JEE PYQ: Complex Numbers Question 13
Question 13 - 2021 (26 Feb Shift 2)
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
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Answer: 48
Solution
$|z+5| \le 4$: disk centered at $(-5,0)$, radius 4. $z(1+i) + \bar{z}(1-i) \ge -10$: $2x - 2y \ge -10$, i.e., $x - y \ge -5$. Maximize $|z+1|^2$ on this region. The farthest point from $(-1,0)$ in the feasible region gives $\alpha + \beta\sqrt{2}$ with $\alpha + \beta = 48$.
BETA
