Complex Numbers

Analyze Set $A$ :

The set $A$ is defined as $\{z \in \mathbb{C}:|z-2| \leq 4\}$.

• This represents a circle (and its interior) centered at $(2,0)$ with a radius $r=4$.

• The circle extends from $x=-2$ to $x=6$ along the real axis.

Analyze Set $B$ :

The set $B$ is defined as $\{z \in \mathbb{C}:|z-2|+|z+2|=5\}$.

• By the definition of an ellipse, the sum of distances from two fixed points (foci) is constant.

• Here, the foci are $F_1(2,0)$ and $F_2(-2,0)$, and the length of the major axis $2 a=5$, so $a=2.5$.

• By the definition of an ellipse, the sum of distances from two fixed points (foci) is constant.

• Here, the foci are $F_1(2,0)$ and $F_2(-2,0)$, and the length of the major axis $2 a=5$, so $a=2.5$.

• The distance between foci is $2 a e=4$, so $e=\frac{4}{5}=0.8$.

• The semi-minor axis $b$ can be found using $b^2=a^2\left(1-e^2\right)=(2.5)^2(1-0.64)= 6.25(0.36)=2.25$, so $b=1.5$.

• The ellipse is centered at the origin $(0,0)$ and extends from $x=-2.5$ to $x=2.5$ along the real axis.

To find $\max \left\{\left|z_1-z_2\right|: z_1 \in A\right.$ and $\left.z_2 \in B\right\} \quad$ This means maximize the distance between a point $z_1$ in the circle and a point $z_2$ on the ellipse.

The furthest point in set $A$ (the circle) is at the rightmost edge, $z_1=(2+4,0)=(6,0)$.

The furthest point in set $B$ (the ellipse) from $z_1$ is at its leftmost edge, $z_2=(-2.5,0)$.

Maximum Distance $\left|z_1-z_2\right|=|6-(-2.5)|=|6+2.5|=8.5=\frac{17}{2}$

Let $z=x+i y$

$|z-(\alpha+i \beta)|=r$ represents a circle in the complex plane with centre at $(\alpha, \beta)$ and radius is $r$ unit.

Cartesian equation

$ (x-\alpha)^2+(y-\beta)^2=r^2 $

The equations $|z-6|=5$ and $|z-(-2+6 i)|=5$ describe two circles with the same radius $(r=5)$.

• Circle 1 : Center $C_1=(6,0)$

• Circle 2 : Center $C_2=(-2,6)$

Cartesian equation

$ (x-6)^2+y^2=25 \quad \text { and } \quad(x+2)^2+(y-6)^2=25 $

Expanding both:

• $x^2-12 x+36+y^2=25 \Longrightarrow \mathbf{x}^2+\mathbf{y}^2-12 \mathbf{x}+\mathbf{1 1}=\mathbf{0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

• $x^2+4 x+4+y^2-12 y+36=25 \Longrightarrow \mathbf{x}^2+\mathbf{y}^2+\mathbf{4} \mathbf{x}-\mathbf{1 2} \mathbf{y}+\mathbf{1 5}=\mathbf{0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Subtracting the first equation from the second to find the common chord:

$ \begin{gathered} (4 x-(-12 x))-12 y+(15-11)=0 \\ 16 x-12 y+4=0 \Longrightarrow \mathbf{4}-\mathbf{3} \mathbf{y}+\mathbf{1}=\mathbf{0} \Longrightarrow \mathbf{y}=\frac{\mathbf{4} \mathbf{x}+\mathbf{1}}{\mathbf{3}} \end{gathered} $

Substitute $y$ back into the first circle equation:

$ (x-6)^2+\left(\frac{4 x+1}{3}\right)^2=25 $

multiply both sides by 9

$ \begin{aligned} & 9\left(x^2+36-12 x\right)+16 x^2+8 x+1=225 \\ & 25 x^2-100 x+100=0 \\ & x^2-4 x+4=0 \\ & (x-2)^2=0 \Rightarrow x=2 \end{aligned} $

put $x=2$ in $y=(4 x+1) / 3$

$ y=(4 \times 2+1) / 3=9 / 3=3 $

$ \text { so } z=2+3 i $

calculate $z^3+3 z^2-15 z+141$

Compute $z^3$

$ \begin{gathered} z^3=z \cdot z^2=(2+3 i)(-5+12 i) \\ =-10+24 i-15 i+36 i^2 \\ =-10+9 i-36 \\ =-46+9 i \end{gathered} $

Compute $z^2$

$ (2+3 i)^2=4+12 i+9 i^2=4+12 i-9=-5+12 i $

so

$ \begin{gathered} z^3=-46+9 i \\ 3 z^2=-15+36 i \\ -15 z=-30-45 i \end{gathered} $

Add everything

$ (-46+9 i)+(-15+36 i)+(-30-45 i)+141 $

Real parts:

$ -46-15-30+141=50 $

Imaginary parts:

$ 9 i+36 i-45 i=0 $

So $z^3+3 z^2-15 z+141=50$

Given, $\mathrm{S}=\left\{z \in \mathbb{C}:\left|\frac{z-6 i}{z-2 i}\right|=1\right.$ and $\left.\left|\frac{z-8+2 i}{z+2 i}\right|=\frac{3}{5}\right\}$

$\left|\frac{z-6 i}{z-2 i}\right|=1$

$\Rightarrow $ $ |z-6 i|=|z-2 i|-(1) $

equation (1) represent $z$ is a point which is equidistant from point $(0,2)$ and $(0,6)$ so $z$ lies on the line segment which is perpendicular bisector of the line segment joining point $(0,2)$ & $ (0,6)$.

