JEE Main 2023 — Mathematics Algebra

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

$$\Rightarrow \frac{{z}^{2}+8iz-15}{{z}^{2}-3iz-2}=\frac{{(x+iy)}^{2}+8i(x+iy)-15}{{(x+iy)}^{2}-3i(x+iy)-2}$$⇒z2−3iz−2z2+8iz−15​=(x+iy)2−3i(x+iy)−2(x+iy)2+8i(x+iy)−15​

$$=\frac{{x}^{2}-{y}^{2}-8y-15+i(2xy+8x)}{{x}^{2}-{y}^{2}+3y-2+i(2xy-3x)}$$=x2−y2+3y−2+i(2xy−3x)x2−y2−8y−15+i(2xy+8x)​

$$=\frac{{x}^{2}-{y}^{2}-8y-15+i(2xy+8x)}{{x}^{2}-{y}^{2}+3y-2+i(2xy-3x)}\times \frac{({x}^{2}-{y}^{2}+3y-2-i(2xy-3x))}{({x}^{2}-{y}^{2}+3y-2-i(2xy-3x))}$$=x2−y2+3y−2+i(2xy−3x)x2−y2−8y−15+i(2xy+8x)​×(x2−y2+3y−2−i(2xy−3x))(x2−y2+3y−2−i(2xy−3x))​

$$=\frac{({x}^{2}-{y}^{2}-8y-15)({x}^{2}-{y}^{2}+3y-2)+(2xy+8x)(2xy-3x)+i(2xy+8x)({x}^{2}-{y}^{2}+3y-2)-i(2xy-3x)({x}^{2}-{y}^{2}-8y-15)}{{({x}^{2}-{y}^{2}+3y-2)}^{2}-{(2xy-3x)}^{2}}$$=(x2−y2+3y−2)2−(2xy−3x)2(x2−y2−8y−15)(x2−y2+3y−2)+(2xy+8x)(2xy−3x)+i(2xy+8x)(x2−y2+3y−2)−i(2xy−3x)(x2−y2−8y−15)​

But $\mathrm{Im}(\frac{{z}^{2}+8iz-15}{{z}^{2}-3iz-2})=0$Im(z2−3iz−2z2+8iz−15​)=0,

$$\Rightarrow -({x}^{2}-{y}^{2}-8y-15)(2xy-3x)+(2xy+8x)({x}^{2}-{y}^{2}+3y-2)=0$$⇒−(x2−y2−8y−15)(2xy−3x)+(2xy+8x)(x2−y2+3y−2)=0

$$\Rightarrow ({x}^{2}-{y}^{2})(2xy+8x-2xy+3x)+(8y+15)(2xy-3x)+(2xy+8x)(3y-2)=0$$⇒(x2−y2)(2xy+8x−2xy+3x)+(8y+15)(2xy−3x)+(2xy+8x)(3y−2)=0

$$\Rightarrow 11{x}^{3}-11x{y}^{2}+16x{y}^{2}-24xy+30xy-45x+6x{y}^{2}-4xy+24xy-16x=0$$⇒11x3−11xy2+16xy2−24xy+30xy−45x+6xy2−4xy+24xy−16x=0

$$\Rightarrow 11{x}^{3}+11x{y}^{2}+26xy-61x=0$$⇒11x3+11xy2+26xy−61x=0

$$\Rightarrow 11{x}^{2}+11{y}^{2}+26y-61=0$$⇒11x2+11y2+26y−61=0

$$\because \alpha \neq 0$$∵α=0

So, put $y=-\frac{13}{11}$y=−1113​, $x=\alpha$x=α

$$11{\alpha }^{2}+11\cdot \frac{{13}^{2}}{{11}^{2}}-26\cdot \frac{13}{11}-61=0$$11α2+11⋅112132​−26⋅1113​−61=0

$$\Rightarrow 121{\alpha }^{2}=840$$⇒121α2=840

$$\Rightarrow 242{\alpha }^{2}=1680$$⇒242α2=1680

Hence this is the required answer.