Mathematics - Complex Number Question with Solution | TestHub

Let $z=x+iy$

Given $|z+5|\le 4$

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

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

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

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

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

Let $P\left(-1,0\right)$

$\boxed{|z+1{|}_{Max.}^2=P{B}^2}$ (where $B$ is in $3^{\mathrm{rd}}$ quadrant) 

for point of intersection 

$\left.\begin{array}{c}{\left(x+5\right)}^2+y^2=16\\ x-y+5=0\end{array}\right\}y=\pm 2\sqrt{2}$

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

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

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

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

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