(3)

Given, |z| = 1

Now, $z_1=e^{i\frac{\mathrm{\pi }}{4}}=\frac{1}{\sqrt{2}}–\frac{i}{\sqrt{2}}=\frac{1–i}{\sqrt{2}}; z_2=e^{i0}=1$;

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

Now, $|z_1{\overline{z}}_2+z_2{\overline{z}}_3+z_3{\overline{z}}_1{|}^2$$=$$|\frac{1–i}{\sqrt{2}}·1+1·\frac{1–i}{\sqrt{2}}+{\left(\frac{1+i}{2}\right)}^2{|}^2$

$=$$|\sqrt{2}+(1–\sqrt{2})i{|}^2$$=5–2\sqrt{2}=\alpha +\beta \sqrt{2}$

$\Rightarrow \alpha =5 \mathrm{and} \beta =–2$          $\therefore {\alpha }^2+{\beta }^2=29$

(4)

Let z = x + iy, then from given equation, we have

(x + iy)(1 + i) + (x – iy)(1 – i) = 4

$\Rightarrow$ x + ix + iy – y + x – ix – iy – y = 4

$\Rightarrow$ 2x – 2y = 4 $\Rightarrow$ x – y = 2

Now, $|z–3|\le 1 \Rightarrow {(x–3)}^2+y^2\le 1$

$\beta$ = Area of shaded region = $\frac{\mathrm{\pi }{\left(1\right)}^2}{4}–\frac{1}{2}\times 1\times 1$

                                           = $(\frac{\mathrm{\pi }}{4}–\frac{1}{2})$ sq. units

$\alpha$ = Area of unshaded region inside the circle

   $=\frac{3}{4}\mathrm{\pi }{\left(1\right)}^2+\frac{1}{2}\times 1\times 1=(\frac{3\mathrm{\pi }}{4}+\frac{1}{2})$ sq. units

 Now, $|\alpha –\beta |$ = difference of area

 = $(\frac{3\mathrm{\pi }}{4}+\frac{1}{2})–(\frac{\mathrm{\pi }}{4}–\frac{1}{2})=\frac{\mathrm{\pi }}{2}+1$.

(1)

Given, $\left|\frac{\overline{z}–i}{2\overline{z}+i}\right|$$=\frac{1}{3} \mathrm{and} z\in C$

$\Rightarrow$$\left|\frac{\overline{z}–i}{\overline{z}+{\displaystyle \frac{i}{2}}}\right|$$=\frac{2}{3} \Rightarrow 3$$|x–iy–i|$$=2$$|x–iy+\frac{i}{2}|$

$\Rightarrow 3$$|x–i(y+1\left)\right|$$=2$$|x–i(y–\frac{1}{2}\left)\right|$

$\Rightarrow 3\sqrt{x^2+{(y+1)}^2}=2\sqrt{{\left(x\right)}^2+{(y–\frac{1}{2})}^2}$

Squaring on both sides, we get

$9(x^2+y^2+1+2y)=4(x^2+y^2+\frac{1}{4}–y)$

$\Rightarrow 9x^2+9y^2+9+18y=4x^2+4y^2+1–4y$

$\Rightarrow 5x^2+5y^2+22y+8=0$

$\Rightarrow x^2+y^2+\frac{22}{5}y+\frac{8}{5}=0$

Centre $\left(C\right)=(0,–\frac{11}{5})$

Now, $\frac{1}{2}$$|0+0+\alpha (0+\frac{11}{5}\left)\right|$$=11$          [Given]

$\Rightarrow \frac{11\alpha }{5}=22 \Rightarrow \alpha =10$          $\therefore {\alpha }^2=100$.

(1)

Let $z=e^{i\theta }$, then $\overline{z}=e^{–i\theta } \Rightarrow \frac{z}{\overline{z}}=e^{i2\theta }$

Also, |z| = 1

$\because$$|\frac{z}{\overline{z}}+\frac{\overline{z}}{z}|$$=1 \Rightarrow$$|e^{i2\theta }+e^{–i2\theta }|$$=1 \Rightarrow$$|\cos 2\theta |$$=\frac{1}{2}$

$\therefore$   Number of solutions = 8.

(2)

We have, $z_1=\sqrt{3}+2\sqrt{2}i$; $\sqrt{3}\left|z_2\right|=\left|z_1\right|$ and $\arg \left(z_2\right)=\arg \left(z_1\right)+\frac{\mathrm{\pi }}{6}$

$\Rightarrow \arg \left(z_2\right)–\arg \left(z_1\right)=\frac{\mathrm{\pi }}{6}$

$\therefore z_2=\frac{\left|z_2\right|}{\left|z_1\right|}·z_1{e}^{i(\mathrm{\pi }/6)}$

            $=\frac{1}{\sqrt{3}}\left[\frac{(\sqrt{3}+2\sqrt{2}i)(\sqrt{3}+i)}{2}\right]$

$\Rightarrow z_2=\frac{1}{2\sqrt{3}}[3–2\sqrt{2}+i(2\sqrt{6}+\sqrt{3}\left)\right]$

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

                      $=\frac{6+4\sqrt{6}i–3+2\sqrt{2}–2\sqrt{6}i–i\sqrt{3}}{2\sqrt{3}}$

                       $=\frac{3+2\sqrt{2}+i(2\sqrt{6}–\sqrt{3})}{2\sqrt{3}}$                      $\therefore$  $|z_1-z_2|$$=$$\left|z_2\right|$

$\Rightarrow$    $△$ABO is isosceles with angles $\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{6}$ and $\frac{2\mathrm{\pi }}{3}$.

