JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Complex Numbers

$P=\{z:|z+2-3i|\le 1\}$: closed disc, centre $(-2,3)$, radius $1$.
$Q$: writing $z=x+iy$, $z(1+i)+\bar{z}(1-i)=(x-y)+i(x+y)+(x-y)-i(x+y)=2(x-y)$, so $Q$ is $2(x-y)\le -8$, i.e. $y\ge x+4$ — the half-plane on or above the line $\ell :x-y+4=0$.

Distance from the disc's centre $(-2,3)$ to $\ell$: $\frac{|-2-3+4|}{\sqrt{2}}=\frac{1}{\sqrt{2}}<1$, so $\ell$ cuts through the disc; since the centre itself satisfies $3\ge -2+4$, $P\cap Q$ is the larger part of the disc.

We need the extremes of $|z-(3-2i)|$, i.e. distance to the fixed point $A=(3,-2)$, over $P\cap Q$.

*Maximum:* Unconstrained, the farthest point of the disc from $A$ lies diametrically opposite $A$ across the centre: direction from centre to $A$ is $\frac{1}{5\sqrt{2}}(5,-5)=\frac{1}{\sqrt{2}}(1,-1)$, so the farthest point is

Check feasibility: $y-x=5+\sqrt{2}\ge 4$ — satisfies $Q$. So this point is the true (constrained) maximum:


*Minimum:* The unconstrained nearest point of the disc to $A$, $\left(-2+\frac{1}{\sqrt{2}},3-\frac{1}{\sqrt{2}}\right)$, gives $y-x=5-\sqrt{2}\approx 3.59<4$ — infeasible (outside $Q$). Since $A$ itself lies on the far side of $\ell$ from the feasible region ($y_A-x_A=-5<4$), the nearest feasible point is the foot of the perpendicular from $A$ onto $\ell$:

This point's distance from the disc's centre is $\sqrt{0.5^2+0.5^2}=\frac{1}{\sqrt{2}}<1$, so it lies inside the disc — a valid point of $P\cap Q$, confirming it is the true minimum:


Matching $\alpha +\beta \sqrt{2}$: $\alpha =31,\beta =5$, so $\alpha +\beta =36$.