Complex Numbers

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

We have, $|z-1| \leq 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} & \text { and }\left|\omega^2 z-1-\omega\right|=a \\ & \qquad \begin{array}{l} \omega^2 z+\omega^2 \mid=a \quad\left[\because \omega^2=-1-\omega\right] \\ \left|\omega^2\right||z+1|=a \\ |z-1+2|=a \\ |z-1|+2 \geq a \end{array} \end{aligned} $

From Eq. (i), we get

$ \begin{aligned} & |z-1|+2=4 \\ \Rightarrow \quad & 0 \leq a \leq 4 \end{aligned} $

We have, $z^3+i z^2+2 i=0$

One root of $z^3+i z^2+2 i=0$ is $i$.

$ \begin{array}{ll} & {\left[\because i^3+i\left(i^2\right)+2 i=-i-i+2 i=0\right]} \\ & z^3+i z^2+2 i=0 \\ \Rightarrow \quad & (z-i)\left(z^2+2 i z-2\right)=0 \\ \Rightarrow \quad & (z-i)\left(z^2+2 i z+i^2+1-2\right)=0 \\ \Rightarrow \quad & (z-i)\left((z+i)^2-1\right)=0 \\ \Rightarrow \quad & (z-i)(z+(i+1))(z+(i-1))=0 \\ \Rightarrow \quad & z=i,-i-1,1-i \end{array} $

$\therefore$ The triangle formed by the vertices $A(0,1), B(-1,-1)$ and $C(1,-1)$.

$ A B=\sqrt{1+4}=\sqrt{5} \text { units } $

$B C=2$ units

$A C=\sqrt{1+4}=\sqrt{5}$ units

$\Rightarrow \triangle A B C$ is an isosceles triangle.

$ [\because A B=A C] $

Since, the non-equal side is less than the equal side, so $A B C$ is not a right-angled triangle.

In the complex plane, any point can be represented in polar form as $(r, \theta) = r(\cos \theta + i \sin \theta)$. Here, $x = (1, \alpha)$, $y = (1, \beta)$, and $z = (1, \gamma)$. Given the condition $x + y + z = 0$, we can express

$ x = \cos \alpha + i \sin \alpha $

Similarly,

$ y = \cos \beta + i \sin \beta $

$ z = \cos \gamma + i \sin \gamma $

These imply:

$ x + y + z = 0 \Rightarrow \cos \alpha + \cos \beta + \cos \gamma = 0 $

$ \sin \alpha + \sin \beta + \sin \gamma = 0 $

From the above, we isolate:

$ \cos \alpha + \cos \beta = -\cos \gamma \quad \text{...(i)} $

$ \sin \alpha + \sin \beta = -\sin \gamma \quad \text{...(ii)} $

Squaring and adding equations (i) and (ii), we get:

$ \begin{align*} (\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2 &= \cos^2 \gamma + \sin^2 \gamma \\ \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta + \sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta &= 1 \end{align*} $

This simplifies to:

$ 2 \cos(\alpha - \beta) = -1 \quad \Rightarrow \quad \cos(\alpha - \beta) = -\frac{1}{2} $

By symmetry,

$ \cos(\beta - \gamma) = -\frac{1}{2} $

and

$ \cos(\gamma - \alpha) = -\frac{1}{2} $

The corresponding sine values are:

$ \sin (\alpha - \beta) = \sin (\beta - \gamma) = \sin (\gamma - \alpha) = \frac{\sqrt{3}}{2} $

Consider $\cos (2\alpha - \beta - \gamma)$:

$ \begin{align*} \cos(2\alpha - \beta - \gamma) &= \cos((\alpha - \beta) - (\gamma - \alpha)) \\ &= \cos(\alpha - \beta) \cos(\gamma - \alpha) + \sin(\alpha - \beta) \sin(\gamma - \alpha) \\ &= \left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \\ &= \frac{1}{4} + \frac{3}{4} \\ &= 1 \end{align*} $

Similarly,

$ \cos(2\beta - \gamma - \alpha) = 1 $

$ \cos(2\gamma - \alpha - \beta) = 1 $

Thus, we find:

