{5} \Rightarrow 2 B=6-\frac{26}{5} \\ & \Rightarrow \quad 2 B=\frac{4}{5} \\ & \Rightarrow \quad B=\frac{2}{5} \\ & \therefore A-B+C=-\frac{1}{5}-\frac{2}{5}+\frac{13}{5}=\frac{10}{5}=2 \end{aligned} $

$ \begin{aligned} &\text { We know that, }\\ &\begin{aligned} & \sinh ^{-1} x=\log \left(x+\sqrt{x^2+1}\right) \\ & \text { So, } \sinh ^{-1} y=\log \left(y+\sqrt{y^2+1}\right) \\ & \Rightarrow \sinh ^{-1} y=x \\ & \quad \sinh x=y \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Here, }\\ &\begin{aligned} 2^{2022}\left(\frac{1}{2}\right. & \left.+\frac{\sqrt{3}}{2} i\right)^{2022}-2^{2022}\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{2022} \\ & =2^{2022} e^{i\left(\frac{\pi}{3}\right)(2022)}-2^{2022} e^{i\left(-\frac{\pi}{6}\right)(2022)} \\ & =2^{2022} e^{i(674) \pi}-2^{2022} e^{i(-337) \pi} \\ & =2^{2022}(1+0 i)-2^{2022}(-1+0 i) \\ & =2^{2022}+2^{2022}=2^{2023} \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Given, }\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^4+\left(\frac{\sqrt{3}-i}{\sqrt{3}+i}\right)^4=r \text { cis } \theta \\ & \Rightarrow \quad \frac{(\sqrt{3}+i)^8+(\sqrt{3}-i)^8}{[(\sqrt{3}-i)(\sqrt{3}+i)]^4}=r \text { cis } \theta \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{\left(2 \operatorname{cis} \frac{\pi}{6}\right)^8+\left(2 \operatorname{cis}\left(-\frac{\pi}{6}\right)\right)^8}{2^8}=r \operatorname{cis} \theta \\ & \begin{aligned} & \Rightarrow \operatorname{cis} \frac{8 \pi}{6}+\operatorname{cis}\left(-\frac{8 \pi}{6}\right)=r \operatorname{cis} \theta \\ & \Rightarrow \cos \frac{8 \pi}{6}+i \sin \frac{8 \pi}{6}+\cos \left(-\frac{8 \pi}{6}\right) +i \sin \left(-\frac{8 \pi}{6}\right)=r \operatorname{cis} \theta \end{aligned} \\ & \Rightarrow 2 \cos \left(\frac{8 \pi}{6}\right)=r \operatorname{cis} \theta \Rightarrow 2\left(-\frac{1}{2}\right)=r \operatorname{cis} \theta \\ & \Rightarrow-1=r \operatorname{cis} \theta \Rightarrow \pm \sqrt{-1}=\sqrt{r \operatorname{cis} \theta} \\ & \Rightarrow \sqrt{r \operatorname{cis} \theta}= \pm i \\ & \Rightarrow \sqrt{r \operatorname{cis} \theta}=\operatorname{cis}\left(\frac{\pi}{2}\right), \operatorname{cis}\left(\frac{3 \pi}{2}\right) \end{aligned} $

$ \begin{aligned} &\text { Given equation, }\\ &\begin{aligned} & \quad|z-2|+|z-2 i|=4 \\ & \Rightarrow \quad|(x+i y)-2|+|(x+i y)-2 i|=4 \\ & \Rightarrow \sqrt{(x-2)^2+y^2}+\sqrt{x^2+(y-2)^2}=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \text { Let, } l=(x-2)^2+y^2=x^2+y^2-4 x+4 \\ & \text { and } m=x^2+(y-2)^2=x^2+y^2-4 y+4 \end{aligned} \end{aligned} $

Eq. (i), can be written as

$ \sqrt{l}+\sqrt{m}=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, $l-m=4 y-4 x$

$ \begin{aligned} \Rightarrow & & (\sqrt{l}+\sqrt{m})(\sqrt{l}-\sqrt{m}) & =4(y-x) \\ \Rightarrow & & 4(\sqrt{l}-\sqrt{m}) & =4(y-x) \\ & \Rightarrow & (\sqrt{l}-\sqrt{m}) & =y-x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

On adding Eqs. (ii) and (iii), we get

$ 2 \sqrt{1}=y-x+4 $

On squaring both sides, we get

$ \begin{gathered} 4 l=(y-x+4)^2 \\ \Rightarrow 4\left(x^2+y^2-4 x+4\right)=y^2+x^2+4^2 -2 x y-8 x+8 y \\ \Rightarrow 3 x^2+2 x y+3 y^2-8 x-8 y=0 \end{gathered} $

$ \begin{aligned} & \text { Given, }(\sqrt{3}-i)^{2 / 5}=\left[2\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)\right]^{2 / 5} \\ & =2^{2 / 5}\left[\cos \left(-\frac{\pi}{6}\right)+i \sin \left(-\frac{\pi}{6}\right)\right]^{2 / 5} \\ & =2^{2 / 5}\left[\cos \left(-\frac{\pi}{6}+6 \pi\right)+i \sin \left(-\frac{\pi}{6}+6 \pi\right)\right]^{2 / 5} \\ & =2^{2 / 5}\left[\cos \left(\frac{35 \pi}{6}\right)+i \sin \left(\frac{35 \pi}{6}\right)\right]^{2 / 5} \\ & =2^{2 / 5}\left[\left(\cos \frac{35 \pi}{6} \times \frac{2}{5}\right)+i \sin \left(\frac{35 \pi}{6} \times \frac{2}{5}\right)\right] \\ & =2^{2 / 5}\left[\cos \left(\frac{7 \pi}{3}\right)+i \sin \left(\frac{7 \pi}{3}\right)\right] \\ & =2^{2 / 5}\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)=2^{2 / 5}\left[\frac{1}{2}+i\left(\frac{\sqrt{3}}{2}\right)\right] \\ & =\frac{2^{2 / 5}}{2}(1+\sqrt{3} i)=2^{-3 / 5}(1+\sqrt{3 i}) \end{aligned} $

