MATHS MCQ

Circle : $(x+2)^2+(y-3)^2=1$

$L_1: x+y-1=0 $

$L_2: x-y+4=0$

On solving circle $\mathrm{L}_1$ we get

$z_1:\left(-2-\frac{1}{\sqrt{2}}, 3+\frac{1}{\sqrt{2}}\right)$

On solving $\mathrm{L}_1$ and $\mathrm{z}_2$ is intersection of line $\mathrm{L}_1 \& \mathrm{~L}_2$ we get $z_2:\left(\frac{-3}{2}, \frac{5}{2}\right)$

$\left|z_1\right|^2+2\left|z_2\right|^2=14+5 \sqrt{2}+17$

$=31+5 \sqrt{2}$

So  $\alpha=31$

$\beta=5$

$\alpha+\beta=36$

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

$ (z+\bar{z})+i(z-\bar{z}) \leq 2 \Rightarrow 2 x+i(2 i y) \leq 2 $

$ \Rightarrow x-y \leq 1 $

$ \operatorname{Im}(z) \geq 0 \Rightarrow y \geq 0$

$Image$

Required area

Area of semi-circle - area of sector A

$ \frac{1}{2} \pi(2)^2-\frac{\pi}{2} $

$ =\frac{3 \pi}{2}$

$ S_2: \operatorname{Im} \text { of } \frac{z+(1-\sqrt{3} i)}{(1-\sqrt{3} i)} \geq 0 $

$ \operatorname{Im} \text { of }\left(\frac{x+i y}{1-\sqrt{3} i}+1\right) \geq 0 $

$ \operatorname{Im} \text { of }\left(\frac{(x+i y)(1+\sqrt{3} i)}{4}\right) \geq 0 $

$ \Rightarrow \sqrt{3} x+y \geq 0 $

$ S_3: x \geq 0 $

$ \text { Area }=\frac{5}{12}\left(\pi(5)^2\right)$

$ \left|z_1\right|^2 2 z_1 \bar{z}_2-2 \bar{z}_1 z_2+4\left|z_2\right|^2 $

$ =4\left(\frac{1}{4}-\frac{\bar{z}_1 z_2}{2}-\frac{z_1 \bar{z}_2}{2}+\left|z_1\right|^2\left|z_2\right|^2\right)$

$\mathrm{z}_1 \overline{\mathrm{z}}_1+2 \mathrm{z}_2 \cdot 2 \overline{\mathrm{z}}_2-\mathrm{z}_1 \overline{\mathrm{z}}_1 2 \mathrm{z}_2 2 \overline{\mathrm{z}}_2-1=0$

$ \left(z, \bar{z}_1-1\right)\left(1-2 z_2 \cdot 2 \bar{z}_2\right)=0 $

$ \left(\left|z_1\right|^2-1\right)\left(\left|2 z_2\right|^2-1\right)=0$

$ \Rightarrow|(x-1)+i y| \leq 1 $

$ \Rightarrow \sqrt{(x-1)^2+y^2} \leq 1 $

$ \Rightarrow(x-1)^2+y^2 \leq 1 ............(1)$

Also $|z-5| \leq|z-5 i|$

$ (x-5)^2+y^2 \leq x^2+(y-5)^2 $

$ -10 x \leq-10 y $

$ \Rightarrow x \geq y.........(2)$

Solving ($1$) and ($2$)

$ \Rightarrow(\mathrm{x}-1)^2+\mathrm{x}^2=1 $

$ \Rightarrow 2 \mathrm{x}^2-2 \mathrm{x}=0 $

$ \Rightarrow \mathrm{x}(\mathrm{x}-1)=0 $

$ \Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}=1 $

$ \mathrm{y}=0 \text { or } \mathrm{y}=1$

$Image$

Given $\mathrm{x}, \mathrm{y} \in \mathrm{I}$

Points $(0,0),(1,0),(2,0),(1,1),(1,-1)$ to find

$\left|z_1\right|^2+\left|z_2\right|^2+\left|z_3\right|^2+\left|z_4\right|^2+\left|z_5\right|^2 $

