JEE Main 2024 — Complex Number Question with Solution

Given: $\left|z+2-3i\right|\le 1$

Putting, $z=x+iy$.

$\Rightarrow {\left(x+2\right)}^2+{\left(y-3\right)}^2\le 1 ...\left(i\right)$

For the circle represented in equation $\left(i\right)$, centre is $\left(-2,3\right)$ and radius, $r=1$.

It is given that, $z\left(1+i\right)+\overline{z}\left(1-i\right)\le -8$

$\Rightarrow \left(x+iy\right)\left(1+i\right)+\left(x-iy\right)\left(1-i\right)\le -8$

$\Rightarrow x+ix-y+iy+x-ix-iy-y\le -8$

$\Rightarrow 2x-2y\le -8$

$\Rightarrow x-y+4=0$

So, the line passing through $\left(3,-2\right)$ and perpendicular to $x-y+4=0$ is given by, $x+y-1=0$.

So, the distance of line $x-y+4=0$ from the centre $\left(-2,3\right)$ is given by,

$d=\left|\frac{-2-3+4}{\sqrt{2}}\right|$

$\Rightarrow d=\frac{1}{\sqrt{2}}$

Now, $x+y-1=0$ can be rewritten as,

$\frac{x+2}{{\displaystyle \frac{-1}{\sqrt{2}}}}=\frac{y-3}{{\displaystyle \frac{1}{\sqrt{2}}}}= CA \text{or} CB$

For $CA=\frac{-1}{\sqrt{2}}$ and $CB=1$, $A\equiv \left(\frac{-3}{2},\frac{5}{2}\right)$ and $B\equiv \left(\frac{-1}{\sqrt{2}}-2, \frac{1}{\sqrt{2}}+3\right)$.

$\Rightarrow {\left|z_1\right|}^2={\left(2+\frac{1}{\sqrt{2}}\right)}^2+{\left(3+\frac{1}{\sqrt{2}}\right)}^2$

$\Rightarrow {\left|z_1\right|}^2=4+\frac{1}{2}+2\sqrt{2}+9+\frac{1}{2}+3\sqrt{2}$

$\Rightarrow {\left|z_1\right|}^2=14+5\sqrt{2}$

$\Rightarrow {\left|z_2\right|}^2=2\left(\frac{9}{4}+\frac{25}{4}\right)$

$\Rightarrow {\left|z_2\right|}^2=17$

$\Rightarrow {\left|z_1\right|}^2+2{\left|z_2\right|}^2=31+5\sqrt{2}$

$\Rightarrow \alpha =31, \beta =5$

$\Rightarrow \alpha +\beta =36$