(2)

For a circle, we have, $|z-z_1{|}^2+|z-z_2{|}^2=|z_1-z_2{|}^2$

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

$\text{Also, }2\lambda =|\alpha -\beta {|}^2$

$\Rightarrow 2\lambda =4(\lambda -1)\Rightarrow 2\lambda =4\lambda -4 \Rightarrow 2\lambda =4\Rightarrow \lambda =2$

$\therefore$  $|\alpha -\beta |$$=2$

(24)

We have, $w=z\overline{z}+k_{1}z+k_{2}iz+\lambda (1+i)$

Put $z=x+iy$

$w=x^2+y^2+k_1(x+iy)+k_{2}i(x+iy)+\lambda (1+i)$

$=x^2+y^2+k_{1}x+k_{1}iy+k_{2}ix-k_{2}y+\lambda +\lambda i$

$=(x^2+y^2+k_{1}x-k_{2}y+\lambda )+i(k_{1}y+k_{2}x+\lambda )$

Given $\text{Re}\left(w\right)=0$

$\Rightarrow x^2+y^2+k_{1}x-k_{2}y+\lambda =0$               $\text{...(i)}$

$\text{Centre}=(\frac{{\displaystyle -k_1}}{{\displaystyle 2}},\frac{{\displaystyle k_2}}{{\displaystyle 2}})$

$\text{Radius}=\sqrt{\frac{k_1^2}{4}+\frac{k_2^2}{4}-\lambda }$

Now, $\frac{-k_1}{2}=1\Rightarrow k_1=-2$

and $\frac{k_2}{2}=2\Rightarrow k_2=4$

Also, $\sqrt{\frac{k_1^2}{4}+\frac{k_2^2}{4}-\lambda }=1$

$\Rightarrow \frac{k_1^2}{4}+\frac{k_2^2}{4}-\lambda =1$

$\Rightarrow 1+4-\lambda =1\Rightarrow \lambda =4$

Put $k_1=-2, k_2=4, \lambda =4$ in (i)

$x^2+y^2-2x-4y+4=0$                $\text{...(ii)}$

Now, $\text{Im}\left(w\right)=0,$ $k_{2}x+k_{1}y+\lambda =0$

$4x-2y+4=0$                                  $\text{...(iii)}$

From (ii) and (iii), $A\equiv (0,2)$ and $B\equiv (0.4,2.8)$

Now, $AB=\sqrt{{(0.4-0)}^2+{(2.8-2)}^2}$

$\Rightarrow A{B}^2={(0.4)}^2+{(0.8)}^2=0.8$

$\therefore$   $30{\left(AB\right)}^2=30(0.8)=24$

(9)

We have $z=1+i$ and $z_1=\frac{1+i\overline{z}}{\overline{z}(1-z)+\frac{1}{z}}$

$=\frac{1+i(1-i)}{(1-i)[1-(1+i\left)\right]+\frac{1}{1+i}}=-1+i$

Now, $\arg \left(z_1\right)=\pi -{\tan }^{-1}\left(1\right)=\pi -\frac{\pi }{4}=\frac{3\pi }{4}$

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