JEE Main 2021 — Mathematics Algebra

Let $z=x+iy$z=x+iy

Given $|z+5|\leq 4$∣z+5∣≤4

$${(x+5)}^{2}+{y}^{2}\leq 16...(1)$$(x+5)2+y2≤16...(1)

$$z(1+i)+\bar{z}(1-i)\geq -10$$z(1+i)+zˉ(1−i)≥−10

$$(z+\bar{z})+i(z-\bar{z})\geq -10$$(z+zˉ)+i(z−zˉ)≥−10

$$x-y+5\geq 0...(2)$$x−y+5≥0...(2)

$$|z+1{|}^{2}=|z-(-1){|}^{2}$$∣z+1∣2=∣z−(−1)∣2

Let $P(-1,0)$P(−1,0)

$|z+1{|}_{Max.}^{2}=P{B}^{2}$∣z+1∣Max.2​=PB2 (where $B$B is in ${3}^{\mathrm{rd}}$3rd quadrant)

for point of intersection

\begin{matrix}{(x+5)}^{2}+{y}^{2}=16 \\ x-y+5=0\end{matrix}}y=\pm 2\sqrt{2}

$$A(2\sqrt{2}-5,2\sqrt{2})B(-2\sqrt{2}-5,-2\sqrt{2})$$A(22​−5,22​)B(−22​−5,−22​)

$$P{B}^{2}={(+2\sqrt{2}+4)}^{2}+{(2\sqrt{2})}^{2}$$PB2=(+22​+4)2+(22​)2

$$|z+1{|}^{2}=8+16+16\sqrt{2}+8$$∣z+1∣2=8+16+162​+8

$$\alpha +\beta \sqrt{2}=32+16\sqrt{2}$$α+β2​=32+162​

$$\alpha =32,\beta =16\Rightarrow \alpha +\beta =48$$α=32,β=16⇒α+β=48