$ =0+1+4+1+1+1+1=9$

$ z \bar{z}-2 i \bar{z}-2 i z+4(-1) $

$ +z \bar{z}+2 z i+2 \bar{z} i+4(-1)=0 $

$ \Rightarrow 2|z|^2=8 \Rightarrow|z|=2 $

$ |z-(6+8 i)|_{\text {maximum }}=10+2=12$

Let $\alpha=\omega$

Now $(1+\alpha)^7=-\omega^{14}=-\omega^2=1+\omega$

$ A=1, B=1, C=0 $

$ \therefore 5(3 A-2 B-C)=5(3-2-0)=5$

$ x=\frac{\sqrt{6} \pm i \sqrt{6}}{2}=\frac{\sqrt{6}}{2}(1 \pm i) $

$ \alpha=\sqrt{3}\left(e^{i \frac{\pi}{4}}\right), \beta=\sqrt{3}\left(e^{-i \frac{\pi}{4}}\right) $

$ \therefore \frac{\alpha^{99}}{\beta}+\alpha^{98}=\alpha^{98}\left(\frac{\alpha}{\beta}+1\right) $

$ =\frac{\alpha^{98}(\alpha+\beta)}{\beta}=3^{49}\left(e^{i 99 \frac{\pi}{4}}\right) \times \sqrt{2} $

$ =3^{49}(-1+i) $

$ =3^n(a+i b) $

$ \therefore n=49, a=-1, b=1$

$ \therefore n+a+b=49-1+1=49$

$(z+1)\left(z^2-z+1\right)+2 z(z+1)=0$

 $(z+1)\left(z^2+z+1\right)=0$

$\begin{aligned} & (\mathrm{z}+1)\left(\mathrm{z}^2+\mathrm{z}+1\right)=0 \\ \Rightarrow \quad & \mathrm{z}=-1, \mathrm{z}=\mathrm{w}, \mathrm{w}^2\end{aligned}$

Now put $\mathrm{z}=\mathrm{w}$

$\begin{array}{ll}\Rightarrow & \mathrm{w}^{1985}+\mathrm{w}^{100}+1 \\ \Rightarrow & \mathrm{w}^2+\mathrm{w}+1=0 \\ & \text { Also, } \mathrm{z}=\mathrm{w}^2 \\ \Rightarrow & \mathrm{w}^{3970}+\mathrm{w}^{200}+1 \\ \Rightarrow & \mathrm{w}+\mathrm{w}^2+1=0\end{array}$

Two common root

$\therefore \frac{2 z-3 i}{2 z+i}+\frac{2 \bar{z}+3 i}{2 \bar{z}-i}=0$

$z=x+i y$

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

Given that $x+y^2=0$

$y^4+y^2-y=3 / 4$

$\frac{2(x+i y)-3 i}{4(x+i t)+2 i}=\frac{2 x+(2 y-3) i}{4 x+(4 y+2) i} \times \frac{4 x-(4 y+2) i}{4 x-(4 y+2) i}$