$ \Sigma \cos (2\alpha - \beta - \gamma) = 1 + 1 + 1 = 3 $

We have, $\arg \left[\frac{(1+i \sqrt{3})(-\sqrt{3}-i)}{(1-i)(-i)}\right]$

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

$ \begin{aligned} \Rightarrow \quad z & =2(1+i) \\ \therefore \arg (z)=\arg (2+2 i) & =\tan ^{-1}(1) \\ & =\tan ^{-1}\left(\tan \frac{\pi}{4}\right) \\ \Rightarrow \quad \arg (z) & =\frac{\pi}{4} \end{aligned} $

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

$ \begin{aligned} & \text { Now, let } \omega=\frac{z-3 i}{z+4}=\frac{x+i y-3 i}{x+i y+4} \\ & =\frac{x+(y-3) i}{(x+4)+i y} \times \frac{(x+4)-i y}{(x+4)-i y} \\ & =\frac{x(x+4)-i x y+(y-3)(x+4) i+y(y-3)}{(x+4)^2+y^2} \\ & \Rightarrow \frac{x(x+4)+y(y-3)}{(x+4)^2+y^2}+i \frac{(y-3)(x+4)-xy}{(x+4)^2+y^2} \end{aligned} $

Now,

$ \begin{aligned} & \arg (\omega)=\tan ^{-1}\left(\frac{(y-3)(x+4)-xy)}{x(x+4)+y(y-3)}\right)= \frac{\pi}{2}\\ & \Rightarrow \frac{(y-3)(x+4)-x y}{x(x+4)+y(y-3)}=\tan \frac{\pi}{2} \\ & \Rightarrow \frac{4 y-3 x-12}{x^2+y^2+4 x-3 y}=\infty \end{aligned} $

So, clearly

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

and $3 x-4 y<0$

Given,

$\alpha_1, \alpha_2, \alpha_3, \alpha_4$ and $\alpha_5$ are roots of the equation

$ \begin{aligned} & x^5-5 x^4+9 x^3-9 x^2+5 x-1=0 \\ \Rightarrow & x^5-x^4-4 x^4+4 x^3+5 x^3-5 x^2 -4 x^2+4 x+x-1=0 \\ \Rightarrow & x^4(x-1)-4 x^3(x-1)+5 x^2(x-1) -4 x(x-1)+1(x-1)=0 \\ \Rightarrow & (x-1)\left[x^4-4 x^3+5 x^2-4 x+1\right]=0 \\ \Rightarrow & (x-1)\left(x^2-3 x+1\right)\left(x^2-x+1\right)=0 \\ \Rightarrow & x=1, \frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}, \frac{1+\sqrt{3} i}{2} \end{aligned} $

$ \begin{aligned} & \text { and } \frac{1-\sqrt{3} i}{2} \\ & \alpha_1=1 \Rightarrow \alpha_1^2=1-\frac{1}{\alpha_1^2}=1 \\ & \alpha_2=\frac{3+\sqrt{5}}{2} \Rightarrow \alpha_2^2=\frac{(3+\sqrt{5})^2}{4} \\ & \Rightarrow \quad \frac{1}{\alpha_2^2}=\frac{4}{(3+\sqrt{5})^2} \\ & \Rightarrow \quad \begin{array}{l} \alpha_3=\frac{3-\sqrt{5}}{2} \Rightarrow \alpha_3^2=\frac{(3-\sqrt{5})^2}{4} \\ \Rightarrow \quad \frac{1}{\alpha_3^2}=\frac{4}{(3-\sqrt{5})^2} \end{array} \end{aligned} $

$ \begin{aligned} \alpha_4 & =\frac{1+\sqrt{3} i}{2} \\ & \Rightarrow \alpha_4^2=\frac{(1+\sqrt{3} i)^2}{4} \\ \Rightarrow & \frac{1}{\alpha_4^2}=\frac{4}{(1+\sqrt{3} i)^2} \\ \alpha_5 & =\frac{1-\sqrt{3} i}{2} \Rightarrow \alpha_5^2=\frac{(1-\sqrt{3} i)^2}{4} \\ & \Rightarrow \quad \frac{1}{\alpha_5^2}=\frac{4}{(1-\sqrt{3} i)^2} \end{aligned} $

$|Z| \leq 3 \Rightarrow$ Complex number Z lies within a circle centered at the origin with radius 3 .

