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

Question

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

Accepted Answer

Given : $|z+2|=z+4(1+i)$

Also, $z=\alpha+i \beta$

$$
\begin{aligned}
& \therefore|z+2|=|\alpha+i \beta+2|=(\alpha+i \beta)+4+4 i \\
& \Rightarrow|(\alpha+2)+i \beta|=(\alpha+4)+i(\beta+4) \\
& \Rightarrow \sqrt{(\alpha+2)^2+\beta^2}=(\alpha+4)+i(\beta+4) \\
& \Rightarrow \beta+4=0 \Rightarrow \beta=-4
\end{aligned}
$$

$$
\begin{aligned}
& \text { Now, }(\alpha+2)^2+\beta^2=(\alpha+4)^2 \\
& \Rightarrow \alpha^2+4+4 \alpha+\beta^2=\alpha^2+16+8 \alpha \\
& \Rightarrow 4+4 \alpha+16=16+8 \alpha \\
& \Rightarrow 4 \alpha=4 \Rightarrow \alpha=1 \\
& \text { So, } \alpha+\beta=-3 \text { and } \alpha \beta=-4 \\
& \therefore \text { Required equation is } \\
& x^2-(-3-4) x+(-3)(-4)=0 \\
& \Rightarrow x^2+7 x+12=0
\end{aligned}
$$

Solutions · 1