Given, $\alpha, \beta, \gamma$ and $\delta$ are the roots of the equation $x^4+x^2+1=0$

Also, given that, $\alpha+\beta=-1, \gamma+\delta=1$, $\alpha^2=\beta$ and $\gamma^2=-\delta$

We know that, roots of the equation $x^4+x^2+1=0$ are $\pm \omega^{1 / 2}$ and $\pm \omega$

Now, it can be conclude that

$ \begin{aligned} & \alpha=-\omega^{1 / 2}, \beta=\omega, \gamma=\omega^{1 / 2} \text { and } \delta=-\omega \\ & \therefore \alpha^{2023}+\beta^{2023}+\gamma^{2022}+\delta^{2022} \\ & =\left(-\omega^{1 / 2}\right)^{2023}+(\omega)^{2023}+\left(\omega^{1 / 2}\right)^{2022}+(-\omega)^{2022} \\ & =-\omega^{1011} \cdot \omega^{1 / 2}+\omega+1+1 \\ & \begin{array}{l} =-(1) \cdot \omega^{1 / 2}+2+\omega=-\omega^{1 / 2}+2+\omega \\ =\omega^2+\omega+2 \quad\left[\because-\omega^{1 / 2}=\omega^2\right] \\ =1 \quad\left[\because 1+\omega+\omega^2=0\right]\\ \\ \end{array} \end{aligned} $

$ \begin{aligned} & (x-2)(x+3)(2 x-1)=0 \\ & \alpha=2, \beta=-3 \text { and } \gamma=\frac{1}{2} \\ & \therefore \alpha^3+\beta^3+\gamma^3=8-27+\frac{1}{8} \\ & \quad=\frac{64-216+1}{8}=-\frac{151}{8} \end{aligned} $

Given, $\alpha, \beta$ and $\gamma$ are the real roots of the equation $18 x^3-15 x^2-4 x+4=0$ such that $\alpha=\beta$ and $\alpha>\gamma$

So, $\alpha+\beta+\gamma=-\frac{(-19)}{18} \Rightarrow 2 \alpha+\gamma=\frac{5}{6}$

$ \begin{array}{ll} \text { and } & \alpha \beta+\beta \gamma+\gamma \alpha \\ \Rightarrow & \alpha^2+2 \alpha \gamma=-\frac{4}{18} \\ \Rightarrow & \alpha^2+2 \alpha\left(\frac{5}{6}-2 \alpha\right)=-\frac{2}{9} \\ \Rightarrow & 27 \alpha^2-15 \alpha-2=0 \end{array} $

$ \Rightarrow \quad \alpha=-\frac{1}{9}, \frac{2}{3} $

If $\alpha=-\frac{1}{9}$, then $\gamma=\frac{5}{6}-2\left(-\frac{1}{9}\right)=\frac{19}{18}$

But $\alpha>\gamma$.

So, $\alpha \neq-\frac{1}{9}$.

If $\alpha=\frac{2}{3}$, then $\gamma=\frac{5}{6}-2\left(\frac{2}{3}\right)=-\frac{1}{2}$

$ \begin{aligned} \therefore \alpha+\beta^2 & +\gamma^3 \\ & =\frac{2}{3}+\left(\frac{2}{3}\right)^2+\left(-\frac{1}{2}\right)^3 \\ & =\frac{2}{3}+\frac{4}{9}-\frac{1}{8}=\frac{71}{72} \end{aligned} $

Given, $\alpha$ is a multiple root of the equation

$ x^5-6 x^4+11 x^3-2 x^2-12 x+8=0 $

The possible rational roots of this equation be $\pm 1, \pm 2, \pm 4$ and $\pm 8$

Let $f(x)=x^5-6 x^4+11 x^3-2 x^2-12 x+8$

$ \begin{aligned} &\text { Now, }\\ &\begin{array}{|c|c|c|c|} \hline \boldsymbol{x} & \boldsymbol{f}(\boldsymbol{x}) & \boldsymbol{x} & \boldsymbol{f}(\boldsymbol{x}) \\ \hline-1 & 0 & -4 & -3240 \\ \hline+1 & 0 & 4 & 120 \\ \hline-2 & -192 & -8 & -63000 \\ \hline+2 & 0 & 8 & 13608 \\ \hline \end{array} \end{aligned} $

Thus, $-1,1$ and 2 be three roots of

$ x^5-6 x^4+11 x^3-2 x^2-12 x+8=0 $

So, $(x+1)(x-1)(x-2)$ or $x^3-2 x^2-x+2$

be a factor of

$ \begin{aligned} & x^5-6 x^4+11 x^3-2 x^2-12 x+8 \\ & \text { Now, } x^5-6 x^4+11 x^3-2 x^2-12 x+8 \\ & \div x^3-2 x^2-x+2=x^2-4 x+4 \end{aligned} $

The roots of equation $x^2-4 x+4=0$ are 2, 2.

Thus, the roots of the equation

$ x^5-6 x^4+11 x^3-2 x^2-12 x+8=0 $

be, $-1,1,2,2$ and 2 .

