If z_{1}, z_{2} are complex numbers such that | vedclass

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(D) Given $\operatorname{Re}(z)=|z-1|$. Let $z=x+iy$. Then $x=\sqrt{(x-1)^2+y^2}$.Squaring both sides, $x^2=(x-1)^2+y^2$ $\Rightarrow x^2=x^2-2x+1+y^2

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$2 \sqrt{3}$

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(D) Given $\operatorname{Re}(z)=|z-1|$. Let $z=x+iy$. Then $x=\sqrt{(x-1)^2+y^2}$.Squaring both sides, $x^2=(x-1)^2+y^2$ $\Rightarrow x^2=x^2-2x+1+y^2$ $\Rightarrow y^2=2x-1$.This represents a parabola $y^2=4a(x-h)$ with $4a=2 \Rightarrow a=\frac{1}{2}$ and vertex $(h,k)=(\frac{1}{2}, 0)$.Points $z_1$ and $z_2$ lie on this parabola. The slope of the chord joining $z_1$ and $z_2$ is $	an(\arg(z_1-z_2)) = 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.For a parabola $y^2=4ax$, the slope of the chord joining points with parameters $t_1$ and $t_2$ is $m = \frac{2}{t_1+t_2}$.Since $y=2at$, $y_1+y_2 = 2a(t_1+t_2) = 2a(\frac{2}{m}) = \frac{4a}{m}$.Substituting $a=\frac{1}{2}$ and $m=\frac{1}{\sqrt{3}}$, we get $y_1+y_2 = \frac{4(1/2)}{1/\sqrt{3}} = 2\sqrt{3}$.Thus, $\operatorname{Im}(z_1+z_2) = y_1+y_2 = 2\sqrt{3}$.

If $z_{1}, z_{2}$ are complex numbers such that $\operatorname{Re}(z_{1})=|z_{1}-1|$, $\operatorname{Re}(z_{2})=|z_{2}-1|$ and $\arg(z_{1}-z_{2})=\frac{\pi}{6}$, then $\operatorname{Im}(z_{1}+z_{2})$ is equal to

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