Question

Let $z$ be those complex numbers which satisfy $|z+5|\le 4$ and $z(1+i)+\bar{z}(1-i)\ge -10,i=\sqrt{-1}$. If the maximum value of $|z+1{|}^2$ is $\alpha +\beta \sqrt{2}$, then the value of $(\alpha +\beta )$ is _____

Answer 1

Answer: 48

$|z+5|\le 4$
$(x+5{)}^2+y^2\le 16\dots ......(1)$
$z(1+i)+\bar{z}(1-i)\ge -10$
$(z+\bar{z})+i(z-\bar{z})\ge -10$
$x-y+5\ge 0(2)$

$|z+1{|}^2=|z-(-1){|}^2$
Let $P(-1,0)$
$|z+1{|}_{Max.}^2=P{B}^2$ (where $B$ is in $3^{\text{rd }}$ quadrant) for point of intersection

$A(2\sqrt{2}-5,2\sqrt{2})B(-2\sqrt{2}-5,-2\sqrt{2})$
$P{B}^2=(+2\sqrt{2}+4{)}^2+(2\sqrt{2}{)}^2$
$|z+1{|}^2=8+16+16\sqrt{2}+8$
$\alpha +\beta \sqrt{2}=32+16\sqrt{2}$
$\alpha =32,\beta =16\Rightarrow \alpha +\beta =48$