So, $\alpha=2$

$ \begin{aligned} \therefore 3 \alpha^2-2 \alpha+1 & =3(2)^2-2(2)+1 \\ & =12-4+1=9 \end{aligned} $

$ \text { } \begin{aligned} & 3^{2023}=3^{4 \times 505+3}=3^3\left(3^4\right)^{505} \\ = & 27\left[(1+80)^{505}\right]=27(1+80 K) \\ = & 27+27 \times 80 K=11+16+16 \lambda \\ = & 11+16(\lambda+1) \end{aligned} $

∴ Required remainder $=11$

$ \sqrt{-5-12 i}= \pm\binom{\sqrt{\frac{\sqrt{(-5)^2+(-12)^2}+(-5)}{2}}-i}{\sqrt{\frac{\sqrt{(-5)^2+(-12)^2}-(-5)}{2}}} $

$ = \pm(\sqrt{4}-i \sqrt{9})= \pm(2-3 i) $

$ \begin{aligned} \sqrt{7+24 i}= \pm\left(\sqrt{\frac{\sqrt{7^2+24^2}+7}{2}}\right. & \left.+i \sqrt{\frac{\sqrt{7^2+24^2}-7}{2}}\right) \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} & = \pm(\sqrt{16}+i \sqrt{9})= \pm(4+3 i) \\ \because & \sqrt{-5-12 i}+\sqrt{7+24 i}=k \end{array}\\ &\text { (a negative real number) }\\ &\begin{array}{lrl} \Rightarrow & -(2-3 i)-(4+3 i) & =k \\ \Rightarrow & k & =-6 \end{array} \end{aligned} $

$\operatorname{amp}\left(\frac{z-3}{z+2 i}\right)=\frac{\pi}{2}$

$ \begin{aligned} & \Rightarrow \mathrm{amp}\left(\frac{(x-3)+i y}{x+i(y+2)} \times \frac{x-i(y+2)}{x-i(y+2)}\right)=\frac{\pi}{2} \\ & \Rightarrow \mathrm{amp} \\ & \begin{array}{r} \left(\frac{x(x-3)+y(y+2)+i(x y-(x-3)(y+2)}{x^2+(y+2)^2}\right) =\frac{\pi}{2} \end{array} \end{aligned} $

Here, $\operatorname{amp}(z)=\frac{\pi}{2}$ means $z$ is purely imaginary

$ \begin{array}{ll} \Rightarrow & \text { Real part }=0 \\ \therefore & x(x-3)+y(y+2)=0 \\ & x^2-3 x+y^2+2 y=0 \\ \Rightarrow & x^2+y^2-3 x+2 y=0 \end{array} $

which represents a circle containing the origin.

$ \begin{array}{r} \frac{z-(2+i)}{z+(1-2 i)}=\frac{(x-2)+i(y-1)}{(x+1)+i(y-2)} \times \frac{(x+1)-i(y-2)}{(x+1)-i(y-2)} \end{array} $

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

$ \begin{aligned} & \because \text { This is purely real. } \\ & \Rightarrow \text { Imaginary point }=0 \\ & \Rightarrow(x+1)(y-1)-(x-2)(y-2)=0 \\ & \qquad[x \neq-1, y \neq 2] \end{aligned} $

$ \begin{aligned} & \Rightarrow x y-x+y-1=x y-2 x-2 y+4 \\ & \Rightarrow-x+2 x+y+2 y-1-4=0 \\ & \Rightarrow x+3 y-5=0, \quad[x \neq-1, y \neq 2] \end{aligned} $

We know that $1, \omega$ and $\omega^2$ are cube roots of unit, where

$ \begin{aligned} & \omega=\frac{-1+i \sqrt{3}}{2} \\ & \omega^2=\frac{-1-i \sqrt{3}}{2} \end{aligned} $

and $\quad \omega^3=1$

$ \begin{array}{lrl} \therefore -2 \omega & =1-i \sqrt{3} \\ \text { and }-2 \omega^2 & =1+i \sqrt{3} \end{array} $

Now, $(1+i \sqrt{3})^{2023}+(1-\sqrt{3} i)^{2023}$

$ \begin{aligned} & =\left(-2 \omega^2\right)^{2023}+(-2 \omega)^{2023} \\ & =(-2)^{2023}\left(\omega^{4046}+\omega^{2023}\right) \\ & =(-2)^{2023}\left(\omega^2+\omega\right) \quad\left[\because \omega^{3 n}=1\right] \\ & =(-2)^{2023}(-1) \quad\left[\omega+\omega^2=-1\right] \\ & =2^{2023} \end{aligned} $

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

$ \begin{aligned} r & =|z|=\sqrt{3+1}=2 \\ \theta & =\arg (z)=-\tan ^{-1} \frac{1}{\sqrt{3}}=-\frac{\pi}{6} \\ \therefore \sqrt{3}-i & =2 \operatorname{cis}\left(-\frac{\pi}{6}\right) \\ \sqrt{3}-i & =2 \operatorname{cis}\left(2 n \pi-\frac{\pi}{6}\right) \\ (\sqrt{3}-i)^{1 / 6} & =2^{1 / 6} \operatorname{cis}\left(\frac{2 n \pi}{6}-\frac{\pi}{36}\right) \\ n & =0,1,2,3,4,5 \\ & =2^{1 / 6} \operatorname{cis} \frac{(12 n-1) \pi}{36} \end{aligned} $

For $n=5$ one of the value is $2^{1 / 6} \mathrm{cis}\left(\frac{59 \pi}{36}\right)$.

On comparing,

$ a x^2-x y-3 y^2-5 x+20 y+c=0 $

with $A x^2+2 H x y+B y^2+2 G x+2 F y+C=0$

we have,

$ \begin{aligned} & A=a, H=-\frac{1}{2}, B=-3 \\ & G=\frac{-5}{2}, F=10, C=c \end{aligned} $

Given, equation represents a pair of straight line, if

$ \begin{gathered} \Delta=\left|\begin{array}{rrr} A & H & G \\ H & B & F \\ G & F & C \end{array}\right|=0 \\ \Rightarrow \quad\left|\begin{array}{rrr} a & -\frac{1}{2} & -\frac{5}{2} \\ -\frac{1}{2} & -3 & 10 \\ -\frac{5}{2} & 10 & c \end{array}\right|=0 \\ \Rightarrow \quad-12 a c-400 a-c+175=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{gathered} $

Given, pair of line passing through $(2,3)$, then

$ 4 a+c+17=0 \Rightarrow c=-4 a-17 $

On substituting value of $c$ in Eq. (i), we get

$ \begin{aligned} & \quad a^2-4 a+4=0 \Rightarrow a=2 \\ & \therefore \quad c=-25 \\ & \text { Now, } a-c=27 \end{aligned} $

