JEE MAIN - Mathematics (2023 - 8th April Morning Shift - No. 16)
If for $$z=\alpha+i \beta,|z+2|=z+4(1+i)$$, then $$\alpha+\beta$$ and $$\alpha \beta$$ are the roots of the equation :
$$x^{2}+2 x-3=0$$
$$x^{2}+3 x-4=0$$
$$x^{2}+x-12=0$$
$$x^{2}+7 x+12=0$$
Explanation
Given : $|z+2|=z+4(1+i)$
Also, $z=\alpha+i \beta$
$$ \begin{aligned} & \therefore|z+2|=|\alpha+i \beta+2|=(\alpha+i \beta)+4+4 i \\\\ & \Rightarrow|(\alpha+2)+i \beta|=(\alpha+4)+i(\beta+4) \\\\ & \Rightarrow \sqrt{(\alpha+2)^2+\beta^2}=(\alpha+4)+i(\beta+4) \\\\ & \Rightarrow \beta+4=0 \Rightarrow \beta=-4 \end{aligned} $$
$$ \begin{aligned} & \text { Now, }(\alpha+2)^2+\beta^2=(\alpha+4)^2 \\\\ & \Rightarrow \alpha^2+4+4 \alpha+\beta^2=\alpha^2+16+8 \alpha \\\\ & \Rightarrow 4+4 \alpha+16=16+8 \alpha \\\\ & \Rightarrow 4 \alpha=4 \Rightarrow \alpha=1 \\\\ & \text { So, } \alpha+\beta=-3 \text { and } \alpha \beta=-4 \\\\ & \therefore \text { Required equation is } \\\\ & x^2-(-3-4) x+(-3)(-4)=0 \\\\ & \Rightarrow x^2+7 x+12=0 \end{aligned} $$
Also, $z=\alpha+i \beta$
$$ \begin{aligned} & \therefore|z+2|=|\alpha+i \beta+2|=(\alpha+i \beta)+4+4 i \\\\ & \Rightarrow|(\alpha+2)+i \beta|=(\alpha+4)+i(\beta+4) \\\\ & \Rightarrow \sqrt{(\alpha+2)^2+\beta^2}=(\alpha+4)+i(\beta+4) \\\\ & \Rightarrow \beta+4=0 \Rightarrow \beta=-4 \end{aligned} $$
$$ \begin{aligned} & \text { Now, }(\alpha+2)^2+\beta^2=(\alpha+4)^2 \\\\ & \Rightarrow \alpha^2+4+4 \alpha+\beta^2=\alpha^2+16+8 \alpha \\\\ & \Rightarrow 4+4 \alpha+16=16+8 \alpha \\\\ & \Rightarrow 4 \alpha=4 \Rightarrow \alpha=1 \\\\ & \text { So, } \alpha+\beta=-3 \text { and } \alpha \beta=-4 \\\\ & \therefore \text { Required equation is } \\\\ & x^2-(-3-4) x+(-3)(-4)=0 \\\\ & \Rightarrow x^2+7 x+12=0 \end{aligned} $$
Comments (0)