$4 x(2 y-3)-2 x(4 y+2)=0$

$x=0 \quad y \neq-\frac{1}{2}$

$\text { Ans. }=3$

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

$\operatorname{Re}(z)=0$

$\frac{1-2 \sin ^2 \theta}{1+\sin ^2 \theta}=0$

$\sin \theta=\frac{ \pm 1}{\sqrt{2}}$

$A=\left\{\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\right\}$

$\text { sum }=4 \pi(\text { Option } 3)$

$z _2= x _2+ iy _2$

$\operatorname{Re}\left(z_1 z_2\right)=x_1 x_2-y_1 y_2=0$

$\operatorname{Re}\left(z_1+z_2\right)=x_1+x_2=0$

$x_1$ and $x_2$ are of opposite sign

$y_1$ and $y_2$ are of opposite sign

$z=x+i y$

$a z=(8 x+14 y)+i(-14 x+8 y)$

$z +\overline{ z }=2 x \quad z -\overline{ z }=2 iy$

Set A: $\frac{2 i(-14 x+8 y)}{i(4 x y-112)}=1$

$(x-4)(y+7)=0$

$x=4 \quad \text { or } \quad y=-7$

Set B: $x^2+(y+3)^2=16$

when $x=4 \quad y=-3$

when $y =-7 \quad x =0$

$\therefore A \cap B=\{4-3 i, 0-7 i\}$

So, $\sum_{z \in A \cap B}(\operatorname{Re} z-\operatorname{Im} z)=4-(-3)+(0-(-7))=14$

$z_1=\frac{1+i \bar{z}}{\bar{z}(1-z)+\frac{1}{z}}$

$z_1=\frac{1+i(1-i)}{(1-i)(1-1-i)+\frac{1}{1+i}}$

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

$=\frac{2+i}{-3 i-1}=\frac{4+2 i}{-3 i-1}$

$=\frac{-(4+2 i)(3 i-1)}{(3 i)^2-(1)^2}$

$\therefore \frac{12}{\pi} \arg \left(z_1\right)=\frac{12}{\pi} \times \frac{3 \pi}{4}=9$

$a z^2+b z+a \bar{z}^2+b \bar{z}=2 a$

$a\left(z^2+\bar{z}^2\right)+ b ( z +\overline{ z })=2 a$

$\operatorname{Re}\left( bz z^2+ az \right)= b$

$b z^2+a z+b \bar{z}^2+ az =2 b$

$b \left( z ^2+\overline{ z }^2\right)+ a ( z +\overline{ z })=2 b$

$(1) \times b-(2) \times(a)$

$\Rightarrow\left(b^2-a^2\right)(z+\bar{z})=0$

$\Rightarrow \quad(z+\bar{z})=0 \quad\left(a^2 \neq b^2\right)$

$(1) \times a-(2) \times(b)$

$\Rightarrow \quad\left(a^2-b^2\right)(z+\bar{z})=2\left(a^2-b^2\right) \quad\left(a^2 \neq b^2\right)$

$z^2+\bar{z}^2=2$

$\Rightarrow(z+\bar{z})^2-2 z \bar{z}=2$

$z \bar{z}=-1$

$\Rightarrow 1+1^2=-1$

$\Rightarrow \text { No solution }$

$\text { But when } a=-b$

$\operatorname{Re}\left(a z^2-a z\right)=a$

$\Rightarrow \quad \operatorname{Re}\left(a\left(x^2-y^2+i 2 x y\right)-a(x+i y)\right)=a$

$\Rightarrow a\left(x^2-y^2\right)-a x=a$

$\Rightarrow x^2-y^2-x=1$

$\Rightarrow x^2-x-1=y^2$

For any real values of $y$ there two values of $x$, hence infinite complex numbers are possible.

$=\alpha+i \beta+4+4 i$

$\sqrt{(\alpha+2)^2+\beta^2}=(\alpha+4)+ i (\beta+4)$

$\beta+4=0$

$(\alpha+2)^2+16=(\alpha+4)^2$

$\beta=-4$

$\alpha^2+4+4 \alpha+16=\alpha^2+16+8 \alpha$

$4=4 \alpha$

$\alpha=1$

$\alpha=1, \beta=-4$

$\alpha+\beta=-3, \alpha \beta=-4$

$\text { Sum of roots }=-7$

$\text { Product of roots }=12$

$x^2+7 x+12=0$

Now $\operatorname{Re}(a+\bar{z}) > \operatorname{Im}(\bar{a}+z)$

$\therefore x _1+ x >- y _1+ y$

$x _1=2, y _1=10, x =-12, y =0$

Given inequality is not valid for these values.

$S 1$ is false.

