JEE Main 2024 — 8 April, Shift 1 — Question 15

$|z+2|=1, \operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$

Let $\mathrm{z}+2=\cos \theta+i \sin \theta$

$\frac{1}{z+2}=\cos \theta-\mathrm{i} \sin \theta$

$\Rightarrow \frac{\mathrm{z}+1}{\mathrm{z}+2}=1-\frac{1}{\mathrm{z}+2}=1-(\cos \theta-\mathrm{i} \sin \theta)$

$=(1-\cos \theta)+\mathrm{i} \sin \theta$

$\operatorname{Im}\left(\frac{z+1}{z+2}\right)=\sin \theta, \sin \theta=\frac{1}{5}$

$\cos \theta= \pm \sqrt{1-\frac{1}{25}}= \pm \frac{2 \sqrt{6}}{5}$

$|\operatorname{Re}(\overline{z+2})|=\frac{2 \sqrt{6}}{5}$