$\therefore$   Area of $△$ABO = $\frac{1}{2}\sqrt{11}\times \frac{\sqrt{11}}{\sqrt{3}}\sin \frac{\mathrm{\pi }}{6}=\frac{11}{4\sqrt{3}}$.

(3)

$AB=\sqrt{{(8–2)}^2+{(2+6)}^2}=\sqrt{100}=10$

$\therefore$$|z_1–z_2{|}_{\min }$$=10–2–1=7$

(4)

$w_1=z_{1}i=(5+4i)i=-4+5i$ 

$w_2=z_2(-i)=(3+5i)(-i)=5-3i$

$w_1-w_2=-9+8i$

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

(1)

$z_0=(\frac{1}{2}+\frac{3i}{2})$ 

$z_1=1+i$

$z_1-z_0=(1+i)-(\frac{1}{2}+\frac{3i}{2})=\frac{1}{2}-\frac{i}{2}=\frac{1-i}{2}$

$|z_1-z_0|$$=\sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{1}{\sqrt{2}}$ 

$\text{As }|z_1-z_0|$$|z_2-z_0|$$=1$

$\Rightarrow \frac{1}{\sqrt{2}}$$|z_2-z_0|$$=1\Rightarrow$$|z_2-z_0|$$=\sqrt{2}$

Slope of line forming  $z_0,z_2=$ $\frac{\frac{3}{2}-1}{\frac{1}{2}-1}=\frac{\frac{1}{2}}{\frac{-1}{2}}=-1$

$\Rightarrow \tan \theta =-1=\tan 135°\Rightarrow \theta =135°$ 

$|z_2-z_0|$$=\sqrt{2}$

$z_2(\frac{1}{2}+\sqrt{2}\cos 135°,\frac{3}{2}+\sqrt{2}\sin 135°)$

or 

$z_2(\frac{1}{2}-\sqrt{2}\cos 135°,\frac{3}{2}-\sqrt{2}\sin 135°)$

$z_2(\frac{-1}{2},\frac{5}{2})$ or $z_2(\frac{3}{2},\frac{1}{2})$

$|z_2{|}^2=\frac{1}{4}+\frac{25}{4}=\frac{26}{4}$ or $|z_2{|}^2=\frac{9}{4}+\frac{1}{4}=\frac{5}{2};$ $\Rightarrow |z_2{|}_{\min }^2=\frac{5}{2}$

(1)

$\text{Centre of circle}$$\left|\frac{z-a}{z-b}\right|$$=k (\ne 1)\text{ is}$

$(\frac{k^{2}b-a}{k^2-1},0)$ and radius = $\frac{k|a-b|}{k^2-1}$ 

Now, centre of circle $\left|\frac{z-2}{z-3}\right|$$=2$ is $(\frac{4\times 3-2}{4-1},0)$ i.e., $(\frac{10}{3},0)$

and radius = $\frac{2\times 1}{3}=\frac{2}{3}$  $\therefore$  $\alpha =\frac{10}{3}, \beta =0, \gamma =\frac{2}{3}$

$3(\alpha +\beta +\gamma )=3(\frac{10}{3}+0+\frac{2}{3})=12$

(2)

We have, curve $C_1:$$\left|z\right|$$=\text{4 ∀ z}\in C$, which is a circle with centre (0, 0) and radius 4, therefore,

$x^2+y^2=16$                 $\cdots \left(i\right)$

$\text{Now, }z=4e^{i\theta }$

$z+\frac{1}{z}=4e^{i\theta }+\frac{1}{4e^{i\theta }}=4e^{i\theta }+\frac{1}{4}{e}^{-i\theta }$

$=4(\cos \theta +i\sin \theta )+\frac{1}{4}(\cos \theta -i\sin \theta )=\frac{17}{4}\cos \theta +i\frac{15}{4}\sin \theta$ 

$\text{To eliminate }\theta , \text{let }\alpha =\frac{17}{4}\cos \theta \text{and} \beta =\frac{15}{4}\sin \theta$

$\frac{{\alpha }^2}{{\left(\frac{17}{4}\right)}^2}+\frac{{\beta }^2}{{\left(\frac{15}{4}\right)}^2}=\frac{{\left(\frac{17}{4}\right)}^2{\cos }^2\theta }{{\left(\frac{17}{4}\right)}^2}+\frac{{\left(\frac{15}{4}\right)}^2{\sin }^2\theta }{{\left(\frac{15}{4}\right)}^2}=1$

$\therefore$   $\text{The curve }{C}_2\text{ is }\frac{x^2}{{\left(\frac{17}{4}\right)}^2}+\frac{y^2}{{\left(\frac{15}{4}\right)}^2}=1$        $\cdots \left(ii\right)$

which is an ellipse with centre (0, 0). 

From (i) and (ii),

Hence, the curves $C_1$ and $C_2$ intersect at 4 points.