Now $\operatorname{Re}(a+\bar{z})<\operatorname{Im}(\bar{a}+z)$

$x _1+ x < - y _1+ y$

$x _1=-2, y _1=-10, x =12, y =0$

Given inequality is not valid for these values.

$S2$ is false.

$x-i y=i\left(x^2-y^2+(2 x y) i+x\right)$

$x =- 2 x x$

$- y =- y ^2+ x ^2+ x$

$\Rightarrow x=0, y=-\frac{1}{2}(\text { from }(1))$

If $x \neq 0$, then $y =0,1$

If $y =-\frac{1}{2}$, then $x =\frac{1}{2},-\frac{3}{2}$

$Z =0+ i 0,0+ i , \frac{1}{2}-\frac{ i }{2},-\frac{3}{2}-\frac{ i }{2}$

Let $z=3+i y$

$\bar{z}=3-i y$

$z_1=\frac{2 i y+\left(9+y^2\right)}{2-3(3+i y)+5(3-i y)}$

$=\frac{9+y^2+i(2 y)}{8-8 i y}$

$=\frac{\left(9+y^2\right)+i(2 y)}{8(1-i y)}$

$\operatorname{Re}\left(z_1\right)=\frac{\left(9+y^2\right)-2 y^2}{8\left(1+y^2\right)}$

$=\frac{9-y^2}{8\left(1+y^2\right)}$

$=\frac{1}{8}\left[\frac{10-\left(1+ y ^2\right)}{\left(1+ y ^2\right)}\right]$

$=\frac{1}{8}\left[\frac{10}{1+ y ^2}-1\right]$

$1+ y ^2 \in[1, \infty]$

$\frac{1}{1+ y ^2} \in(0,1]$

$\frac{10}{1+ y ^2} \in(0,10]$

$\frac{10}{1+ y ^2}-1 \in(-1,9]$

$\operatorname{Re}\left( z _1\right) \in\left(\frac{-1}{8}, \frac{9}{8}\right]$

$\alpha=\frac{-1}{8}, \beta=\frac{9}{8}$

$24(\beta-\alpha)=24\left(\frac{9}{8}+\frac{1}{8}\right)=30$

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

Apply polar form,

$r \cos \theta=\frac{\sqrt{3}-1}{2}r$

$\sin \theta=\frac{\sqrt{3}+1}{2}$

Now, $\tan \theta=\frac{\sqrt{3}+1}{\sqrt{3}-1}$

So, $\quad \theta=\frac{5 \pi}{12}$

$\Rightarrow x + y =7$

$z \overline{ z }+4+2 i ( z -\overline{ z })=4( z \overline{ z }+1+ i (\overline{ z }- z ))$

$3 z \overline{ z }-6 i ( z -\overline{ z })=0$

$x ^2+ y ^2-2 i (2 iy )=0$

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

$\Rightarrow w =\frac{17}{4} \cos \theta+ i \frac{15}{4} \sin \theta$

So locus of $w$ is ellipse $\frac{x^2}{\left(\frac{17}{4}\right)^2}+\frac{y^2}{\left(\frac{15}{4}\right)^2}=1$

