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4-1.Complex numbers
hard
If for $z=\alpha+i \beta,|z+2|=z+4(1+i)$, then $\alpha+\beta$ and $\alpha \beta$ are the roots of the equation
A
$x^2+7 x+12=0$
B
$x^2+3 x-4=0$
C
$x^2+2 x-3=0$
D
$x ^2+ x -12=0$
(JEE MAIN-2023)
Solution
$|z+2|=|\alpha+i \beta+2|$
$=\alpha+i \beta+4+4 i$
$\sqrt{(\alpha+2)^2+\beta^2}=(\alpha+4)+ i (\beta+4)$
$\beta+4=0$
$(\alpha+2)^2+16=(\alpha+4)^2$
$\beta=-4$
$\alpha^2+4+4 \alpha+16=\alpha^2+16+8 \alpha$
$4=4 \alpha$
$\alpha=1$
$\alpha=1, \beta=-4$
$\alpha+\beta=-3, \alpha \beta=-4$
$\text { Sum of roots }=-7$
$\text { Product of roots }=12$
$x^2+7 x+12=0$
Standard 11
Mathematics