Given that

$ \left(z=\sin \left(\frac{6 \pi}{5}\right)+i\left(1+\cos \frac{6 \pi}{5}\right)\right) $

Thus, clearly $\left(\sin \frac{6 \pi}{5}, 1+\left(\cos \frac{6 \pi}{5}\right)\right)$ lies in the 2 nd quadrant of complex plane. Hence, its argument is

$ \begin{aligned} & \arg (z)=\pi-\tan ^{-1}\left|\left(\frac{y}{z}\right)\right| \\ & \quad\left(\because z=x+i y, \forall x<0 \text { and } y^{}\right. \geqslant 0)\end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\pi-\tan ^{-1}\left|\frac{1+\cos \frac{6 \pi}{5}}{\sin \frac{6 \pi}{5}}\right| \\ & =\pi-\tan ^{-1}\left|\frac{1-\cos \frac{\pi}{5}}{\sin \frac{\pi}{5}}\right| \\ & =\pi-\tan ^{-1}\left|\frac{2 \sin ^2 \frac{\pi}{10}}{2 \sin \frac{\pi}{10} \cos \frac{\pi}{10}}\right| \\ & =\pi-\tan ^{-1}\left|\frac{\sin \frac{\pi}{10}}{\cos \frac{\pi}{10}}\right| \\ & =\pi-\tan ^{-1}\left(\tan \frac{\pi}{10}\right) \\ & =\pi-\frac{\pi}{10}\left(\because-\frac{\pi}{2} \leq \tan ^{-1} x \leq \frac{1}{2}\right) \\ & =\frac{10 \pi-\pi}{10}=\frac{9 \pi}{10} \end{aligned}\\ &\text { Thus, } \arg (z)=\frac{9 \pi}{10} \end{aligned} $

Given,

$ \begin{aligned} & x+i y =\sqrt{\frac{3+i}{1+3 i}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow & \quad \overline{x+i y} =\sqrt{\frac{\overline{3+i}}{1+3 i}} \\ \Rightarrow & x-i y =\sqrt{\frac{3-i}{1-3 i}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

On multiplying Eqs. (i) and (ii), we get

$ \begin{aligned} & x^2+y^2=\sqrt{\frac{(3+i)(3-i)}{(1+3 i)(1-3 i)}}=\sqrt{\frac{9+1}{1+9}}=1 \\ \Rightarrow & \left(x^2+y^2\right)^2=(1)^2=1 \end{aligned} $

Now, Img $\left(\frac{2 z+1}{i z+1}\right)=-2$

Now, let $z=x+i y$

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

$ =\frac{\begin{array}{r} (2 x+1)(1-y)+i\left(2 y-2 y^2\right) -i\left(2 x^2+x\right)+2 x y \end{array}}{(1-y)^2+x^2} $

From Eq. (i), we get

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

which is a straight line.

If $ i = \sqrt{-1} $, then consider evaluating $ (1+i)^{10} + (1-i)^{10} $.

Solution

We know that $ i^2 = -1 $.

First, let's compute $ (1+i)^2 $ and $ (1-i)^2 $:

Calculate $ (1+i)^2 $:

$ (1+i)^2 = (1^2 + 2 \cdot 1 \cdot i + i^2) = 1 + 2i + i^2 = 1 + 2i - 1 = 2i $

Calculate $ (1-i)^2 $:

$ (1-i)^2 = (1^2 - 2 \cdot 1 \cdot i + i^2) = 1 - 2i + i^2 = 1 - 2i - 1 = -2i $

Now, raise each result to the 5th power:

$ \{(1+i)^2\}^5 = (2i)^5 $

$ (2i)^5 = 2^5 \cdot i^5 = 32 \cdot i^5 $

Since $ i^5 = i^4 \cdot i = 1 \cdot i = i $, it follows that:

$ (2i)^5 = 32i $

$ \{(1-i)^2\}^5 = (-2i)^5 $

$ (-2i)^5 = (-2)^5 \cdot i^5 = -32 \cdot i^5 $

Similarly, $ i^5 = i $, thus:

$ (-2i)^5 = -32i $

Finally, sum these results:

$ (1+i)^{10} + (1-i)^{10} = 32i - 32i = 0 $

Thus, the final answer is $ 0 $.

To solve the problem, we have the condition that the magnitude of the sum of two complex numbers, $ z_1 $ and $ z_2 $, equals the sum of their magnitudes:

$ \left| z_1 + z_2 \right| = \left| z_1 \right| + \left| z_2 \right| $

We start by squaring both sides of this equation:

$ \begin{aligned} & \left( \left| z_1 + z_2 \right| \right)^2 = \left( \left| z_1 \right| + \left| z_2 \right| \right)^2 \\ & \left| z_1 \right|^2 + \left| z_2 \right|^2 + 2 \left| z_1 \right| \left| z_2 \right| \cos(\theta_1 - \theta_2) = \left| z_1 \right|^2 + \left| z_2 \right|^2 + 2 \left| z_1 \right| \left| z_2 \right| \end{aligned} $

Upon simplifying, we have:

$ 2 \left| z_1 \right| \left| z_2 \right| \cos(\theta_1 - \theta_2) = 2 \left| z_1 \right| \left| z_2 \right| $

This implies:

$ \cos(\theta_1 - \theta_2) = 1 $

The condition $\cos(\theta_1 - \theta_2) = 1$ indicates that:

$ \theta_1 - \theta_2 = 0 $

Therefore, the difference in amplitude (or argument) between $ z_1 $ and $ z_2 $ is:

$ \arg(z_1) - \arg(z_2) = 0 $

Thus, the two complex numbers have the same direction or angle in the complex plane.