Locus of $z$ is circle $x ^2+ y ^2=16$

So intersect at $4$ points

$=x^2+y^2-4 x+4=4 x^2+4 y^2-24 x+36$

$=3 x^2+3 y^2-20 x+32=0$

$=x^2+y^2-\frac{20}{3} x +\frac{32}{3}=0$

$=(\alpha, \beta)=\left(\frac{10}{3}, 0\right)$

$\gamma=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3}$

$3(\alpha, \beta, \gamma)=3\left(\frac{10}{3}+\frac{2}{3}\right)$

$=12$

$|1+a i l=| b+i l$

$a^2+1=b^2+1 \Rightarrow a=\pm b \Rightarrow b=-a \quad \text { as } a b < 0$

$(a, b) \text { lies on }| z -1|=|2 z |$

$la + ib -1 \mid=2 l a + ibl$

$( a -1)^2+ b ^2=4\left( a ^2+ b ^2\right)$

$( a -1)^2= a ^2=4\left(2 a ^2\right)$

$1-2 a =6 a ^2 \Rightarrow 6 a ^2+2 a -1=0$

$a =\frac{-2 \pm \sqrt{28}}{12}=\frac{-1 \pm \sqrt{7}}{6}$

$a =\frac{\sqrt{7}-1}{6}$ and $b =\frac{1-\sqrt{7}}{6}$

${[ a ]=0}$

$\therefore \frac{1+[ a ]}{4 b }=\frac{6}{4(1-\sqrt{7})}=-\left(\frac{1+\sqrt{7}}{4}\right.$

$\text { or }[ a ]=0$

Similarly it is not matching with $a =\frac{-1-\sqrt{7}}{6}$

No answer is matching.

$\left|z-z_1\right|^2+\left|z-z_2\right|^2=\left|z_1-z_2\right|^2$

$r =\frac{\left|z_1-z_2\right|}{2}=\frac{|\alpha-\beta|}{2}=\sqrt{\lambda-1}$

$2 \lambda=|\alpha-\beta|^2$

$|\alpha-\beta|=2 \sqrt{\lambda-1}$

$|\alpha-\beta|^2=4 \lambda-4=2 \lambda$

$\lambda=2$

$\Rightarrow|\alpha-\beta|^2=4$

$|\alpha-\beta|=2$

$W _2= z _2(- i )=(3+5 i )(- i )=5-3 i$

$W _1- W _2=-9+8 i$

$\text { Principal argument }=\pi-\tan ^{-1}\left(\frac{8}{9}\right)$

$\operatorname{Re}(w)=x^2+y^2+k_1 x-k_2 y+\lambda=0$

$\text { Centre } \equiv\left(\frac{-k_1}{2}, \frac{k_2}{2}\right) \equiv(1,2)$

$\Rightarrow k_1=-2, k_2=4$

$\text { radius }=1 \Rightarrow \lambda=4$

$\operatorname{Im}=k_1 y+k_2 x+\lambda=0$

$\therefore 2 x-y+2=0$

$d=\frac{2}{\sqrt{5}}$

$\frac{1^2}{4}=1-\frac{4}{5}=\frac{1}{5}$

$\therefore 301^2=24$

$2^{200}\left(\cos \frac{\pi}{3}- i \sin \frac{\pi}{3}\right)^{200}=2^{199}( p + iq )$ $2\left(-\frac{1}{2}- i \frac{\sqrt{3}}{2}\right)= p + iq$

$p =-1, q =-\sqrt{3}$

$\alpha= p + q + q ^2=2-\sqrt{3}$

$\beta= p - q + q ^2=2+\sqrt{3}$

$\alpha+\beta=4$

$\alpha \cdot \beta=1$

equation $x^2-4 x+1=0$

$\left(\frac{1+ z }{1+\bar{z}}\right)^3=\left(\frac{1+ z }{1+\frac{1}{ z }}\right)^3= z ^3$

$\Rightarrow\left( i \left(\cos \frac{2 \pi}{9}- i \sin \frac{2 \pi}{9}\right)\right)^3$

$=- i \left(\cos \frac{2 \pi}{3}- i \sin \frac{2 \pi}{3}\right)=- i \left(\frac{-1}{2}- i \frac{\sqrt{3}}{2}\right)$

$\Rightarrow \frac{-1}{2}(\sqrt{3}-i)$

$=\sqrt{3} e ^{ \pm \frac{3 \pi i }{4}}$

Required expression

$=\frac{(\sqrt{3})^{23}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{14}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{15}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{10}\left(2 \cos \frac{30 \pi}{4}\right)}$

