Q48 / 90 · JEE Main 2024 (08 Apr Shift 1) · Mathematics
Let $z$ be a complex number such that $|z+2|=1$ and $\operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$. Then the value of $|\operatorname{Re}(\overline{z+2})|$ is
- Correct answerA$\frac{2 \sqrt{6}}{5}$
- B$\frac{24}{5}$
- C$\frac{1+\sqrt{6}}{5}$
- D$\frac{\sqrt{6}}{5}$
Solution
$|z+2|=1, \operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$
Let $z+2=\cos \theta+i \sin \theta$ $\begin{aligned} & \frac{1}{z+2}=\cos \theta-i \sin \theta \\ & \Rightarrow \frac{z+1}{z+2}=1-\frac{1}{z+2}=1-(\cos \theta-i \sin \theta) \\ & =(1-\cos \theta)+\operatorname{isin} \theta \\ & \operatorname{Im}\left(\frac{z+1}{z+2}\right)=\sin \theta, \sin \theta=\frac{1}{5} \\ & \cos \theta= \pm \sqrt{1-\frac{1}{25}}= \pm \frac{2 \sqrt{6}}{5} \\ & |\operatorname{Re}(\overline{z+2})|=\frac{2 \sqrt{6}}{5} \end{aligned}$
Let $z+2=\cos \theta+i \sin \theta$ $\begin{aligned} & \frac{1}{z+2}=\cos \theta-i \sin \theta \\ & \Rightarrow \frac{z+1}{z+2}=1-\frac{1}{z+2}=1-(\cos \theta-i \sin \theta) \\ & =(1-\cos \theta)+\operatorname{isin} \theta \\ & \operatorname{Im}\left(\frac{z+1}{z+2}\right)=\sin \theta, \sin \theta=\frac{1}{5} \\ & \cos \theta= \pm \sqrt{1-\frac{1}{25}}= \pm \frac{2 \sqrt{6}}{5} \\ & |\operatorname{Re}(\overline{z+2})|=\frac{2 \sqrt{6}}{5} \end{aligned}$
Hence, the correct answer is option: 'A'.