Given, $1+i^2+i^4+i^6+\ldots+i^{2024}$

$ \begin{aligned} & =1+\left(i^2+i^4+i^6+\ldots+1012 \text { terms }\right) \\ & =1+(-1+1-1+1-\ldots+1-1+1) \end{aligned} $

i.e. 506 terms will be +1 and 506 terms will be $-1=1+0=1$

To solve the problem, let's first express the given complex number:

$ z = \frac{1 + i \cos \theta}{1 - 2 i \cos \theta} $

To simplify this expression and ensure it is purely real, we'll multiply the numerator and denominator by the conjugate of the denominator:

$ z = \frac{(1 + i \cos \theta)(1 + 2 i \cos \theta)}{(1 - 2 i \cos \theta)(1 + 2 i \cos \theta)} $

Recall that multiplying by the conjugate will result in a real denominator:

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

Simplifying the denominator, since $(2i \cos \theta)^2 = -4 \cos^2 \theta$, we get:

$ 1 - (2i \cos \theta)^2 = 1 + 4 \cos^2 \theta $

Thus, the expression for $ z $ becomes:

$ z = \frac{1 + 3i \cos \theta - 2 \cos^2 \theta}{1 + 4 \cos^2 \theta} $

For $ z $ to be purely real, the imaginary part must be zero. The imaginary part is:

$ \operatorname{Im}(z) = \frac{3 \cos \theta}{1 + 4 \cos^2 \theta} = 0 $

This implies that $ \cos \theta = 0 $.

Substituting $ \theta = \frac{\pi}{2} $, we note:

$\cos \theta = 0$

$\sin \theta = 1$

Now, substitute these into the expression $ \cos^3 \theta + \sin^2 \theta + \cos \theta + 1 $:

$ \cos^3 \theta + \sin^2 \theta + \cos \theta + 1 = 0^3 + 1^2 + 0 + 1 = 2 $

Therefore, the final result is 2.

$T_{10}=(\cos \theta+i \sin \theta)^9$

Using De-Moivre theorem,

$ \begin{aligned} & \cos (9 \theta)+i \sin (9 \theta) \\ = & \cos \left(\frac{9 \pi}{6}\right)+i \sin \left(\frac{9 \pi}{6}\right) \\ = & \cos \left(\frac{3 \pi}{2}\right)+i \sin \left(\frac{3 \pi}{2}\right) \\ = & 0+i(-1) \\ = & -i \end{aligned} $

To find the minimum value of $|z|^2$ where $z=(\alpha+i\beta)(2+7i)$ is purely imaginary, follow these steps:

First, express $z$ in terms of real and imaginary components:

$ z = (\alpha + i \beta)(2 + 7i) = 2\alpha + 7\alpha i + 2\beta i - 7\beta = (2\alpha - 7\beta) + (7\alpha + 2\beta)i $

Since $z$ is purely imaginary, its real part must be zero:

$ 2\alpha - 7\beta = 0 \implies \alpha = \frac{7}{2} \beta $

Given that $\alpha$ and $\beta$ are integers, $\beta$ must be such that $\alpha$ is also an integer. Therefore, $\beta$ should be chosen so that $\frac{7}{2}\beta$ is an integer. This implies $\beta$ must be even, so assume $\beta = 2k$ for some integer $k$.

Substituting $\beta = 2k$ into $\alpha = \frac{7}{2}\beta$, we get $\alpha = 7k$.

Next, calculate $|z|^2$:

$ |z|^2 = (7\alpha + 2\beta)^2 = (7(7k) + 2(2k))^2 = (49k + 4k)^2 = (53k)^2 $

To minimize $|z|^2$, we minimize $k$. The smallest positive integer value for $k$ is 1. Therefore, when $k = 1$, $|z|^2 = 53^2 = 2809$.

Thus, the minimum value of $|z|^2$ is:

$ \boxed{2809} $

          $ x^2-2 x+2=0 $

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

$ \begin{aligned} &\text { }\\ &\begin{aligned}Taking\,\,\, \alpha & =1+i, \beta=1-i \\ \alpha^2 & =(1+i)^2=1+(i)^2+2 i \\ & =1-1+2 i=2 i \\ \beta^2 & =(1-i)^2=1+(i)^2-2 i \\ & =1-1-2 i=-2 i \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} Now, \alpha^{2020} & +\beta^{2020} \\ & =\left(\alpha^2\right)^{1010}+\left(\beta^2\right)^{1010} \\ & =(2 i)^{1010}+(-2 i)^{1010} \\ & =2^{1010}\left[i^{1010}+(-i)^{1010}\right] \\ & =2^{1010}\left[2\left(i^{1010}\right)\right]=2^{1011}(i)^2 \\ & =2^{1011}(-1)=-2^{1011} \end{aligned} $

$ \begin{aligned} z & =\frac{-1-i \sqrt{3}}{2} \\ z & =\omega^2 \cdot\left(z^k+\frac{1}{z^k}\right)^2=\left(\omega^{2 k}+\frac{1}{\omega^{2 k}}\right)^2 \\ & =\omega^{4 k}+\frac{1}{\omega^{4 k}}+2 \\ & =\omega^{4 k}+\frac{\omega^{2 k}}{\omega^{4 k} \cdot \omega^{2 k}}+2 \\ & =\omega^{4 k}+\frac{\omega^{2 k}}{\omega^{6 k}}+2 \\ & =\omega^{4 k}+\omega^{2 k}+2 \quad\left[\because \omega^{6 k}=\left(\omega^3\right)^{2 k}=(1)^{2 k}=1\right]\\ & =\omega^k+\omega^{2 k}+2\,\,\,\,\,\, \left[\because \omega^{3 k}=1\right]\end{aligned} $

$ \begin{aligned} & \text { } \begin{aligned}Now, \sum_{k=1}^{2022} & \omega^k+\omega^{2 k}+2 \\ = & {\left[\omega+\omega^2+\omega^3+\ldots .+\omega^{2022}\right] } \\ & +\left[\omega^2+\omega^4+\omega^6+\ldots+\omega^{4044}\right] \end{aligned} \\ & \begin{aligned} \frac{\omega \cdot\left(\omega^{2022}-1\right)}{(\omega-1)}+\frac{\omega^2\left(\left(\omega^2\right)^{2022}-1\right)}{\omega^2-1}+\sum_{k=1}^{2022} 2 \end{aligned} \\ & =\frac{\omega(1-1)}{(\omega-1)}+\frac{\omega^2(1-1)}{\omega^2-1}+2 \times(1+1+\ldots .+1) \\ & =0+0+2 \times 2022 \\ & =4044 \end{aligned} $

For $x \in[0,2 \pi]$

$\sin x+i \cos 2 x$ and $\cos x-i \sin 2 x$ are conjugate to each other.