$(\sqrt{3})^8=81$

$x=\frac{\sqrt{2} \pm \sqrt{2-8}}{2}=\frac{\sqrt{2} \pm \sqrt{6} i}{2}$

$\alpha=\frac{\sqrt{2}+\sqrt{6} i}{2}=\sqrt{2} e^{\frac{i \pi}{3}}$ $\beta=\sqrt{2} e^{\frac{-i \pi}{3}}$

$\alpha^{14}=2^7 e^{\frac{i 14 \pi}{3}}=128\left[e^{\frac{i 2 \pi}{3}}\right]$

$\beta^{14}=128\left[e^{\frac{-i 2 \pi}{3}}\right]$

$\alpha^{14}+\beta^{14}=128(2) \cos \left(\frac{2 \pi}{3}\right)=-128$

$\Rightarrow 1+\frac{(11 i z-13)}{\left(z^2-3 i z-2\right)} \in R$

Put $z=\alpha-\frac{13}{11} i$

$\Rightarrow\left(z^2-3 i z-2\right)$ is imaginary

Put $z=x+i y$

$\Rightarrow\left(x^2-y^2+2 x y i-3 i x+3 y-2\right) \in \text { Imaginary }$

$\Rightarrow \operatorname{Re}\left(x^2-y^2+3 y-2+(2 x y-3 x) i\right)=0$

$\Rightarrow x^2-y^2+3 y-2=0$

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

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

$x^2=(y-1)(y-2) \therefore z=\alpha-\frac{13}{11} i$

Put $x=\alpha, y=\frac{-13}{11}$

$\alpha^2=\left(\frac{-13}{11}-1\right)\left(\frac{-13}{11}-2\right)$

$\alpha^2=\frac{(24 \times 35)}{121}$

$242 \alpha^2=48 \times 35=1680$

$\Rightarrow\left|z_2-z_{\circ}\right|=\sqrt{2} ; \text { centre }\left(\frac{1}{2}, \frac{3}{2}\right)$

$z_{0}\left(\frac{1}{2}, \frac{3}{2}\right) \text { and } z_1(1,1)$

$\tan \theta=-1 \Rightarrow \theta=135^{\circ}$

$z_2\left(\frac{1}{2}+\sqrt{2} \cos 135^{\circ}, \frac{3}{2}+\sqrt{2} \sin 135^{\circ}\right)$

or

$\left(\frac{1}{2}-\sqrt{2} \cos 135^{\circ}, \frac{3}{2}-\sqrt{2} \sin 135^{\circ}\right)$

$\Rightarrow z_2\left(-\frac{1}{2}, \frac{5}{2}\right) \text { or } z_2\left(\frac{3}{2}, \frac{1}{2}\right)$

$\Rightarrow\left|z_2\right|^2=\frac{26}{4}, \frac{5}{2}$

$\Rightarrow\left|z_2\right|_{\min }^2=\frac{5}{2}$

$z _{1}=- iz _{2}$

$\arg \left(\frac{ z _{1}}{\overline{ z }_{2}}\right)=\pi$

$\arg \left(- i \frac{ z _{2}}{\overline{ z }_{2}}\right)=\pi \quad \arg \left( z _{2}\right)=\theta$

$-\frac{\pi}{2}+\theta+\theta=\pi$

$2 \theta=\frac{3 \pi}{2}$

$\arg \left( z _{2}\right)=\theta=\frac{3 \pi}{4}, \arg z _{1}=\frac{\pi}{4}$

$=2\left({ }^{5} C_{0} 2^{5}+{ }^{5} C_{2} 2^{3}(3 i)^{2}+{ }^{5} C_{4} 2^{1}(3 i)^{4}\right)$

$=2(32+10 \times 8(-9)+5 \times 2 \times 81)=244$

$E: \frac{(x-2)^{2}}{9}+\frac{y^{2}}{5}=1$

Lower Extremity of vertical diameter of circle $\rightarrow(4,1)$

Put in ellipse $\Rightarrow \frac{(4-2)^{2}}{9}+\frac{1}{5}-1$

$=\frac{4}{9}+\frac{1}{5}-1$

$=\frac{29}{45}-1<0$

Two Solutions

Answer (C)

For minimising, $z$ should lie on $A B$ and $A B=5 \sqrt{2}$

$(A B)^{2}=18+2 p^{2}$

$p=\pm 4$

$\Rightarrow\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)=0$

$\Rightarrow x=\pm \;\omega, \pm \;\omega^{2}$ where $\omega=1^{1 / 3}$ and imaginary.

