If $z_{1}, z_{2}$ are complex numbers such that $\operatorname{Re}\left(z_{1}\right)=\left|z_{1}-1\right|$, $\operatorname{Re}\left(z_{2}\right)=\left|z_{2}-1\right|$ and $\arg \left(z_{1}-z_{2}\right)=\frac{\pi}{6}$, then $\operatorname{Im}\left(z_{1}+z_{2}\right)$ is equal to :
Correct Option: , 2
Let $z_{1}=x_{1}+i y_{1}$ and $z_{2}=x_{2}+i y_{2}$'
$\because\left|z_{1}-1\right|=\operatorname{Re}\left(z_{1}\right)$
$\Rightarrow\left(x_{1}-1\right)^{2}+y_{1}^{2}=x_{1}^{2}$
$\Rightarrow y_{1}^{2}-2 x_{1}+1=0$ .....(i)
$\left|z_{2}-1\right|=\operatorname{Re}\left(z_{2}\right) \Rightarrow\left(x_{2}-1\right)^{2}+y_{2}^{2}=x_{2}^{2}$
$\Rightarrow y_{2}^{2}-2 x_{2}+1=0$ ....(ii)
$y_{1}^{2}-y_{2}^{2}-2\left(x_{1}-x_{2}\right)=0$
$\Rightarrow \tan ^{-1}\left(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\right)=\frac{\pi}{6}$
$\Rightarrow \frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{1}{\sqrt{3}}$ $\left[\right.$ From, $\left.\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{2}{y_{1}+y_{2}}\right]$
$\therefore y_{1}+y_{2}=2 \sqrt{3} \Rightarrow \operatorname{lm}\left(z_{1}+z_{2}\right)=2 \sqrt{3}$
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