$ \begin{aligned} \Rightarrow \quad \sin x+i \cos 2 x & =\overline{\cos x-i \sin 2 x} \\ \sin x+i \cos 2 x & =\cos x+i \sin 2 x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

On comparing both side of Eq. (i), we get

$ \begin{array}{rlrlrl} & & \sin x & =\cos x ; & \sin 2 x=\cos 2 x \\ \Rightarrow & & x & =\frac{\pi}{4}, \frac{5 \pi}{4} ; & \tan 2 x=1 \\ \Rightarrow & & 2 x & =\frac{\pi}{4}, \frac{5 \pi}{4}, \frac{9 \pi}{4}, \frac{13 \pi}{4} \\ \Rightarrow & & x & =\frac{\pi}{8}, \frac{5 \pi}{8}, \frac{9 \pi}{8}, \frac{13 \pi}{8} \end{array} $

But there is no common value of $x$ which satisfy both the condition. So, $x=\phi$

Given, $|x+i y|=\sqrt{x^2+y^2}$

$ \begin{aligned}Since,\,\ & (1-\sqrt{3} i)^9+(\sqrt{3}+i)^9 \\ & \quad=i^{-9}\left(\frac{1}{i}-\sqrt{3}\right)^9+(\sqrt{3}+i)^9 \\ & {\left[\because i^{-9}=\frac{1}{i^9}=\frac{1}{i^8 \cdot i}=\frac{1}{\left(i^2\right)^4 i}\right.} \left.=\frac{1}{(-1)^4 i}=\frac{1}{i}=-i\right] \end{aligned} $

$ \begin{aligned} Now,\,\,&\left|(1-\sqrt{3} i)^9+(\sqrt{3}+i)^9\right| \\ &=\left|(i+\sqrt{3})^9(i+1)\right| \\ &=|i+1|\left|(i+\sqrt{3})^9\right| \\ &=|i+1||i+\sqrt{3}|^9 \\ &=\left(\sqrt{1^2+1^2}\right)^{\left(\sqrt{1^2+(\sqrt{3})^2}\right)^9} \\ &=\sqrt{2}(\sqrt{4})^9 \\ &=2^{\frac{1}{2}} \cdot 2^9=2^{\frac{19}{2}} \end{aligned} $

If $x^3=1$

Then, $\quad x=1, \omega, \omega^2$ and $x^4=1$

$ \begin{array}{lr} \Rightarrow & x^4-1=0 \\ \Rightarrow & x \pm 1, \pm i \end{array} $

So, roots are $1, i,-1,-i$

i.e. $\quad \alpha=i$

$\left[\because 1, \alpha, \alpha^2, \alpha^3\right.$ are fourth roots of unity]

Now, $\alpha+\alpha \omega-\alpha^3 \omega^2=\alpha+\alpha \omega-\frac{\omega^2}{\alpha}$

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

$ \begin{aligned} &\text { Given, } z=\alpha+i \beta \text {, }\\ &\begin{array}{ll} |z|-z=1+2 i ;|z|=\sqrt{\alpha^2+\beta^2} \\ \sqrt{\alpha^2+\beta^2}-\alpha-i \beta=1+2 i \\ \Rightarrow \quad \beta=-2 \\ \text { and } \sqrt{\alpha^2+4}-\alpha=1 \end{array} \end{aligned} $

$ \begin{array}{lc} \Rightarrow & \alpha^2+4=(1+\alpha)^2 \\ & \quad[\because \text { on squaring both sides }] \\ \Rightarrow & \alpha^2+4=\alpha^2+1+2 \alpha \\ \Rightarrow & 2 \alpha=3 \\ \Rightarrow & \alpha=\frac{3}{2} \end{array} $

$ \quad \begin{aligned}\therefore\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, z \bar{z} & =|z|^2=\alpha^2+\beta^2 \\ & =\left(\frac{3}{2}\right)^2+(-2)^2 \\ & =\frac{9}{4}+4=\frac{25}{4} \end{aligned} $

Given, $i z^2-2(i+1) z+(2-i)=0$

$\therefore-i$ and $\alpha$ are the roots of above equation.

$ \begin{aligned} & & -i \times \alpha & =\frac{2-i}{i} \\ \Rightarrow & & \alpha & =\frac{2-i}{-i^2} \\ \Rightarrow & & \alpha & =2-i \end{aligned} $

Clearly, $\tan \theta=-\frac{1}{2}, \gamma=\sqrt{5}$

$ \sin \theta=\frac{-1}{\sqrt{5}}, \cos \theta=\frac{2}{\sqrt{5}} $

[ ∵ from the polar form]

$ \begin{aligned} & \cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta=\frac{4}{5}-\frac{1}{5}=\frac{3}{5} \\ & \cos 6 \theta=4 \cos ^3 2 \theta-3 \cos 2 \theta \\ &=4\left(\frac{3}{5}\right)^3-3 \times \frac{3}{5}=\frac{108}{125}-\frac{9}{5} \\ &=\frac{108-225}{125}=-\frac{117}{125} \\ & \therefore \quad 5^3 \cos 6 \theta=5^3 \times-\left(\frac{117}{125}\right)=-117 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } z^n=1 \\ & z^n-1=(z-1)\left(z-\alpha_1\right)\left(z-\alpha_2\right)\left(z-\alpha_3\right) \\ & \ldots\left(z-\alpha_{n-1}\right) \end{aligned} \end{aligned} $