So $\alpha^{1011}+\alpha^{2002}-\alpha^{3033}=1+1-1=1$

Hence $\alpha^{8}=\beta^{8}$

$\left|\alpha^{8}+\beta^{8}\right|=\left|2 \alpha^{8}\right|=2\left|\alpha^{2}\right|^{4}$

$=2 \sqrt{5}^{4}=50$

Consider $z=x+i y$

$2 x=(i+1)\left(x^{2}-y^{2}+2 x y i\right)$

$\Rightarrow 2 x=x^{2}-y^{2}-2 x y \text { and } x^{2}-y^{2}+2 x y=0$

$\Rightarrow 2 x=-4 x y$

$\Rightarrow x=0 \text { or } y=\frac{-1}{2}$

Case $1: x=0 \Rightarrow y=0$ here $z=0$

$\text { Case } 2: y=\frac{-1}{2}$

$\Rightarrow 4 x^{2}-4 x-1=0$

$(2 x-1)^{2}=2$

$2 x-1=\pm \sqrt{2}$

$x=\frac{1 \pm \sqrt{2}}{2}$

Here $z =\frac{1+\sqrt{2}}{2}-\frac{ i }{2}$ or $z =\frac{1-\sqrt{2}}{2}-\frac{ i }{2}$

Sum of squares of modulus of $z$

$=0+\frac{(1+\sqrt{2})^{2}+1}{4}+\frac{(1-\sqrt{2})^{2}+1}{4}=\frac{8}{4}=2$

$\left|z_{2}+\right| z_{2}-1||\left(\bar{z}_{2}+\left|z_{2}-1\right|\right)=\left(z_{2}-\left|z_{2}+1\right|\right)\left(\bar{z}_{2}-\left(z_{2}+1\right)\right)$

$z_{2}\left|\bar{z}_{2}+12_{2}-1\right|-\left(\bar{z}_{2}-\left|z_{2}+1\right|\right)+\bar{z}_{2}\left(\left|z_{2}-1\right|+\left|z_{2}+1\right|\right)$

$=\left|z_{2}+1\right|^{2}=\left|z_{2}-1\right|^{2}$

${\left[z_{2}+\bar{z}_{2}\right)\left(\left|z_{2}-1\right|\right)+\left(z_{2}+1 \mid\right)=2\left(z_{2}+\bar{z}_{2}\right) }$

$\left(z_{2}+\bar{z}_{2}\right)\left(\left|z_{2}-1\right|+\left|z_{2}+1\right|-2\right)=0$

$z_{2}+\bar{z}_{2}=0$ or $\left|z_{2}-1\right|+\left|z_{2}+1\right|-2=0$

$\therefore z_{2}+\bar{z}_{2}=0 \text { or }\left|z_{2}-1\right|+\left|z_{2}+1\right|-2=0$

$\therefore z _{2}$ lie on imaginary axis. Or on real axis with in $[-1,1]$

Also $\left|z_{1}-3\right|=\frac{1}{2}$ lie on circle having centre 3 and radius $\frac{1}{2}$

Clearly $\left|z_{1}-z_{2}\right| \min =\frac{5}{2}-1=\frac{3}{2}$

|| $z \left|-\frac{1}{| z |}\right| \leq\left| z -\frac{1}{ z }\right| \leq| z |+\frac{1}{| z |}$

