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

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Complex Numbers

z_{1}, z_{2}

are complex numbers such that

Re (z_{1}) = ∣ z_{1} - 1∣, Re (z_{2}) = ∣ z_{2} - 1∣

and

ar g (z_{1} - z_{2}) = \frac{π}{6}

, then

Im (z_{1} + z_{2})

is equal to:

Answer: (A)

The condition

Re (z) = ∣ z - 1∣

z = x + i y

reads

x = (x - 1)^{2} + y^{2}

, which squares to

y^{2} = 2 x - 1

; both

z_{1}

and

z_{2}

lie on this parabola, so

y_{1}^{2} = 2 x_{1} - 1

and

y_{2}^{2} = 2 x_{2} - 1

. Subtracting gives

(y_{1} - y_{2}) (y_{1} + y_{2}) = 2 (x_{1} - x_{2})

. The argument condition gives

tan \frac{π}{6} = \frac{y _{1} - y _{2}}{x _{1} - x _{2}} = \frac{1}{3}

, so

\frac{x _{1} - x _{2}}{y _{1} - y _{2}} = 3

. Hence

y_{1} + y_{2} = \frac{2 ( x _{1} - x _{2} )}{y _{1} - y _{2}} = 23

, which is

Im (z_{1} + z_{2}) = 23

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option