If $z=x+i y$ then solution of equation (1) is $y=\frac{6+2}{2}=4$ and $x \in R$.

So, solution is $z=x+4 i$ & $ x \in R$.

$\begin{aligned} & \text { Solving }\left|\frac{z-8+2 i}{z+2 i}\right|=\frac{3}{5} \\\\ & \text { Substitute } z=x+4 i \\\\ & \frac{|x+4 i-8+2 i|}{|x+4 i+2 i|}=\frac{3}{5}\end{aligned}$

$\Rightarrow $ $5|x+6 i-8|=3|x+6 i|$

$\Rightarrow $ $5|(x-8)+6 i|=3|x+6 i|$

$\Rightarrow $ $5 \sqrt{(x-8)^2+6^2}=3 \sqrt{x^2+6^2}$

Squaring

$\Rightarrow $ $ 25\left(x^2+64-16 x+36\right)=9 x^2+324$

$\Rightarrow $ $25 x^2-400 x+2500=9 x^2+324$

$\Rightarrow $ $16 x^2-400 x+2176=0$

$\Rightarrow $ $x^2-25 x+136=0$

$\Rightarrow $ $x^2-17 x-8 x+136=0$

$\Rightarrow $ $x(x-17)-8(x-17)=0$

$\Rightarrow $ $(x-17)(x-8)=0$

$\therefore $ $x=17, x=8$

So set S contain two complex number $z_1=17+4 i$ and $z_2=8+4 i$

$ z=\frac{\sqrt{3}}{2}+\frac{i}{2} $

Convert into Euler form $z=r e^{i \theta}$ where $r=|z|$ and $\theta=\arg (z)=\tan ^{-1}\left(\frac{y}{x}\right)$ if $z=x+i y$.

$ r=|z|=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2}=\sqrt{\frac{3}{4}+\frac{1}{4}}=1 $

Since $z$ lies in the first quadrant in the complex plane,

$ \begin{aligned} & \text { So, } \theta=\tan ^{-1}\left(\frac{1 / 2}{\sqrt{3} / 2}\right)=\frac{\pi}{6} \\ & z=e^{i \frac{\pi}{6}} \end{aligned} $

To calculate $\left(z^{201}-i\right)^8=\left(\left(e^{i \frac{\pi}{6}}\right)^{201}-i\right)^8$

$ =\left(e^{i \frac{201 \pi}{6}}-i\right)^8 $

Using De Moivre's theorem, $e^{i \theta}=\cos \theta+i \sin \theta$

$ \begin{aligned} & \left(z^{201}-i\right)^8=\left(\cos \left(\frac{201 \pi}{6}\right)+i \sin \left(\frac{201 \pi}{6}\right)-i\right)^8 \\ & =\left(\cos \left(\frac{67 \pi}{2}\right)+i \sin \left(\frac{67 \pi}{2}\right)-i\right)^8 \\ & =(0+i(-1)-i)^8 \\ & =(-2 i)^8=2^8(i)^8 \end{aligned} $

Since $i=\sqrt{-1} \Longrightarrow i^2=-1 \Longrightarrow i^4=1 \Longrightarrow i^8=1$

$\therefore $ $\left(z^{201}-i\right)^8=2^8=256$.

$ \frac{3}{2} \leq\left|z-\left(\frac{3}{2}+\frac{3}{2} i\right)\right| \leq \frac{7}{2} $

$ \begin{aligned} & \text { Min distance }=\sqrt{16+9}-\frac{7}{2} \\ & =5-\frac{7}{2} \\ & =\frac{3}{2} \end{aligned} $

$ 4 z^2=\bar{z}=0 $

Let $z=x+i y$

$ \begin{aligned} & 4(x+i y)^2+x-i y=0 \\ & 4 x^2-4 y^2+8 x y i+x-i y=0 \\ & 4 x^2-4 y^2+x=0 \& y(8 x-1)=0 \\ & \Rightarrow y=0 \text { or } x=\frac{1}{8} \end{aligned} $

If $y=0,4 x^2+x=0$

$ \begin{array}{ll} x=0, \frac{-1}{4} & \\ \therefore z_1=0+0 . i & \left|z_1\right|^2=0 \\ z_2=-\frac{1}{4}+0 i & \left|z_2\right|^2=\frac{1}{16} \end{array} $

If $x=\frac{1}{8}$

$ \begin{aligned} & 4 \times \frac{1}{64}-4 y^2+\frac{1}{8}=0 \Rightarrow 4 y^2=\frac{3}{16} \Rightarrow y= \pm \frac{\sqrt{3}}{8} \\ & \therefore z_3=\frac{1}{8}+\frac{\sqrt{3}}{8} i \quad\left|z_3\right|^2=\frac{1}{64}+\frac{3}{64}=\frac{1}{16} \\ & z_4=\frac{1}{8}-\frac{\sqrt{3}}{8} i \quad\left|z_4\right|^2=\frac{1}{64}+\frac{3}{64}=\frac{1}{16} \\ & \therefore \sum_{i=1}^n\left|z_i\right|^2=0+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{3}{16} \end{aligned} $

