Question

If $z_1,z_2$ are complex numbers such that $Re(z_1)=|z_1-1|,Re(z_2)=|z_2-1|$ and $arg(z_1-z_2)=\frac{\pi }{6}$, then Im $(z_1+z_2)$ is equal to:

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $2\sqrt{3}$

Answer 1

Answer: (D)

$Re(z)=|z-1|$
$\Rightarrow x=\sqrt{(x-1{)}^2+(y-0{)}^2}$$(x>0)$
$\to y^2=2x-1=4\cdot \frac{1}{2}\left(x-\frac{1}{2}\right)$
$\Rightarrow$ a parabola with focus $(1,0)$ & directrix as imaginary axis
$\therefore$ Vertex $=\left(\frac{1}{2},0\right)$

$A\left(z_1\right)$ & $B\left(z_2\right)$ are two points on it such that
slope of $AB=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$
$\left(\arg \left(z_1-z_2\right)=\frac{\pi }{6}\right)$
for $y^2=4ax$
Let $A\left(a{t}_1^2,2a{t}_1\right)$ & $B\left(a{t}_2^2,2a{t}_2\right)$
$m_{AB}=\frac{2}{t_1+t_2}=\frac{4a}{y_1+y_2}=\frac{1}{\sqrt{3}}$
$($ Here $a=\frac{1}{2})$
$\Rightarrow y_1+y_2=4a\sqrt{3}=2\sqrt{3}$