Let z be those complex number which
Question:

Let $z$ be those complex number 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

Solution:

Given, $|z+5| \leq 4$

$\Rightarrow(x+5)^{2}+y^{2} \leq 16 \ldots(1)$

Also, $z(1+i)+\bar{z}(1-i) \geq-10$

$\Rightarrow x-y \geq-5 \ldots(2)$

From (1) and (2) Locus of $z$ is the shaded region in the diagram.

$|z+1|$ represents distance of $^{\prime} z^{\prime}$ from $Q(-1,0)$ Clearly ‘ $\mathrm{p}^{\prime}$ is the required position of ‘ $\mathrm{z}^{\prime}$ when $|\mathrm{z}+1|$ is maximum.

$\therefore P \equiv(-5-2 \sqrt{2},-2 \sqrt{2})$

$\left.\therefore(\mathrm{PQ})^{2}\right|_{\max }=32+16 \sqrt{2}$

$\Rightarrow \alpha=32$

$\Rightarrow \beta=16$

Thus, $\alpha+\beta=48$

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