$\left| r -\frac{1}{ r }\right| \leq 2 \leq r +\frac{1}{ r }$

$\left| r -\frac{1}{ r }\right| \leq 2 \& r +\frac{1}{ r } \geq 2$ always true

$r -\frac{1}{ r } \geq-2 \& r -\frac{1}{ r } \leq 2$

$r ^{2}-1 \leq 2 r$

$r ^{2}-2 r \leq 1$

$( r -1)^{2} \leq 2$

$r -1 \leq \sqrt{2}$

$\left.\therefore z \right|_{\max }=1+\sqrt{2}$

$|x+(y-1) i|=|x+(y+5) i|$

$x^{2}+(y-1)^{2}=x^{2}+(y+5)^{2}$

$(y-1)^{2}-(y+5)^{2}=0$

$(2 y+4)(-6)=0$

$y=-2$

$\therefore x^{2}+(-2)^{2}=4$

$x=0$

$Z \equiv(0,-2)$, check options

Let $z = x + iy$

$z ^{2}= x ^{2}- y ^{2}+2 ixy$

$\bar{z}=x-i y$

$z^{2}+\bar{z}=x^{2}-y^{2}+x+i(2 x y-y)=0$

$x^{2}+x-y^{2}=0 \text { \& } 2 x y-y=0$

$y=0$ or $x=\frac{1}{2}$

If $y =0 ; x =0,-1$

If $x=\frac{1}{2} ; y=\frac{\sqrt{3}}{2}, \frac{-\sqrt{3}}{2}$

$\sum_{z \in S}(\operatorname{Re}(z)+\operatorname{Im}(z)=\left(0-1+\frac{1}{2}+\frac{1}{2}\right)$

$+0+0+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2})$

$4 ix +(1+ i ) y =0$ and

$8\left(\cos \frac{2 \pi}{3}+ i \sin \frac{2 \pi}{3}\right) x +\overline{ a } y =0$

$\left|\begin{array}{cc}4 i \quad 1+ i \\ 8 e ^{ i 2 \pi / 3} \quad \overline{ a }\end{array}\right|=0$

$\Rightarrow 4 i \overline{ a }-(1+ i ) 8 e ^{ i 2 \pi / 3}=0$

$\Rightarrow 4 i (\alpha+ i \beta)-8(1+ i )\left(\frac{-1+ i \sqrt{3}}{2}\right)=0$

$\Rightarrow i \alpha-\beta+1+\sqrt{3}+ i (1-\sqrt{3})=0$

$\Rightarrow \beta=\sqrt{3}+1$

$\alpha=\sqrt{3}-1$

So, $\frac{\alpha}{\beta}$

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

represent pt. i/s circle of radius $1 $ centred at $(3,0)$

$z(4+3 i)+\bar{z}(4-3 i) \leq 24$

$(x+i y)(4+3 i)+(x-i y)(4-3 i) \leq 24$

$4 x+3 x i+4 i y-3 y+4 x-3 i x-4 i y-3 y \leq 24$

$8 x-6 y \leq 24$

$4 x-3 y \leq 12$

minimum of $(0,4)$ from circle $=\sqrt{3^{2}+4^{2}}-1=4$ will lie along line joining $(0,4) \&(3,0)$

$\therefore$ equation line

$\frac{x}{3}+\frac{y}{4}=1 \Rightarrow 4 x+3 y=12 \ldots \text { (i) }$

equation circle $(x-3)^{2}+y^{2}=1 \ldots$ $(ii)$

$\left(\frac{12-3 y}{4}-3\right)^{2}+y^{2}=1$

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

$\frac{25 y^{2}}{16}=1 \Rightarrow y=\pm \frac{4}{5}$

for minimum distance $y=\frac{4}{5}$

$\therefore x=\frac{12}{5}$

$\therefore 25(\alpha+\beta)=25\left(\frac{4}{5}+\frac{12}{5}\right)$

$=16 \times 5=80$

$B=\{z \in A:|z-(1-i)|=1\}$

$A \cap B$ has infinite set