Let S=\z in mathbbC-\i, 2 i\: fracz^2+8 i z-15z^2-3 i z-2 in mathbbR\. If alpha-(13 / 11) i in S, alpha in mathbbR-\0\, then 242 alpha^2 is equal to _________.

Put z=x+i y lm((z^2+8 i z-15 / z^2-3 i z-2))=0 \ => -(x^2-y^2-8 y-15)(2 x y-3 x)+ (2 x y+8 x)(x^2-. .y^2+3 y-2)=0 \ => (x^2-y^2)(2 x y+8 x-2 x y+3 x)+ (8 y+15)(2 x y- 3 x)+(2 x y+8 x)(3 y-2)=0 \ => 11 x^3-11 x y^2+16 x y^2-24 x y+30 x y- 45 x+ 6 x y^2 -4 x y+24 x y-16 x=0 \ => 11 x^3+11 x y^2+26 x y-61 x=0 \ => (11 x^2+11 y^2+26 y-61)=0. \ Because alpha ≠ 0, => x=0 (neglected) \ Put y=-(13 / 11), x=alpha \ 11 alpha^2+11 * (13^2 / 11^2)-26 * (13 / 11)-61=0 \ => 121 alpha^2=840 \ => 242 alpha^2=1680

Let Sz Ci 2 i frac z28 i z15z23 i z2 R If alphafrac 131... | null (Mathematics)

Put

z=x+i y

\begin{aligned} & \operatorname{lm}\left(\frac{z^2+8 i z-15}{z^2-3 i z-2}\right)=0 \\\\ & \Rightarrow-\left(x^2-y^2-8 y-15\right)(2 x y-3 x)+ \\& (2 x y+8 x)\left(x^2-\right. \left.y^2+3 y-2\right)=0 \\\\ & \Rightarrow\left(x^2-y^2\right)(2 x y+8 x-2 x y+3 x)+ \\& (8 y+15)(2 x y- 3 x)+(2 x y+8 x)(3 y-2)=0 \\\\ & \Rightarrow 11 x^3-11 x y^2+16 x y^2-24 x y+30 x y- \\& 45 x+ 6 x y^2 -4 x y+24 x y-16 x=0 \\\\ & \Rightarrow 11 x^3+11 x y^2+26 x y-61 x=0 \\\\ & \Rightarrow\left(11 x^2+11 y^2+26 y-61)=0\right. \\\\ & \because \alpha \neq 0, \Rightarrow x=0 \text { (neglected) } \\\\ & \text { Put } y=-\frac{13}{11}, \quad x=\alpha \\\\ & 11 \alpha^2+11 \cdot \frac{13^2}{11^2}-26 \cdot \frac{13}{11}-61=0 \\\\ & \Rightarrow 121 \alpha^2=840 \\\\ & \Rightarrow 242 \alpha^2=1680 \end{aligned}

Explanation

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