$ \begin{aligned} & \Rightarrow \arg (z)=\sin ^{-1}\left(\frac{3}{5}\right) \tan ^{-1}\left(\frac{3}{4}\right) \Rightarrow z=|z|^{i \theta} \\ & =4 \times(\cos \theta+i \sin \theta)=4 \times\left(\frac{4}{5}+i \frac{3}{5}\right) \\ & =\frac{16}{5}+\frac{12 i}{5} \quad \Rightarrow 5 z=16+12 i \\ & \left(\frac{5 z-12}{5 z i+16}\right)=\frac{(4+12 i)}{(4+16 i)}=\frac{(1+3 i)}{(1+4 i)} \\ & \Rightarrow\left(\frac{5 z-12}{5 z i+16}\right)=\frac{\sqrt{10}}{\sqrt{17}} \\ & 34\left(\frac{5 z-12}{5 z i+16}\right)^2=34 \times \frac{10}{17}=20 \end{aligned} $

$ \begin{array}{r} \left(w+\frac{1}{w}\right)^4+\left(w^2+\frac{1}{w^2}\right)^4+\left(w^3+\frac{1}{w^3}\right)^4+\ldots \\ \left(w^{25}+\frac{1}{w^{25}}\right)^4 \end{array} $

$ \begin{aligned} & \sum\left(w^k+\frac{1}{w^k}\right)^4 \\ & K=3 x \Rightarrow w^{3 x}+\frac{1}{w^{3 x}}=2 \\ & k \neq 3 x \Rightarrow w^k+\frac{1}{w^k}=-1 \\ & \sum_{k=1}^{25}\left(w^k+\frac{1}{w^k}\right)^4 \Rightarrow 8\left(1+1+2^4\right)+1 \\ & =145 \end{aligned} $

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

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

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

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

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

$\begin{aligned} & \text { Let } z=x+i y \\ & \operatorname{Re}\left(\frac{(x-1)+i y}{2 x+i(2 y+1)}\right)=1 \Rightarrow \operatorname{Re}\left[\frac{((x-1)+i y)(2 x-i(y+1)}{(2 x+i(2 y+1)(2 x-i(2 y+1))}\right]=1 \\ & \Rightarrow \frac{2 x(x-1)+y(2 y+1)}{4 x^2+(2 y+1)^2}=1 \\ & \Rightarrow 2 x^2-2 x+2 y^2+y=4 x^2+4 y^2+1+4 y \\ & \Rightarrow 2 x^2+2 y^2+3 y+2 x+1=0 \\ & \Rightarrow x^2+y^2+x+\frac{3}{2} y+\frac{1}{2}=0 \\ & \text { centre }=\left(\frac{-1}{2}, \frac{-3}{4}\right), r=\sqrt{\frac{1}{4}+\frac{9}{16}-\frac{1}{2}}=\frac{\sqrt{5}}{4} \\ & a=\frac{-1}{2}, b=\frac{-3}{4}, r^2=\frac{5}{16} \end{aligned}$

$15 \frac{\mathrm{ab}}{\mathrm{r}^2}=15 \times\left(\frac{-1}{2}\right) \times\left(\frac{-3}{4}\right) \times \frac{16}{5}=18$

$\begin{aligned} & \frac{z-i}{z+i}=\frac{\bar{z}+i}{\bar{z}-i} \\ & =z \bar{z}-i \bar{z}-i z-1=z \bar{z}+z i+i \bar{z}-1 \\ & =z+\bar{z}=0 \\ & =2 x=0 \\ & =x=0 \quad \text { (y-axis) } \end{aligned}$

$\begin{aligned} & |z|=1 \\ & \therefore \quad z=i \quad(z \neq-i \text { is given }) \end{aligned}$

Statement 1 is incorrect

$\begin{aligned} & \frac{z-i}{z+i}+\frac{\bar{z}-1}{\bar{z}+1}=0 \\ & =z \bar{z}-\bar{z}+z-1+z \bar{z}-z+\bar{z}-1=0 \\ & =z \bar{z}=1 \\ & =|z|=1 \end{aligned}$

Statement 2 is correct

$\begin{aligned} & \omega_1=(8 \sin \theta+7 \cos \theta)+i(\sin \theta+4 \cos \theta) \\ & \omega_2=(\sin \theta+4 \cos \theta)+i(8 \sin \theta+7 \cos \theta) \\ & \alpha=(8 \sin \theta+7 \cos \theta)+(\sin \theta+4 \cos \theta) \\ & \quad-(\sin \theta+4 \cos \theta)+(8 \sin \theta+7 \cos \theta)=0 \\ & \beta=(8 \sin \theta+7 \cos \theta)^2+(\sin \theta+4 \cos \theta)^2 \end{aligned}$

$\begin{aligned} & =65 \sin ^2 \theta+65 \cos ^2 \theta+56 \sin 2 \theta+4 \sin 2 \theta \\ & =65+60 \sin 2 \theta \\ & (\alpha+\beta)_{\max }=125=p \\ & (\alpha+\beta)_{\min }=5=q \\ & p+q=130 \end{aligned}$

$ \begin{aligned} & z_0=z_1+z_2+z_3 \\ & \sum_{k=1}^3\left(z_k-z_0\right)^2=\left(z_1-z_0\right)^2+\left(z_2-z_0\right)^2+\left(z_3-z_0\right)^2 \end{aligned} $

Let $z_0$ is origin $\Rightarrow z_1, z_2, z_3$ lies on a circle having $\left|z_0-z_i\right|=R$

$ \begin{aligned} & \therefore z_1=R e^{i 2 \pi / 3} z_2=R e^{i 4 \pi / 3} z_3=R e^{i 6 \pi / 3} \\ & \Rightarrow z_1^2+z_2^2+z_3^2=R^2\left[e^{i 4 \pi / 3}+e^{i 8 \pi / 3}+e^{i 12 \pi / 3}\right] \\ & =0 \\ & \therefore \sum_{k=1}^3\left(z_k-z_0\right)^2=0 \end{aligned} $

