JEE Main 2023 — Complex Number Question with Solution

Put $z=x+iy$ in the given expression and equate the imaginary part to zero.

$\Rightarrow \frac{z^2+8iz-15}{z^2-3iz-2}=\frac{{\left(x+iy\right)}^2+8i\left(x+iy\right)-15}{{\left(x+iy\right)}^2-3i\left(x+iy\right)-2}$

$=\frac{x^2-y^2-8y-15+i\left(2xy+8x\right)}{x^2-y^2+3y-2+i\left(2xy-3x\right)}$

$=\frac{x^2-y^2-8y-15+i\left(2xy+8x\right)}{x^2-y^2+3y-2+i\left(2xy-3x\right)}\times \frac{\left(x^2-y^2+3y-2-i\left(2xy-3x\right)\right)}{\left(x^2-y^2+3y-2-i\left(2xy-3x\right)\right)}$

$=\frac{\left(x^2-y^2-8y-15\right)\left(x^2-y^2+3y-2\right)+\left(2xy+8x\right)\left(2xy-3x\right)+i\left(2xy+8x\right)\left(x^2-y^2+3y-2\right)-i\left(2xy-3x\right)\left(x^2-y^2-8y-15\right)}{{\left(x^2-y^2+3y-2\right)}^2-{\left(2xy-3x\right)}^2}$

But $\mathrm{Im}\left(\frac{z^2+8iz-15}{z^2-3iz-2}\right)=0$,

$\Rightarrow -\left(x^2-y^2-8y-15\right)\left(2xy-3x\right)+\left(2xy+8x\right)\left(x^2-y^2+3y-2\right)=0$

$\Rightarrow \left(x^2-y^2\right)\left(2xy+8x-2xy+3x\right)+\left(8y+15\right)\left(2xy-3x\right)+\left(2xy+8x\right)\left(3y-2\right)=0$

$\Rightarrow 11x^3-11x{y}^2+16x{y}^2-24xy+30xy-45x+6x{y}^2-4xy+24xy-16x=0$

$\Rightarrow 11x^3+11x{y}^2+26xy-61x=0$

$\Rightarrow 11x^2+11y^2+26y-61=0$

$\because \alpha \ne 0$

So, put $y=-\frac{13}{11}$, $x=\alpha$
$11{\alpha }^2+11·\frac{{13}^2}{{11}^2}-26·\frac{13}{11}-61=0$
$\Rightarrow 121{\alpha }^2=840$

$\Rightarrow 242{\alpha }^2=1680$

Hence this is the required answer.