Let S = z ∈ C : z - 2 ≤ 1 , z 1 + i + z ¯ 1 - i ≤ 2 . Let z - 4   i attains…

Given $\left|z-2\right|\le 1$

${\left(x-2\right)}^2+y^2\le 1 \cdots \left(i\right)$ represents interior region of a circle with centre $\left(2,0\right)$ and radius $1$

Now $z\left(1+i\right)+\overline{z}\left(1-i\right)\le 2$

Puting $z=x+iy$, we get

$\left(x+iy\right)\left(1+i\right)+\left(x-iy\right)\left(1-i\right)\le 2$

$\Rightarrow x-y+i\left(x+y\right)+x-y-i\left(x+y\right)\le 2$

$\therefore x-y\le 1 \dots \left(i\right)$

Let $A\left(1,0\right)$ and $B\left(2,1\right)$ be the points on the line $x-y=1$ and the circle ${\left(x-2\right)}^2+y^2=1$

$\left|z-4i\right|$ represents the distance from $P\left(0,4\right)$ to any point in the region $S$

Here $PA=\sqrt{17}, PB=\sqrt{13}$ 

So $A\left(1,0\right)$ is the point representing $z_2$

Let $D\left(2+\cos \theta ,0+\sin \theta \right)$ be the point representing $z_1$

$\therefore m_{CP}=\tan \theta =-2$

$\cos \theta =-\frac{1}{\sqrt{5}},\sin \theta =\frac{2}{\sqrt{5}}$

$\therefore D\equiv \left(2-\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}\right)$

$\Rightarrow z_1=\left(2-\frac{1}{\sqrt{5}}\right)+\frac{2i}{\sqrt{5}}$

So $\left|z_1\right|=\frac{25-4\sqrt{5}}{5} \& z_2=1$

$\therefore {\left|z_2\right|}^2=1$

$\therefore 5\left({\left|z_1\right|}^2+{\left|z_2\right|}^2\right)=30-4\sqrt{5}$

$\therefore \alpha =30$

$\beta =-4$

Hence $\alpha +\beta =26$