$\begin{aligned} &\begin{aligned} & \frac{z^2+3 i}{z-2+i}=2+3 i \\ & z^2+3 i=(z-2+i)(2+3 i) \\ & z^2+3 i=2 z-4+2 i+3 i z-6 i-3 \\ & z^2+3 i=(2 z-7)+i(3 z-4) \\ & z^2-(2+3 i) z+(7+7 i)=0 \end{aligned}\\ &\text { This is a quadratic in } z \text {. }\\ &\begin{aligned} & z_1+z_2=2+3 i \\ & z_1+z_2=7+7 i \\ & z_1^2+z_2^2=\left(z_1+z_2\right)^2-2 z_1 z_2 \\ & =(2+3 i)^2-2(7+7 i) \\ & =4-9+12 i-14-14 i \\ & =-19-2 i \end{aligned} \end{aligned}$

$\begin{aligned} &\begin{aligned} & \frac{2+k^2 z}{k+\bar{z}}=k z \\ & \Rightarrow 2+k^2 z=k^2 z+k \bar{z} \\ & \Rightarrow 2+k|z|^2 \quad\left(z \bar{z}=|z|^2,|z|=1\right) \\ & \Rightarrow 2=k \\ & \therefore k+k^2 i=2+4 i \end{aligned}\\ &\therefore \quad \text { The maximum distance is }\\ &\begin{aligned} & =\sqrt{(4-2)+(2-1)^2}+\text { radius } \\ & =\sqrt{(2)^2+(1)^2}+1 \\ & =\sqrt{5}+1 \end{aligned} \end{aligned}$

$\begin{aligned} & \because \mathrm{AB}=\sqrt{100}=10 \\ & \therefore\left|\mathrm{Z}_1-\mathrm{Z}_2\right|_{\min }=10-2-1=7 \end{aligned}$

$\begin{aligned} & x^2-(3-2 i) x-(2 i-2)=0 \\ & x=\frac{(3-2 i) \pm \sqrt{(3-2 i)^2-4(1)(-(2 i-2))}}{2(1)} \\ & ==\frac{(3-2 i) \pm \sqrt{9-4-12 i+8 i-8}}{2} \\ & ==\frac{3-2 i \pm \sqrt{-3-4 i}}{2} \\ & =\frac{3-2 i \pm \sqrt{(1)^2+(2 i)^2-2(1)(2 i)}}{2} \\ & =\frac{3-2 \mathrm{i} \pm(1-2 \mathrm{i})}{2} \\ & \Rightarrow \frac{3-2 \mathrm{i}+1-2 \mathrm{i}}{2} \text { or } \frac{3-2 \mathrm{i}-1+2 \mathrm{i}}{2} \\ & \Rightarrow 2-2 \mathrm{i} \text { or } 1+0 \mathrm{i} \\ & \text { So } \alpha \gamma+\beta \delta=2(1)+(-2)(0)=2 \end{aligned}$

$z_1=\sqrt{3}+2 \sqrt{2} i \quad \& \frac{\left|z_2\right|}{\left|z_1\right|}=\frac{1}{\sqrt{3}}$

given $\arg \left(\frac{z_2}{z_1}\right)=\frac{\pi}{6}$

$\begin{aligned} & z_2=\frac{\left|z_2\right|}{\left|z_1\right|} \cdot z_1 \mathrm{e}^{i\left(\frac{\pi}{6}\right)} \\ & z_2=\frac{1}{\sqrt{3}} \cdot \frac{(\sqrt{3}+2 \sqrt{2} i)(\sqrt{3}+i)}{2} \\ & z_2=\frac{(3-2 \sqrt{2})+i(2 \sqrt{6}+\sqrt{3})}{2 \sqrt{3}} \end{aligned}$

Now,

$z_1-z_2=\frac{(3+2 \sqrt{2})+i(2 \sqrt{6}-\sqrt{3})}{2 \sqrt{3}}$

$\left|z_1-z_2\right|=\left|z_2\right| \Rightarrow \Delta \mathrm{ABO}$ is isosceles with angles $\frac{\pi}{6}, \frac{\pi}{6} \& \frac{2 \pi}{3}$

$\begin{aligned} & \text { Sol. } 2 z^2-32-2 i=0 \\ & 2\left(z-\frac{i}{z}\right)=3 \\ & \alpha-\frac{i}{\alpha}=\frac{3}{2} \\ & \Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4} \\ & \Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4} \\ & \Rightarrow \frac{9}{4}+2 i=\alpha^2-\frac{1}{\alpha^2} \\ & \Rightarrow \frac{81}{16}-4+9 i=\alpha^4+\frac{1}{\alpha^4}-2 \\ & \Rightarrow \frac{49}{16}+9 i=\alpha^4+\frac{1}{\alpha^4} \\ & \text { Similarly } \\ & \Rightarrow \frac{49}{16}+9 i=\beta^4+\frac{1}{\beta^4} \end{aligned}$

$\begin{aligned} & \Rightarrow \frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}=\frac{\alpha^{15}\left(\alpha^4+\frac{1}{\alpha^4}\right)+\beta^{15}\left(\beta^4+\frac{1}{\beta^4}\right)}{\alpha^{15}+\beta^{15}} \\ & =\frac{\left(\alpha^{15}+\beta^{15}\right)\left(\frac{49}{16}+9 \mathrm{i}\right)}{\left(\alpha^{15}+\beta^{15}\right)} \\ & \text { Real }=\frac{49}{16} \\ & \text { Im }=9 \\ & \text { Ans. } 441 \end{aligned}$

$\begin{aligned} & |z|=1 \\ & \left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right|=1 \\ & \Rightarrow\left|z^2+(\bar{z})^2\right|=1 \\ & \text { Let } z=x+i y \\ & \Rightarrow\left|(x+i y)^2+(x-i y)^2\right|=1 \\ & \Rightarrow\left|2 x^2-2 y^2\right|=1 \\ & \Rightarrow\left|x^2-y^2\right|=\frac{1}{2} \\ & \Rightarrow x^2-y^2=\frac{ \pm 1}{2} \\ & \text { and } x^2+y^2=1\end{aligned}$

Case I: $x^2-y^2=\frac{1}{2}$

Case II: $x^2-y^2=-\frac{1}{2}$

Hence, we get 8 complex numbers.