$\frac{-\pi}{2} \leq \arg (z) \leq \frac{\pi}{2}$. This restricts z to the right half of the complex plane.

The intersection of these region is a semicircle with radius 3 . The area of semicircle is given by half the area of the full circle

The area of circle with radius $r$ is $=\pi r^2$

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\pi(3)^2=9 \pi $

Half of this area $=\frac{9 \pi}{2}$

So, area of region $=\frac{9 \pi}{2}$ sq units.

To determine the locus of the complex number $ Z $, where $ Z = x + i y $ (with $ x $ and $ y $ as real numbers), we analyze the given condition:

$ \arg \left(\frac{Z-1}{Z+1}\right) = \frac{\pi}{4} $

This equation implies that the complex number $ A = \frac{Z-1}{Z+1} $ forms an angle of $ \frac{\pi}{4} $ with the positive real axis. Therefore, $ A $ lies on a line making an angle of $ \frac{\pi}{4} $.

To simplify, we express:

$ \frac{Z-1}{Z+1} = e^{i \frac{\pi}{4}} = \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} $

Thus, we have:

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

This calculation indicates that the complex number $ \frac{Z-1}{Z+1} $ is on the unit circle. Therefore, we equate:

$ \left|\frac{Z-1}{Z+1}\right| = 1 \Rightarrow |Z-1| = |Z+1| $

This describes the locus as the perpendicular bisector of the line segment joining 1 and -1, which corresponds to the imaginary axis $ x = 0 $. Thus, the locus is a straight line on the imaginary axis.

To find the values of $(8i)^{\frac{1}{3}}$, we utilize De Moivre's theorem. Express $8i$ in polar form as $8e^{i(\frac{\pi}{2} + 2k\pi)}$, then take the cube root:

$ (8i)^{\frac{1}{3}} = 8^{\frac{1}{3}} e^{i\left(\frac{\theta + 2k\pi}{3}\right)} = 2e^{i\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right)} $

Next, calculate this for different values of $k$:

For $k = 0$:

$ 2e^{i\left(\frac{\pi}{6}\right)} = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = \sqrt{3} + i $

For $k = 1$:

$ 2e^{i\left(\frac{\pi}{6} + \frac{2\pi}{3}\right)} = 2e^{i\left(\frac{5\pi}{6}\right)} = 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right) = 2\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = -\sqrt{3} + i $

For $k = 2$:

$ 2e^{i\left(\frac{\pi}{6} + \frac{4\pi}{3}\right)} = 2e^{i\left(\frac{3\pi}{2}\right)} = 2(0 - i) = -2i $

Therefore, all the values of $(8i)^{\frac{1}{3}}$ are $\sqrt{3} + i$, $-\sqrt{3} + i$, and $-2i$, or in other words, $\pm \sqrt{3} + i$, $-2i$.

$ x^9-x^5+x^4-1=0 $

$ \begin{array}{lrl} \Rightarrow & x^5\left(x^4-1\right)+1\left(x^4-1\right) & =0 \\ \Rightarrow & \left(x^5+1\right)\left(x^4-1\right) & =0 \\ \Rightarrow & x^5+1=0 \text { or } x^4-1=0 \\ \Rightarrow & x^5=-1 \text { or } x^4=1 \\ \Rightarrow & x^5=-1 \text { or } x=1,-1, i,-i \end{array} $

$\therefore$ Number of real roots $(n)=3$

Number of complex roots $(m)=2$

Number of complex roots having argument in 2nd quadrant $(k)=1$

$ \therefore m \cdot n \cdot k=3(2)(1)=6 $

We have,

$ \begin{aligned} & \frac{(1-i)^3}{(2-i)(3-2 i)}=\frac{1^3-i^3-3.1 . i(1-i)}{\left(6-4 i-3 i+2 i^2\right)} \\ & \quad\left[\because(a-b)^3=a^3-b^3-3 a b(a-b)\right] \\ &= \frac{1-(-i)-3 i+3 i^2}{(6-7 i-2)}=\frac{1+i-3 i-3}{4-7 i} \\ &= \frac{-2-2 i}{4-7 i} \\ &= \frac{-2-2 i}{4-7 i} \times \frac{4+7 i}{4+7 i} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[rationalising]\\ &= \frac{-(2+2 i)(4+7 i)}{4^2-(7 i)^2} \\ &= \frac{-(8+14 i+8 i-14)}{16+49}=\frac{6-22 i}{65} \end{aligned} $

On comparing with $Z=x+i y$, we get

Imaginary part is $\frac{-22}{65}$.

