(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 $\overline{z}=e^{–i\theta } \Rightarrow \frac{z}{z}=e^{i2\theta }$

Also, |z| = 1

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

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

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

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

$\Rightarrow$    $△$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)

$AB=\sqrt{{(8–2)}^2+{(2+6)}^2}=\sqrt{100}=0$

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