Let z be those complex numbers which satisfy |z + 5| ≤ 4 and z(1 +...

Let $z = x + iy$

Given $| z + 5 | \le 4$

${\left(x + 5\right)}^{2} + {y}^{2} \le 16 . . . \left(1\right)$

$z\left(1 + i\right) + \overset{¯}{z}\left(1 - i\right) \ge - 10$

\left(z + \overset{¯}{z}\right)$+ i$\left(z - \overset{¯}{z}\right)$\ge - 10$

$x - y + 5 \ge 0 . . . \left(2\right)$

$| z + 1{|}^{2} = | z - \left(- 1\right){|}^{2}$

Let $P\left(- 1, 0\right)</p><p>$| z + 1{|}_{Max .}^{2} = P{B}^{2}$(where$B$is in${3}^{rd}quadrant) $</p><p>for point of$intersection $</p><p>$\left$$\begin{array}{ll}{\left(x + 5\right)}^{2} + {y}^{2} = 16 \\ x - y + 5 = 0\end{array}$$\right\}y = \pm 2\sqrt{2}

A \left(2\sqrt{2} - 5, 2\sqrt{2}\right)B\left(- 2\sqrt{2} - 5, - 2\sqrt{2}\right)

P{B}^{2} = {\left(+ 2\sqrt{2} + 4\right)}^{2} + {\left(2\sqrt{2}\right)}^{2}

$| z + 1{|}^{2} = 8 + 16 + 16\sqrt{2} + 8$

$\alpha + \beta\sqrt{2} = 32 + 16\sqrt{2}$

$\alpha = 32, \beta = 16 ⇒ \alpha + \beta = 48$