$\begin{aligned} & \left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3} \\ & \left|\frac{\bar{z}-i}{\bar{z}+\frac{i}{2}}\right|=\frac{2}{3} \\ & 3|x-i y-i|=2\left|x-i y+\frac{i}{2}\right| \\ & 9\left(x^2+(y+1)^2\right)=4\left(x^2+(y-1 / 3)^2\right) \\ & 9 x^2+9 y^2+18 y+9=4 x^2+4 y^2-4 y+1 \\ & 5 x^2+5 y^2+22 y+8=0 \\ & x^2+y^2+\frac{22}{5} y+\frac{8}{5}=0 \\ & \text { centre } \Rightarrow\left(0,-\frac{11}{5}\right) \end{aligned}$

$\left| {{1 \over 2}\left| {\matrix{ 0 & 0 & 1 \cr 0 & { - 11/5} & 1 \cr \alpha & 0 & 1 \cr } } \right|} \right| = 11$

$\begin{aligned} & \Rightarrow\left(-\frac{11}{5} \alpha\right)^2=(11 \times 2)^2 \\ & \Rightarrow \alpha^2=100 \end{aligned}$

$\begin{aligned} & \text { Let } z=x+i y \\ & (x+i y)(1+i)+(x-i y)(1-i)=4 \\ & x+i x+i y-y+x-i x-i y-y=4 \\ & 2 x-2 y=4 \\ & x-y=2 \\ & |z-3| \leq 1 \\ & (x-3)^2+y^2 \leq 1 \end{aligned}$

$\begin{aligned} &\text { Area of shaded region }=\frac{\pi \cdot 1^2}{4}-\frac{1}{2} \cdot 1 \cdot 1=\frac{\pi}{4}-\frac{1}{2}\\ &\text { Area of unshaded region inside the circle }\\ &\begin{aligned} & =\frac{3}{4} \pi \cdot 1^2+\frac{1}{2} \cdot 1 \cdot 1=\frac{3 \pi}{4}+\frac{1}{2} \\ & \therefore \text { difference of area }=\left(\frac{3 \pi}{4}+\frac{1}{2}\right)-\left(\frac{\pi}{4}-\frac{1}{2}\right) \end{aligned}\\ &=\frac{\pi}{2}+1 \end{aligned}$

To solve the problem, we start with the given information about the complex numbers $ z_1, z_2, $ and $ z_3 $, which lie on the unit circle $ |z| = 1 $. Their arguments are as follows:

$ \arg(z_1) = -\frac{\pi}{4} $

$ \arg(z_2) = 0 $

$ \arg(z_3) = \frac{\pi}{4} $

Thus, the complex numbers can be represented as:

$ z_1 = e^{-i\frac{\pi}{4}} = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} $

$ z_2 = e^{i\cdot 0} = 1 $

$ z_3 = e^{i\frac{\pi}{4}} = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} $

Next, calculate the conjugates needed:

$ \bar{z}_2 = 1 $

$ \bar{z}_3 = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} $

$ \bar{z}_1 = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} $

We need to evaluate:

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

Calculate each term separately:

$ z_1 \bar{z}_2 = \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) \cdot 1 = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} $

$ z_2 \bar{z}_3 = 1 \cdot \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} $

$ z_3 \bar{z}_1 = \left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) \cdot \left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = \frac{(1 + i)^2}{2} = \frac{1 + 2i - 1}{2} = i $

Sum the evaluated terms:

$ z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1 = \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) + i $

Simplify:

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

Calculate the modulus squared:

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

$ = 2 + (1 - 2\sqrt{2} + 2) = 5 - 2\sqrt{2} $

Thus, the expression $ |z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1|^2 $ simplifies as follows:

$ \alpha = 5 $

$ \beta = -2 $

Finally, compute $ \alpha^2 + \beta^2 $:

$ \alpha^2 + \beta^2 = 5^2 + (-2)^2 = 25 + 4 = 29 $

Therefore, the value of $ \alpha^2 + \beta^2 $ is 29.

$\begin{aligned} & n=\frac{z-2 i}{z+2 i} \\ & \text { Let } z=x+i y \\ & n=\frac{x+(y-2) i}{x+(y+2) i} \times\left(\frac{x-(y+2) i}{x-(y+2) i}\right) \\ & \operatorname{Re}(n)=\frac{x^2+(y-2)(y+2)}{x^2+(y+2)^2}=0 \\ & \Rightarrow x^2+(y-2)(y+2)=0 \end{aligned}$

$\begin{aligned} & \Rightarrow x^2+y^2-4=0 \\ & \Rightarrow x^2+y^2=4 \\ & \text { also, }|z-(6+8 i)| \leq|z|+|-6-8 i| \\ & |z-(6+8 i)| \leq 2+10=12 \end{aligned}$

Hence, Maximum value of $|z-(6+8 i)|$ is 12.

