QJEEM24Maf567: JEE Main 2024 Mathematics - Class 11 JEEM24Maf567 Solution: Let be a complex number such that and . Then the value of is

$\begin{aligned}& |z+2|=1 \\& \operatorname{Im}_m\left(\frac{z+1}{z+2}\right)=\frac{1}{5} \\& |\operatorname{Re}(\overline{z+2})|=?\end{aligned}$

Let $z=x+i y$

$\begin{aligned}& \because|z+2|=1 \Rightarrow(x+2)^2+y^2=1 \quad \ldots(1) \\& I_m\left(\frac{z+1}{z+2}\right)=\frac{1}{5} \Rightarrow I_m\left(\frac{x+i y+1}{x+i y+2}\right)=\frac{1}{5} \\& \Rightarrow I_m\left|\frac{[(x+1)+i y][(x+)-i y]}{(x+2)^2+y^2}\right|=\frac{1}{5}\end{aligned}$

$\frac{y(x+2)-y(x+)}{(x+2)^2+y^2}=\frac{1}{5}\quad \text{.... (2)}$

$\Rightarrow y=\frac{1}{5}$

Substituting in equation (1)

$\begin{aligned}& (x+2)^2+\frac{1}{25}=1 \\& (x+2)^2=\frac{24}{25} \\& \Rightarrow x=-2 \pm \frac{\sqrt{24}}{5} \\& |\operatorname{Re}(\overline{x+i y+2})| \\& =x+2= \pm \frac{\sqrt{24}}{5}=\frac{2 \sqrt{6}}{5}\end{aligned}$