If z _ 1 , z _ 2 are complex numbers such that Re ( z _ 1 ) = | z _ 1 - 1| and Re ( z _ 2 ) = | z _ 2 - 1| and arg( z _ 1 - z _ 2 ) = (π)/(6) , then Im( z _ 1 + z _ 2 ) is equal to :

Correct answer: A. 2 3 Let z _ 1 = x _ 1 + i y _ 1 and z _ 2 = x _ 2 + i y _ 2 Given | z _ 1 - 1| = Re( z _ 1 ) ⇒ ( x _ 1 - 1) ^ 2 + y _ 1 ^ 2 = x _ 1 ^ 2 ⇒ y _ 1 ^ 2 - 2 x _ 1 + 1 = 0 . . . . . . . . . . . . . . (1) | z _ 2 - 1| = Re ( z _ 2 ) ⇒ ( x _ 2 - 1) ^ 2 + y _ 2 ^ 2 = x _ 2 ^ 2 ⇒ y _ 2 ^ 2 - 2 x _ 2 + 1 = 0 . . . . . . . . . . . (2) Subtracting equation (2) from equation (1) , we get y _ 1 ^ 2 - y _ 2 ^ 2 - 2( x _ 1 - x _ 2 ) = 0 ⇒ ( y _ 1 - y _ 2 )( y _ 1 + y _ 2 ) = 2( x _ 1 - x _ 2 ) ⇒ y _ 1 + y _ 2 = 2( x _ 1 - x _ 2 y _ 1 - y _ 2 ) . . . . . . . . . . . . (3) Also given, arg ( z _ 1 - z _ 2 )=(π)/(6) ⇒ ^ - 1 ( y _ 1 - y _ 2 x _ 1 - x _ 2 ) = (π)/(6) ⇒ y _ 1 - y _ 2 x _ 1 - x _ 2 = 1 3 . . . . . . . . . (4) Putting equation (4) in equation (3), we get y _ 1 + y _ 2 = 2 3 ⇒ Im ( z _ 1 + z _ 2 ) = 2 3

JEE Main 20203 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 3 September
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2020 · 3 September · Evening Shift · Mathematics · Complex Number

If z _ 1 , z _ 2 are complex numbers such that Re ( z _ 1 ) = | z _...

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

Options

A $2\sqrt{3}$
B $\dfrac{\sqrt{3}}{2}$
C $\dfrac{1}{\sqrt{3}}$
D $\dfrac{2}{\sqrt{3}}$

Correct answer

A. $2\sqrt{3}$

Step-by-step solution

Let ${z}_{1} = {x}_{1} + i{y}_{1} and {z}_{2} = {x}_{2} + i{y}_{2}$

Given \left|{z}_{1} - 1\right| = Re\left({z}_{1}\right)

$⇒ {\left({x}_{1} - 1\right)}$ ^{2} + {y}_{1}^{2} = {x}_{1}^{2} $$ ⇒ {y}_{1}^{2} - 2{x}_{1} + 1 = 0 . . . . . . . . . . . . . . \left(1\right) $\left|{z}_{2} - 1\right|$ = Re \left({z}_{2}\right) $$ ⇒ {\left({x}_{2} - 1\right)}^{2} + {y}_{2}^{2} = {x}_{2}^{2} $$ ⇒ {y}_{2}^{2} - 2{x}_{2} + 1 = 0 . . . . . . . . . . . \left(2\right) $Subtracting equation \left(2\right)$ from equation \left(1\right) $, we get$ {y}_{1}^{2} - {y}_{2}^{2} - 2\left({x}_{1} - {x}_{2}\right) = 0 $$ ⇒ \left({y}_{1} - {y}_{2}\right)\left({y}_{1} + {y}_{2}\right) = 2\left({x}_{1} - {x}_{2}\right) $$ ⇒ {y}_{1} + {y}_{2} = 2\left(\dfrac{{x}_{1} - {x}_{2}}{{y}_{1} - {y}_{2}}\right) . . . . . . . . . . . . \left(3\right)

Also given, $arg \left({z}_{1} - {z}_{2}\right)=\dfrac{\pi}{6}$ ⇒ $\tan^{- 1}\left(\dfrac{{y}_{1} - {y}_{2}}{{x}_{1} - {x}_{2}}\right) = \dfrac{\pi}{6}$

$⇒ \dfrac{{y}_{1} - {y}_{2}}{{x}_{1} - {x}_{2}} = \dfrac{1}{\sqrt{3}} . . . . . . . . . \left(4\right)Putting equation \left(4\right)$ in $equation \left(3\right), we get$ {y}_{1} + {y}_{2} = 2\sqrt{3} ⇒ Im \left({z}_{1} + {z}_{2}\right) $= 2\sqrt{3}$

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