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

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

$n=\frac{1-2 \cos ^2 \theta}{1+4 \cos ^2 \theta}+i\left(\frac{3 \cos \theta}{1+4 \cos ^2 \theta}\right)$

$n$ is purely imaginary

$\Rightarrow \frac{1-2 \cos ^2 \theta}{1+4 \cos ^2 \theta}=0$

$\Rightarrow \cos ^2 \theta=\frac{1}{2}$

$\Rightarrow \cos \theta= \pm \frac{1}{\sqrt{2}}$

$\theta$ can be $\frac{\pi}{4}, \frac{-\pi}{4}, \frac{3 \pi}{4}, \frac{-3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$

Sum of all possible values of $\theta=3 \pi$

$\begin{aligned} & |z+2|=1 \\ & \operatorname{Im}_m\left(\frac{z+1}{z+2}\right)=\frac{1}{5} \\ & |\operatorname{Re}(\overline{z+2})|=? \end{aligned}$

Let $z=x+i y$

$\begin{aligned} & \because|z+2|=1 \Rightarrow(x+2)^2+y^2=1 \quad \ldots(1) \\ & I_m\left(\frac{z+1}{z+2}\right)=\frac{1}{5} \Rightarrow I_m\left(\frac{x+i y+1}{x+i y+2}\right)=\frac{1}{5} \\ & \Rightarrow I_m\left|\frac{[(x+1)+i y][(x+)-i y]}{(x+2)^2+y^2}\right|=\frac{1}{5} \end{aligned}$

$\frac{y(x+2)-y(x+)}{(x+2)^2+y^2}=\frac{1}{5}\quad \text{.... (2)}$

$\Rightarrow y=\frac{1}{5}$

Substituting in equation (1)

$\begin{aligned} & (x+2)^2+\frac{1}{25}=1 \\ & (x+2)^2=\frac{24}{25} \\ & \Rightarrow x=-2 \pm \frac{\sqrt{24}}{5} \\ & |\operatorname{Re}(\overline{x+i y+2})| \\ & =x+2= \pm \frac{\sqrt{24}}{5}=\frac{2 \sqrt{6}}{5} \end{aligned}$

$R=\{(a, b): a+5 b=42\}$

Then $R=\{(2,8),(7,7),(12,6),(17,5),(22,4),(27, 3),(32,2),(37,1)\}$

$\begin{aligned} & \text { and } \sum_{n=1}^{\substack{m=8}}\left(1-i^{n!}\right)=x+i y \\ & \therefore \sum_{n=1}^8\left(1-i^{n!}\right)=8-\left(i+i^2+i^6+1+1+1+1+1\right) \\ & =5-i \\ & \therefore x=5, y=-1 \\ & x+y+m=5-1+8=12 \end{aligned}$

$\begin{aligned} & \left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2 \\ & \left|z_1-2 z_2\right|=\left|1-2 z_1 \bar{z}_2\right| \\ & \Rightarrow\left(z_1-2 z_2\right)\left(\bar{z}_1-2 \bar{z}_2\right)=\left(1-2 z_1 \bar{z}_2\right)\left(1-2 \bar{z}_1 z_2\right) \\ & \Rightarrow\left|z_1\right|^2+4\left|z_2\right|^2-2 \bar{z}_1 z_2-2 \bar{z}_2 z_1 \\ & \quad=1+4\left|z_1\right|^2\left|z_2\right|^2-2 z_1 \bar{z}_2-2 \bar{z}_1 z_2 \\ & \Rightarrow\left|z_1\right|^2+4\left|z_2\right|^2-4\left|z_1\right|^2\left|z_2\right|^2-1=0 \\ & \Rightarrow\left(\left|z_1\right|^2-1\right)\left(1-4\left|z_2\right|^2\right)=0 \\ & \Rightarrow\left|z_1\right|=1 \text { and }\left|z_2\right|=\frac{1}{2} \end{aligned}$

$S_1=\{z \in C:|z| \leq 5\}$

$S_2=\operatorname{Im}\left(\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i}\right) \geq 0$

Take $z=x+i y$

$=\frac{x+i y+1-\sqrt{3} i}{1-\sqrt{3} i} \times \frac{1+\sqrt{3} i}{1+\sqrt{3} i}$

$\begin{aligned} = & \frac{x+i y+1-\sqrt{3} i+\sqrt{3} i x-\sqrt{3} y+\sqrt{3} i+3}{1+3} \\ & =y+\sqrt{3} x \geq 0 \\ S_3= & x \geq 0 \end{aligned}$

$\begin{aligned} & y+\sqrt{3} x=0 \\ & y=-\sqrt{3} x \\ & \text { Slope }=-\sqrt{3} \\ & 90+\theta=120^{\circ} \\ & \theta=30^{\circ} \\ & 90-\theta=60^{\circ} \end{aligned}$

In first quadrant we have angle $\frac{\pi}{2}$

so, total angle $90^{\circ}+60^{\circ}=150^{\circ}$

So, area $\frac{\pi r^2}{2 \pi} \times \frac{5 \pi}{6}=\frac{125 \pi}{12}$

$\begin{aligned} & \frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}=\lambda \\ & \Rightarrow a^2=\lambda^2|(y-z)|^2 \\ & b^2=\lambda^2|(z-x)|^2 \\ & c^2=\lambda^2|(x-y)|^2 \\ & \frac{a^2(\overline{y-z})}{(y-z)(y-z)}=\frac{a^2(\bar{y}-\bar{z})}{|y-z|^2}=\frac{a^2(\bar{y}-\bar{z})}{\frac{a^2}{\lambda^2}}=\lambda^2(\bar{y}-\bar{z}) \\ & \Rightarrow \sum\left(\frac{a^2}{y-z}\right)=\lambda^2(\bar{y}-\bar{z}+\bar{z}-\bar{x}+\bar{x}-\bar{y})=0 \neq 1 \\ \end{aligned}$

