Question

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

Answer 1

$|z+5| \leq 4$
$(x+5)^{2}+y^{2} \leq 16 \,\,\,\, \ldots ......(1)$
$z(1+i)+\bar{z}(1-i) \geq-10$
$(z+\bar{z})+i(z-\bar{z}) \geq-10$
$x-y+5 \geq 0 \,\,\,\,\, (2)$

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

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