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  • If for <img alt="\mathrm{z}=\alpha+\mathrm{i} \beta,|\mathrm{z}+2|=\mathrm{z}+4(1+\mathrm{i})" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bz%7D%3D%5Calpha&plus;%5Cmathrm%7B

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

Option: 1

x^2+3 x-4=0


Option: 2

x^2+7 x+12=0


Option: 3

x^2+x-12=0


Option: 4

x^2+2 x-3=0


Answers (1)

\begin{aligned} & |z+2|=|\alpha+\mathrm{i} \beta+2| \\ & =\alpha+i \beta+4+4 i \end{aligned}

\begin{array}{ll} \sqrt{(\alpha+2)^2+\beta^2}=(\alpha+4)+i(\beta+4) & \beta+4=0 \\ (\alpha+2)^2+16=(\alpha+4)^2 & \beta=-4 \end{array}

\begin{aligned} & \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 \end{aligned}

Sum of roots = – 7

Product of roots = 12

x^2+7 x+12=0

Posted by

Sumit Saini

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