Let S = \ z ∈ C: |z - 2| ≤ 1, z(1 + i) + ¯ z (1 - i) ≤ 2\ . Let |z...

Given $\left|z - 2\right| \le 1$

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

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

Puting $z = x + iy$, we get

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

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

$∴ x - y \le 1 \ldots \left(i\right)$

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$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/a90ce137-3cc0-4687-8ab8-8e17a3497c06-image.png" style="width: 257px; height: 200px;" /></p><p>Here$PA = \sqrt{17}, PB = \sqrt{13}$</p><p>So$A\left(1, 0\right)$is the point representing${z}_{2}$</p><p>Let$D\left(2 + cos\theta, 0 + sin\theta\right)$be the point representing${z}_{1}$</p><p>$∴ {m}_{CP} = tan\theta = - 2$</p><p>$cos\theta = - \dfrac{1}{\sqrt{5}}, sin\theta = \dfrac{2}{\sqrt{5}}$</p><p>$∴ D ≡ \left(2 - \dfrac{1}{\sqrt{5}}, \dfrac{2}{\sqrt{5}}\right)$</p><p>$⇒ {z}_{1} = \left(2 - \dfrac{1}{\sqrt{5}}\right) + \dfrac{2i}{\sqrt{5}}$</p><p>So \left|{z}_{1}\right| = \dfrac{25 - 4\sqrt{5}}{5} \text{ and } {z}_{2} = 1$

$∴ {\left|{z}_{2}\right|}^{2} = 1$

$∴ 5\left({\left|{z}_{1}\right|}^{2} + {\left|{z}_{2}\right|}^{2}\right) = 30 - 4\sqrt{5}$

$∴ \alpha = 30$

$\beta = - 4$

Hence $\alpha + \beta = 26$