Sum of $n$th roots of unity $=0$

$ \begin{aligned} 1+\alpha_1+\alpha_2+\alpha_3+\ldots+\alpha_{n-1} & =0 \\ \Rightarrow \quad \alpha_1+\alpha_2+\alpha_3+\ldots+\alpha_{n-1} & =-1 \end{aligned} $

Also, the coefficient of $z^{n-1}$ is zero, so sum of product of zeroes taken two at a time $=0$

$ \begin{aligned} \Rightarrow \alpha_1+\alpha_2+\ldots & +\alpha_{n-1} \\ & +\sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j=0 \end{aligned} $

$ \Rightarrow \,\,\,\,\, - 1 + \sum\limits_{1 \le i < j \le n - 1}^{} {} {a_i}{a_j} = 0$

$ \Rightarrow \,\,\,\,\,\sum\limits_{1 \le i < j \le n - 1}^{} {} {a_i}{a_j} = 1$

Given, 2- $i$ is one of the roots of the equation

$ x^4-9 x^3+31 x^2-49 x+30=0 $

As coefficient of $x^4, x^3, x^2, x$ and constant are real numbers and $2-i$ is one of the root, then $2+i$ is the another root of the equation.

$ \begin{aligned} & & (x-2-i)(x-2+i) & =0 \\ \Rightarrow & & (x-2)^2-i^2 & =0 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad x^2-4 x+4+1=0 \\ & \Rightarrow \quad x^2-4 x+5=0 \end{aligned}\\ &\text { Now, } \end{aligned} $

$ \begin{array}{ll} \therefore & x^2-5 x+6=0 \\ \Rightarrow & x^2-3 x-2 x+6=0 \\ \Rightarrow & x(x-3)-2(x-3)=0 \\ \Rightarrow & (x-2)(x-3)=0 \\ \Rightarrow & x=2,3 \\ \therefore & \alpha=2, \beta=3 \\ & 2 \alpha-\beta=2 \times 2-3=1 \end{array} $

Given, $e^{i t}=\cos t+i \sin t$ and $e^{-i t}=\cos t-i \sin t$

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

$ \begin{aligned} = & \frac{1}{2}\left[e^x[\cos y+i \sin y-\cos y+i \sin y]\right. \left.+e^{-x}[\cos y-i \sin y-\cos y-i \sin y]\right] \\ = & \frac{1}{2}\left[e^x(2 i \sin y)+e^{-x}[-2 i \sin y]\right. \\ = & 2 i \sin y\left[\frac{e^x-e^{-x}}{2}\right] \\ = & 2 i \sin y \sinh x \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &(2 x-y+1)+i(x-2 y-1)=2-3 i \end{aligned} $

$ \begin{array}{llrl} \Rightarrow & & 2 x-y+1 & =2 \\ \text { and } & & x-2 y-1 & =-3 \\ \Rightarrow & & 2 x-y & =1 \\ \text { and } & & x & =2 y-2 \end{array} $

Now, substitute value of $x=2 y-2$ in equation $2 x-y=1$, we get

$ \begin{aligned} & & 2(2 y-2)-y & =1 \\ \Rightarrow & & 4 y-4-y & =1 \end{aligned} $

$ \begin{aligned} &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3 y=5 \Rightarrow y=\frac{5}{3}\\ &\text { Now, substitute } y=\frac{5}{3} \text { in } x=2 y-2\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{10}{3}-2=\frac{4}{3} \end{aligned} $

$ \begin{aligned} &\text { Multiplicative inverse of }(x-i y)=\frac{1}{x-i y}\\ &\begin{aligned} & =\frac{1}{\frac{4}{3}-i \frac{5}{3}}=\frac{3}{4-5 i} \times \frac{(4+5 i)}{(4+5 i)} \\ & =\frac{12}{4^2+5^2}+\frac{15}{4^2+5^2} i=\frac{12}{41}+\frac{15}{41} i \end{aligned} \end{aligned} $

$ \begin{aligned} & (-1)=\cos \pi+i \sin \pi=\operatorname{cis} \pi \\ & \begin{aligned} (-1)^{1 / 4} & =(\cos \pi+i \sin \pi)^{1 / 4} \\ & =\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4} \\ & {\left[\because \cos \frac{\pi}{4} \text { is not }-(\mathrm{ve})\right] } \end{aligned} \end{aligned} $

$ \begin{aligned} & =\operatorname{cis} \frac{3 \pi}{4}(-i)^{1 / 2}=\left(\cos \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2}\right)^{1 / 2} \\ & =\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)=\operatorname{cis} \frac{3 \pi}{4} \\ \therefore & \alpha=\frac{3 \pi}{4} \Rightarrow \tan \alpha=\tan \frac{3 \pi}{4}=-1 \end{aligned} $

The roots of the given equations are $\sqrt{3}+\sqrt{2} i$ and $\sqrt{3}-\sqrt{2}$.

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

$ \begin{aligned} & \Rightarrow \quad x^2+3-2 \sqrt{3} x=-2 \Rightarrow x^2+5=2 \sqrt{3} x \\ & \Rightarrow \quad\left(x^2+5\right)^2=(2 \sqrt{3} x)^2 \\ & \Rightarrow \quad x^4+25+10 x^2=12 x^2 \\ & \Rightarrow \quad x^4-2 x^2+25=0 \end{aligned} $

$ \begin{array}{ll} \text { Also, } & x=\sqrt{3}-\sqrt{2} \\ \Rightarrow & x^2=(\sqrt{3}-\sqrt{2})^2 \\ \Rightarrow & x^2=3+2-2 \sqrt{6} \end{array} $

$ \begin{aligned} &\begin{array}{lr} \Rightarrow & x^2-5=-2 \sqrt{6} \\ \Rightarrow & \left(x^2-5\right)^2=(-2 \sqrt{6})^2 \\ \Rightarrow & x^4+25-10 x^2=24 \\ \Rightarrow & x^4-10 x^2+1=0 \end{array}\\ &\text { ∴ The required equation is }\\ &\left(x^4-2 x^2+25\right)\left(x^4-10 x^2+1\right)=0 \end{aligned} $