To find the square root of $7 + 24i$, set it equal to $x + yi$, where $x, y \in \mathbb{R}$.

Squaring both sides gives:

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

This leads to:

$ x^2 - y^2 + 2xyi = 7 + 24i $

By equating the real and imaginary parts, we obtain:

$ x^2 - y^2 = 7 \quad \text{(equation 1)} $

$ 2xy = 24 \quad \Rightarrow \quad xy = 12 \quad \text{(equation 2)} $

From equation 2, solve for $y$:

$ y = \frac{12}{x} $

Substitute $y = \frac{12}{x}$ into equation 1:

$ x^2 - \left(\frac{12}{x}\right)^2 = 7 $

Simplifying the equation, we get:

$ x^2 - \frac{144}{x^2} = 7 $

$ x^4 - 144 = 7x^2 $

$ x^4 - 7x^2 - 144 = 0 $

This can be factored as:

$ (x^2 - 16)(x^2 + 9) = 0 $

Since $x$ is real, we discard $x^2 + 9 = 0$ as it does not lead to real solutions for $x$. Therefore:

$ x^2 = 16 \quad \Rightarrow \quad x = \pm 4 $

For $x = 4$, $y = 3$.

For $x = -4$, $y = -3$.

Thus, the square roots of $7 + 24i$ are $\pm(4 + 3i)$.

Therefore, the correct solution is $4 + 3i$.

We have,

$ Z=\cos \theta+i \sin \theta, \theta \neq(2 n+1) \frac{\pi}{2} $

Now, $\frac{1+Z^{2 n}}{1-Z^{2 n}}=\frac{1+(\cos \theta+i \sin \theta)^{2 n}}{1-(\cos \theta+i \sin \theta)^{2 n}}$

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

[By De Moivre's theorem]

$ \begin{aligned} & =\frac{1+\left(2 \cos ^2 n \theta-1\right)+2 i \sin n \theta \cos n \theta}{1-\left(1-2 \sin ^2 n \theta\right)-2 i \sin n \theta \cos n \theta} \\ & \quad\left[\cos 2 A=2 \cos ^2 A-1=1-2 \sin ^2 A\right] \\ & \quad=\frac{2 \cos n \theta[\cos n \theta+i \sin n \theta]}{2 \sin n \theta[\sin n \theta-i \cos n \theta]} \\ & \quad=-i \cot n \theta\left[\frac{\cos n \theta+i \sin n \theta}{\cos n \theta+i \sin n \theta}\right] \\ & \quad=-i \cot n \theta . \end{aligned} $

When calculating the complex conjugate of the expression $(4-3i)(2+3i)(1+4i)$, follow these steps:

First, compute the product:

$ (4-3i)(2+3i) = 8 + 12i - 6i - 9i^2 = 8 + 6i + 9 = 17 + 6i $

Here, we used the fact that $i^2 = -1$.

Next, multiply this result by $(1+4i)$:

$ (17+6i)(1+4i) = 17 + 68i + 6i + 24i^2 = 17 + 74i - 24 $

Simplifying this, we get:

$ -7 + 74i $

To find the complex conjugate of $-7 + 74i$, change the sign of the imaginary part:

$ -7 - 74i $

So, the complex conjugate of the expression is $-7 - 74i$.

To determine the locus of $ z $, given that the amplitude (also known as the argument) of $ (z-2) $ is $\frac{\pi}{2}$, we proceed as follows:

Let $ z = x + i y $, where $ x $ and $ y $ are real numbers. Given the condition:

$ \arg(z-2) = \frac{\pi}{2} $

Substituting $ z = x + iy $ into the expression, we have:

$ \arg((x-2) + iy) = \frac{\pi}{2} $

This implies:

$ \tan^{-1}\left(\frac{y}{x-2}\right) = \frac{\pi}{2} $

From the property of arctangent, we know:

$ \frac{y}{x-2} \to \infty $

This scenario occurs when $ x-2 = 0 $. Hence,

$ x = 2 $

Thus, the locus of $ z $ is a vertical line at $ x = 2 $, which is parallel to the Y-axis, only considered for $ y > 0 $.

