If for z=alpha+i beta,|z+2|=z+4(1+i), then al | vedclass

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Question

$|z+2|=|\alpha+i \beta+2|$

$=\alpha+i \beta+4+4 i$

$\sqrt{(\alpha+2)^2+\beta^2}=(\alpha+4)+ i (\beta+4)$

$\beta+4=0$

$(\alpha+2)^2+16=(\al

Vedclass · Accepted Answer

$x^2+7 x+12=0$

Vedclass · Answer

$|z+2|=|\alpha+i \beta+2|$

$=\alpha+i \beta+4+4 i$

$\sqrt{(\alpha+2)^2+\beta^2}=(\alpha+4)+ i (\beta+4)$

$\beta+4=0$

$(\alpha+2)^2+16=(\alpha+4)^2$

$\beta=-4$

$\alpha^2+4+4 \alpha+16=\alpha^2+16+8 \alpha$

$4=4 \alpha$

$\alpha=1$

$\alpha=1, \beta=-4$

$\alpha+\beta=-3, \alpha \beta=-4$

$	ext { Sum of roots }=-7$

$	ext { Product of roots }=12$

$x^2+7 x+12=0$

If for $z=\alpha+i \beta,|z+2|=z+4(1+i)$, then $\alpha+\beta$ and $\alpha \beta$ are the roots of the equation

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