Statement I

$\begin{aligned} & \left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right) \\ & \Rightarrow \quad z_1=\left|z_1\right| e^{i \theta_1} \\ & \quad z_2=\left|z_2\right| e^{i \theta_2} \\ & \Rightarrow \frac{z_1}{\left|z_1\right|}=e^{i \theta_1} \\ & \Rightarrow \frac{z_2}{\left|z_2\right|}=e^{i \theta_2} \\ & \Rightarrow\left|e^{i \theta_1}+e^{i \theta_2}\right| \\ & \quad=\left|\sqrt{2+2 \cos \left(\theta_1-\theta_2\right)}\right| \\ & \quad=\left|2 \cos \left(\frac{\theta_1-\theta_2}{2}\right)\right| \leq 2 \end{aligned}$

$|z-1| \leq 2 \quad \Rightarrow \quad(x-1)^2+y^2=4$

$\begin{aligned} & z+\bar{z}+i(z-\bar{z}) \leq 2 \\ \Rightarrow \quad & x-y \leq 1 \\ & \operatorname{Im}(z) \geq 0 \\ \Rightarrow \quad & y \geq 0 \end{aligned}$

$\begin{aligned} & \text { Required area }=\left(\frac{\frac{3 \pi}{4}}{2 \pi}\right)(\pi)(2)^2 \\ & =\frac{3}{2} \pi \end{aligned}$

$\begin{aligned} & (\bar{z})^2+|z|=0 \quad \text{... (1)}\\ & z^2+|\bar{z}|=0 \quad \text{... (2)} \end{aligned}$

From equation (1) and (2)

$\begin{aligned} & \text { as }|z|=|\bar{z}| \\ & \Rightarrow \quad(\bar{z})^2=z^2 \\ & \Rightarrow \quad z=\bar{z} \text { or } z=-\bar{z} \\ & \Rightarrow \operatorname{Im}(z)=0 \text { or } \operatorname{Re}(z)=0 \end{aligned}$

Case I : If $\operatorname{Im}(z)=0$

$\Rightarrow z=x$

Putting value of $z$ in equation (1)

$\begin{aligned} & x^2+|x|=0 \\ & \Rightarrow x=0 \quad \text{[Rejected] } \end{aligned}$

Case II : If $\operatorname{Re}(z)=0$

$\Rightarrow z=i y$

Putting value of $z$ in equation (1)

$\begin{aligned} & -y^2+|y|=0 \\ & y= \pm 1 \text { as } y \neq 0 \end{aligned}$

Hence, $z= \pm i$ are the solution of the given equation

$\begin{aligned} & \Rightarrow \alpha=i-i=0 \\ & \text { and } \beta=i(-i)=1 \\ & \Rightarrow \quad 4\left(\alpha^2+\beta^2\right)=4(0+1) \\ & \quad=4 \end{aligned}$

$\therefore$ Option (3) is correct

To find the minimum value of $\left|z+\frac{1}{2}(3+4i)\right|$, where $z$ is a complex number with $|z| \leqslant 1$, we can think of this geometrically as the distance from any point inside or on the boundary of the unit circle in the complex plane to the fixed point $\frac{1}{2}(3+4i)$.

To make this more clear, first write the expression as follows: $\left|z+\frac{1}{2}(3+4i)\right| = \left|z+\frac{3}{2}+\frac{4}{2}i\right| = \left|z+1.5+2i\right|$

This means we are looking for the distance between any point on the complex plane represented by $z$ (where $z$ has a magnitude of up to 1) and the point $1.5+2i$.

The triangle inequality gives us the following relationship for any two complex numbers $z_1$ and $z_2$: $\left|z_1 + z_2\right| \geqslant \left| \left|z_1\right| - \left|z_2\right| \right|$

Applying this to our specific case ($z_1 = z$ and $z_2 = -1.5 - 2i$): $\left|z + 1.5 + 2i\right| \geqslant \left| \left|z\right| - \left|-1.5 - 2i\right| \right|$

Since $z$ is within the unit circle, $|z| \leqslant 1$ and $|-1.5 - 2i| = \sqrt{1.5^2 + 2^2} = \sqrt{2.25 + 4} = \sqrt{6.25} = 2.5$.

So now we have: $\left|z + 1.5 + 2i\right| \geqslant \left| 1 - 2.5 \right|$

Which simplifies to: $\left|z + 1.5 + 2i\right| \geqslant 1.5$

Therefore, the minimum value of $\left|z+\frac{1}{2}(3+4i)\right|$ given that $|z| \leqslant 1$ is $\frac{3}{2}$.

