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Q.
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:
JEE MainJEE Main 2020Complex Numbers and Quadratic Equations
Solution:
${Re}(z)=|z-1|$
$\Rightarrow x=\sqrt{(x-1)^{2}+(y-0)^{2}}$$(x >\, 0)$
$\rightarrow y^{2}=2 x - 1=4 \cdot \frac{1}{2}\left(x-\frac{1}{2}\right)$
$\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 $A \left( at _{1}^{2}, 2 at _{1}\right)$ & $B \left( at _{2}^{2}, 2 at _{2}\right)$
$ m _{ AB }=\frac{2}{ t _{1}+ t _{2}}=\frac{4 a }{ y _{1}+ y _{2}}=\frac{1}{\sqrt{3}}$
$\left(\right.$ Here $\left. a =\frac{1}{2}\right)$
$\Rightarrow y _{1}+ y _{2}=4 a \sqrt{3}=2 \sqrt{3}$
