# If $z_{1},z_{2}$ are complex numbers such that $Re(z_{1})=\left | z_{1}-1 \right |,\; \; Re(z_{2})=\left | z_{2}-1 \right |$ and  $arg(z_{1}-z_{2})=\frac{\pi }{6},$ then $Im(z_{1}+z_{2})$  is equal to : Option: 1   Option: 2  Option: 3  Option: 4

Re(z) = |z – 1|

$\\\Rightarrow x=\sqrt{(x-1)^{2}+(y-0)^{2}} \quad(x>0) \\ \Rightarrow y^{2}=2 x-1=4 \times\frac{1}{2}\left(x-\frac{1}{2}\right)$

a parabola with focus (1, 0) & directrix as imaginary axis.

$\therefore \quad \text { Vertex }=\left(\frac{1}{2}, 0\right)$

$\mathrm{A}\left(\mathrm{z}_{1}\right) \& \mathrm{~B}\left(\mathrm{z}_{2}\right) \text { are two points on it such that slope of } \mathrm{AB}=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$

$\left(\arg \left(z_{1}-z_{2}\right)=\frac{\pi}{6}\right)$

$\\ \text { for } y^{2}=4 a x \\ \text { Let } A\left(a t_{1}^{2}, 2 a t_{1}\right) \& B\left(a t_{2}^{2}, 2 a t_{2}\right) \\ m_{A B}=\frac{2}{t_{1}+t_{2}}=\frac{4 a}{y_{1}+y_{2}}=\frac{1}{\sqrt{3}} \\ \left(\text { Here } a=\frac{1}{2}\right) \\ \Rightarrow y_{1}+y_{2}=4 a \sqrt{3}=2 \sqrt{3}$

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