4-1.Complex numbers
hard

$z=\alpha+i \beta$ માટે જો $|z+2|=z+4(1+i)$ હોય, તો $\alpha+\beta$ અને $\alpha \beta$ એ $.........$ સમીકરણ ના બીજ છે.

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

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