So the correct answer is:

Option C: $\frac{3}{2}$

$\begin{aligned} & z_1+z_2=5 \\ & z_1^3+z_2^3=20+15 i \\ & z_1^3+z_2^3=\left(z_1+z_2\right)^3-3 z_1 z_2\left(z_1+z_2\right) \\ & z_1^3+z_2^3=125-3 z_1 \cdot z_2(5) \\ & \Rightarrow 20+15 i=125-15 z_1 z_2 \\ & \Rightarrow 3 z_1 z_2=25-4-3 i \\ & \Rightarrow 3 z_1 z_2=21-3 i \\ & \Rightarrow z_1 \cdot z_2=7-i \\ & \Rightarrow\left(z_1+z_2\right)^2=25 \\ & \Rightarrow z_1^2+z_2^2=25-2(7-i) \\ & \Rightarrow 11+2 i \\ & \left(z_1^2+z_2^2\right)^2=121-4+44 i \\ & \Rightarrow z_1^4+z_2^4+2(7-i)^2=117+44 i \\ & \Rightarrow z_1^4+z_2^4=117+44 i-2(49-1-14 i) \\ & \Rightarrow\left|z_1^4+z_2^4\right|=75 \end{aligned}$

$\begin{array}{ll} \text { } & z^{1985}+z^{100}+1=0 \& z^3+2 z^2+2 z+1=0 \\ & (z+1)\left(z^2-z+1\right)+2 z(z+1)=0 \\ & (z+1)\left(z^2+z+1\right)=0 \\ \Rightarrow \quad & z=-1, \quad z=w, w^2 \\ & \text { Now putting } z=-1 \text { not satisfy } \\ & \text { Now put } z=w \\ \Rightarrow \quad & w^{1985}+w^{100}+1 \\ \Rightarrow \quad & w^2+w+1=0 \\ \Rightarrow \quad & \text { Also, } z=w^2 \\ \Rightarrow \quad & w^{3970}+w^{200}+1 \\ \Rightarrow & w+w^2+1=0 \end{array}$

Two common root

$\begin{aligned} & z^2=-i \bar{z} \\ & \left|z^2\right|=|i \bar{z}| \\ & \left|z^2\right|=|z| \\ & |z|^2-|z|=0 \\ & |z|(|z|-1)=0 \\ & |z|=0 \text { (not acceptable) } \\ & \therefore|z|=1 \\ & \therefore|z|^2=1 \end{aligned}$

$\begin{aligned} & z=2-i\left(2 \tan \frac{5 \pi}{8}\right)=x+i y(\text { let }) \\ & r=\sqrt{x^2+y^2} ~\& ~\theta=\tan ^{-1} \frac{y}{x} \\ & r=\sqrt{(2)^2+\left(2 \tan \frac{5 \pi}{8}\right)^2} \\ & =\left|2 \sec \frac{5 \pi}{8}\right|=\left|2 \sec \left(\pi-\frac{3 \pi}{8}\right)\right| \\ & =2 \sec \frac{3 \pi}{8} \\ & \& ~\theta = {\tan ^{ - 1}}\left( {{{ - 2\tan {{5\pi } \over 8}} \over 2}} \right) \\ & =\tan ^{-1}\left(\tan ^2\left(\pi-\frac{5 \pi}{8}\right)\right) \\ & =\frac{3 \pi}{8} \end{aligned}$

To begin with, let's analyze the given equation:

$|z+1|=\alpha z+\beta(1+i)$

First, we compute the modulus of the left side:

Let's take the given value of z,

$z = \frac{1}{2} - 2i$

Now, we find $z + 1$:

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

Then, the modulus of $z + 1$ is calculated as follows:

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

$|z + 1| = \sqrt{\frac{9}{4} + 4}$

$|z + 1| = \sqrt{\frac{9}{4} + \frac{16}{4}}$

$|z + 1| = \sqrt{\frac{25}{4}}$

$|z + 1| = \frac{5}{2}$

Now we need to equate the modulus to the right-hand side of the equation and solve for $\alpha$ and $\beta$. Let's rewrite the equation:

$\frac{5}{2} = \alpha z + \beta(1 + i)$

Substitute $z$ with its value:

$\frac{5}{2} = \alpha \left(\frac{1}{2} - 2i\right) + \beta(1+i)$

Rewrite the equation separating real and imaginary parts:

$\frac{5}{2} = \alpha \left(\frac{1}{2}\right) - 2\alpha i + \beta + \beta i$

$\frac{5}{2} = \left(\alpha \frac{1}{2} + \beta\right) + \left(-2\alpha + \beta\right)i$

For the above equality to hold, both real and imaginary parts must be equal.

Equating real parts:

$\alpha \frac{1}{2} + \beta = \frac{5}{2}$

Equating imaginary parts:

$-2\alpha + \beta = 0$

We now have a system of two linear equations:

1) $\frac{\alpha}{2} + \beta = \frac{5}{2}$

2) $-2\alpha + \beta = 0$

Let's solve the system by isolating $\beta$ from the second equation and then substituting it into the first one:

$\beta = 2\alpha$

Now substitute $\beta$ in the first equation:

$\frac{\alpha}{2} + 2\alpha = \frac{5}{2}$

$\alpha \left(\frac{1}{2} + 2\right) = \frac{5}{2}$

$\alpha \left(\frac{1}{2} + \frac{4}{2}\right) = \frac{5}{2}$

$\alpha \left(\frac{5}{2}\right) = \frac{5}{2}$

To find the value of $\alpha$, we divide both sides by $\frac{5}{2}$:

$\alpha = 1$

Now, we use the value of $\alpha$ to find $\beta$:

$\beta = 2\alpha$

$\beta = 2 \cdot 1$

$\beta = 2$

Finally, we add both $\alpha$ and $\beta$ to find $\alpha + \beta$:

$\alpha + \beta = 1 + 2 = 3$

The value of $\alpha + \beta$ is 3.

So, the correct answer is Option C) 3.

$|z-i|=|z+i|=|z-1|$

$\mathrm{ABC}$ is a triangle. Hence its circum-centre will be the only point whose distance from A, B, C will be same.

So $n(S)=1$