Therefore, the locus of $ z $ is a vertical line $ x = 2 $ for $ y > 0 $.

If $\omega$ is the cube root of unity.

$ \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}=? $

Now, we write as

$ \begin{aligned} & A=a+b \omega+c \omega^2 \\ & B=b+c \omega+a \omega^2 \\ & C=c+a \omega+b \omega^2 \end{aligned} $

Find common denominator for the fractions.

$ \frac{A}{C}+\frac{A}{B}=\frac{A \cdot B+A \cdot C}{B \cdot C}=\frac{A(B+C)}{B C} $

Now, we know that according to given condition.

$ \begin{gathered} \omega^3=1 \\ 1+\omega+\omega^2=0 \\ B+C=\left(b+c \omega+a \omega^2\right)+\left(c+a \omega+b \omega^2\right) \\ b+c \omega+a \omega^2+c+a \omega+b \omega^2 \\ =\left(b+\omega^2 b\right)+(c \omega+c)+\left(a \omega^2+a \omega\right) \\ =b\left(1+\omega^2\right)+c(1+\omega)+a\left(\omega^2+\omega\right) \\ \quad\left(\because \text { using } 1+w+w^2=0\right) \\ 1+\omega^2=-\omega, 1+\omega=-\omega^2 \end{gathered} $

$ \begin{aligned} B+C & =b(-\omega)+C\left(-\omega^2\right)+a(-1) \\ & =-a-b \omega-c \omega^2 \end{aligned} $

Thus, $B+C=-A$

Because, $A=a+b \omega+c \omega^2$

Substituting this into the expression

$ \frac{A(B+C)}{B C}=\frac{A(-A)}{B C}=\frac{-A^2}{B C} $

The denominator $(B C)$ does not change the overall simplification that each term involves $\bar{A}, B$ and $C$ with symmetrical properties.

Thus, $\frac{A}{C}+\frac{A}{B}=-1$

$ \begin{aligned} &\text { Given, }(3+i) \text { is a root of }\\ &\begin{aligned} & x^2+a x+b=0 \\ &(3+i)^2+a(3+i)+b=0 \\ & 9-1+6 i+a(3+i)+b=0 \\ & 8+6 i+3 a+a i+b=0 \\ &(8+3 a+b)+i(6+a)=0 \\ & 3 a+b=-8 \\ & a+6=0 \Rightarrow a=-6 \end{aligned} \end{aligned} $

Given, $z_1=10+6 i, z_2=4+6 i$

and $\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi}{4}$

Let $z=x+i y$

Now, $\frac{z-z_1}{z-z_2}=\frac{(x+i y)-(10+6 i)}{(x+i y)-(4+6 i)}$

$ \begin{aligned} & =\frac{(x-10)+i(y-6)}{(x-4)+i(y-6)} \\ & =\frac{(x-10)+i(y-6)}{(x-4)+i(y-6)} \times \frac{(x-4)-i(y-6)}{(x-4)-i(y-6)} \\ & =\frac{(x-10)(x-4)+(y-6)^2+i[-(x-10)(y-6)+(y-6)(x-4)]}{(x-4)^2+(y-6)^2} \end{aligned} $

Since, $\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi}{4}$

i.e. $\tan \theta=1$

$ \therefore \frac{(y-6)(x-4)-(x-10)(y-6)}{(x-10)(x-4)+(y-6)^2}=1 $

$ \Rightarrow \frac{x y-6 x-4 y+24-x y+6 x+10 y-60}{x^2-14 x+40+y^2-12 y+36}=1 $

$ \Rightarrow 6 y-36=x^2+y^2-12 y-14 x+76 $

$ \Rightarrow x^2+y^2-18 y-14 x+112=0 $

$ \Rightarrow \quad(x-7)^2+(y-9)^2=18 $

$ \Rightarrow|z-(7+9 i)|=3 \sqrt{2} $