$ \begin{aligned} & \text {}\text { }\frac{x+i(x-2)}{3+i}-i=\frac{2 y+i(1-3 y)}{i-3} \\ & \Rightarrow \frac{x+(x-2) i}{3+i}+\frac{2 y+(1-3 y) i}{3-i}=i \end{aligned} $

$ \Rightarrow \frac{\{(4 x-2)+i(2 x-6)\}+\{(9 y-1)+i(3-7 y)\}}{(3+i)(3-i)}=i $

$ \Rightarrow(4 x+9 y-3)+i(2 x-7 y-3) $

$ =0+10 i $

On comparing the real and imaginary parts

$ \begin{aligned} & 4 x+9 y-3=0 \text { and } 2 x-7 y-3=10 \\ & \Rightarrow \quad 4 x+9 y=3 \\ & 2 x-7 y=13 \end{aligned} $

On subtracting Eq. (ii) $\times 2$ from Eq. (i),

From Eq. (i), $x=3$

$ \therefore \quad x+y=2 $

$ \begin{aligned} & z_1=e^{i \alpha}, z_2=e^{i \beta}, z_3=e^{i \gamma} \\ & \frac{1}{z_1}=e^{-i \alpha}, \frac{1}{z_2}=e^{-i \beta}, \frac{1}{z_3}=e^{-i \gamma} \end{aligned} $

According to the questions,

$ z_1+z_2+z_3=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Now, $\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}$

$ \begin{aligned} = & (\cos \alpha-i \sin \alpha)+(\cos \beta-i \sin \beta) +(\cos \gamma-i \sin \gamma) \\ = & \Sigma \cos \alpha-i \Sigma \sin \alpha=0-i \cdot 0=0 \\ \Rightarrow & z_1 z_2+z_2 z_3+z_3 z_1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & z_1^2+z_2^2+z_3^2 \\ = & \left(z_1+z_2+z_3\right)^2-2\left(z_1 z_2+z_2 z_3+z_3 z_1\right) \\ \Rightarrow & z_1^2+z_2^2+z_3^2=0-0 \end{aligned} $

[From Eqs. (i) and (ii)]

$ \begin{aligned} & \Rightarrow e^{i 2 \alpha}+e^{i 2 \beta}+e^{i 2 \gamma}=0 \\ & \Rightarrow \cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma=0 \\ & \text { and } \sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

Now, $z_1 z_2+z_2 z_3+z_3 z_1=0$

$ \begin{aligned} & \Rightarrow \quad e^{i(\alpha+\beta)}+e^{i(\beta+\gamma)}+e^{i(\gamma+\alpha)}=0 \\ & \Rightarrow \quad \Sigma \cos (\alpha+\beta)=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

and $\Sigma \sin (\alpha+\beta)=0$

From Eqs. (iii) and (iv),

$ \begin{aligned} & \cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma \\ & =\cos (\alpha+\beta)+\cos (\beta+\gamma)+\cos (\gamma+\alpha) \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \text {Let } x=(-32 i)^{\frac{2}{5}}=(-1024)^{\frac{1}{5}} \\ & \therefore \quad x^5=-1024 \Rightarrow\left(\frac{-x}{4}\right)^5=1 \\ & \therefore \quad \frac{-x}{4}=\operatorname{cis}\left(\frac{2 n \pi}{5}\right), n=0,1,2,3,4 \end{aligned}\\ &\text { For } n=4 \text {, }\\ &x=-4 \operatorname{cis} \frac{8 \pi}{5}=4 \operatorname{cis} \frac{3 \pi}{5} \end{aligned} $

Given, $\sqrt{(-3+4 i)(8+6 i)}$

$ =\sqrt{-48+14 i}=\sqrt{2} \sqrt{-24+7 i} $

We know that,

$ \begin{aligned} & \sqrt{z}= \pm\left(\sqrt{\frac{|z|+\operatorname{Re}(z)}{2}}+i \sqrt{\frac{|z|-\operatorname{Re}(z)}{2}}\right) \text { for } \\ & \operatorname{Im}(z)>0 \\ & \therefore \sqrt{-24+7 i} \\ & = \pm\left(\sqrt{\frac{25-24}{2}}+i \sqrt{\frac{25+24}{2}}\right) \\ & = \pm \frac{1}{\sqrt{2}}(1+7 i) \\ & \therefore \sqrt{(-3+4 i)(8+6 i)}= \pm(1+7 i) \end{aligned} $

$ \begin{aligned} &\text { We know that, }\\ &\begin{aligned} & \omega=\frac{-1}{2}+i \frac{\sqrt{3}}{2} \text { and } \omega^2=\frac{-1}{2}-i \frac{\sqrt{3}}{2} \\ \Rightarrow & -2 i \omega=\sqrt{3}+i \text { and } 2 i \omega^2=\sqrt{3}-i \\ & \left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^m=1 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad\left(\frac{-2 i \omega}{2 i \omega^2}\right)^m=1 \Rightarrow(-\omega)^m=1 \\ & \Rightarrow \quad(-1)^m \omega^m=1 \end{aligned} $

$\Rightarrow m$ is even number and multiple of 3 . Clearly, 2028 is even and multiple of 3.

Here, 1, $\omega$ and $\omega^2$ are the cube root of unity, $n \in N, n>2$

$ \begin{array}{ll} \therefore \quad & 1+\omega+\omega^2=0, \omega^3=1 \\ & x^n-x=0 \rightarrow 1+\omega \text { is a root. } \end{array} $

$ \begin{aligned} & \text { So, }(1+\omega)^n-(1+\omega)=0 \\ & \Rightarrow(1+\omega)^n=(1+\omega) \\ & \quad(1+\omega)^{n-1}=1 \Rightarrow\left(-\omega^2\right)^{n-1}=1 \end{aligned} $

Least value of $n$ must be 7 .