Let S=z C-i,2i:z^2+8iz-15z^2-3iz-2 R.

Question

Let $S={z\in C-{i,2i}:\frac{{z}^{2}+8iz-15}{{z}^{2}-3iz-2}\in R}$. If $\alpha -\frac{13}{11}i\in S,\alpha \in R-{0}$, then

$242{\alpha }^{2}$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 1680

Accepted Answer

Put $z=x+iy$

$\begin{gathered} lm(\frac{{z}^{2}+8iz-15}{{z}^{2}-3iz-2})=0 \ \Rightarrow -({x}^{2}-{y}^{2}-8y-15)(2xy-3x)+(2xy+8x)({x}^{2}-{y}^{2}+3y-2)=0 \ \Rightarrow ({x}^{2}-{y}^{2})(2xy+8x-2xy+3x)+(8y+15)(2xy-3x)+(2xy+8x)(3y-2)=0 \ \Rightarrow 11{x}^{3}-11x{y}^{2}+16x{y}^{2}-24xy+30xy-45x+6x{y}^{2}-4xy+24xy-16x=0 \ \Rightarrow 11{x}^{3}+11x{y}^{2}+26xy-61x=0 \ \Rightarrow (11{x}^{2}+11{y}^{2}+26y-61)=0 \ \because \alpha 
e 0,\Rightarrow x=0 \ y=-\frac{13}{11},\, x=\alpha \ 11{\alpha }^{2}+11\cdot \frac{{13}^{2}}{{11}^{2}}-26\cdot \frac{13}{11}-61=0 \ \Rightarrow 121{\alpha }^{2}=840 \ \Rightarrow 242{\alpha }^{2}=1680 \end{gathered}$

Solutions · 1