∣z+2∣=1,Im(z+2z+1)=51
Let z+2=cosθ+isinθ $\begin{aligned}
& \frac{1}{z+2}=\cos \theta-i \sin \theta \
& \Rightarrow \frac{z+1}{z+2}=1-\frac{1}{z+2}=1-(\cos \theta-i \sin \theta) \
& =(1-\cos \theta)+\operatorname{isin} \theta \
& \operatorname{Im}\left(\frac{z+1}{z+2}\right)=\sin \theta, \sin \theta=\frac{1}{5} \
& \cos \theta= \pm \sqrt{1-\frac{1}{25}}= \pm \frac{2 \sqrt{6}}{5} \
& |\operatorname{Re}(\overline{z+2})|=\frac{2 \sqrt{6}}{5}
\end{aligned}$