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Let $\mathrm{z}$ be a complex number such that $|\mathrm{z}+2|=1$ and $\operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$. Then the value of $|\operatorname{Re}(\overline{z+2})|$ is :
$\frac{\sqrt{6}}{5}$
$\frac{1+\sqrt{6}}{5}$
$\frac{24}{5}$
$\frac{2 \sqrt{6}}{5}$
Solution
$ |z+2|=1, \operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5} $
$ \text { Let } z+2=\cos \theta+i \sin \theta $
$ \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}$
