JEE MAIN - Mathematics (2020 - 3rd September Evening Slot - No. 16)
If z1
, z2
are complex numbers such that
Re(z1) = |z1 – 1|, Re(z2) = |z2 – 1| , and
arg(z1 - z2) = $${\pi \over 6}$$, then Im(z1 + z2 ) is equal to :
Re(z1) = |z1 – 1|, Re(z2) = |z2 – 1| , and
arg(z1 - z2) = $${\pi \over 6}$$, then Im(z1 + z2 ) is equal to :
$${{\sqrt 3 } \over 2}$$
$${1 \over {\sqrt 3 }}$$
$${2 \over {\sqrt 3 }}$$
$${2\sqrt 3 }$$
Explanation
Let $${z_1} = {x_1} + i{y_1},\,{z_2} = {x_2} + i{y_2}$$
Given Re(z1) = |z1 – 1|
$$ \therefore $$ x1 = |(x1 - 1) + iy1|
$$ \Rightarrow $$ x1 = $$\sqrt {{{\left( {{x_1} - 1} \right)}^2} + y_1^2} $$
$$ \Rightarrow $$ $${x_1}^2 = {({x_1} - 1)^2} + {y_1}^2$$
$$ \Rightarrow {y_1}^2 - 2{x_1} + 1 = 0$$
Also given Re(z2) = |z2 – 1|
$$ \therefore $$ x2 = |(x2 - 1) + iy2|
$$ \Rightarrow $$ $${x_2}^2 = {({x^2} - 1)^2} + {y_2}^2$$
$$ \Rightarrow $$ $$y_2^2 - 2{x_2} - 1 = 0$$
Performing equation (1) - (2),
$$({y_1}^2 - {y_2}^2) + 2({x_2} - {x_1}) = 0$$
$$ \Rightarrow $$ $$({y_1} + {y_2})({y_1} - {y_2}) = 2({x_1} - {x_2})$$
$$ \Rightarrow $$ $${y_1} + {y_2} = 2\left( {{{{x_1} - {x_2}} \over {{y_1} - {y_2}}}} \right)$$
Now given, $$\arg ({z_1} - {z_2}) = {\pi \over 6}$$
$$ \Rightarrow $$ $${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 6}$$
$$ \Rightarrow {{{y_1} - {y_2}} \over {{x_1} - {x_2}}} = {1 \over {\sqrt 3 }}$$
$$ \therefore $$ $${y_1} + {y_2} = 2\sqrt 3 $$
Given Re(z1) = |z1 – 1|
$$ \therefore $$ x1 = |(x1 - 1) + iy1|
$$ \Rightarrow $$ x1 = $$\sqrt {{{\left( {{x_1} - 1} \right)}^2} + y_1^2} $$
$$ \Rightarrow $$ $${x_1}^2 = {({x_1} - 1)^2} + {y_1}^2$$
$$ \Rightarrow {y_1}^2 - 2{x_1} + 1 = 0$$
Also given Re(z2) = |z2 – 1|
$$ \therefore $$ x2 = |(x2 - 1) + iy2|
$$ \Rightarrow $$ $${x_2}^2 = {({x^2} - 1)^2} + {y_2}^2$$
$$ \Rightarrow $$ $$y_2^2 - 2{x_2} - 1 = 0$$
Performing equation (1) - (2),
$$({y_1}^2 - {y_2}^2) + 2({x_2} - {x_1}) = 0$$
$$ \Rightarrow $$ $$({y_1} + {y_2})({y_1} - {y_2}) = 2({x_1} - {x_2})$$
$$ \Rightarrow $$ $${y_1} + {y_2} = 2\left( {{{{x_1} - {x_2}} \over {{y_1} - {y_2}}}} \right)$$
Now given, $$\arg ({z_1} - {z_2}) = {\pi \over 6}$$
$$ \Rightarrow $$ $${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 6}$$
$$ \Rightarrow {{{y_1} - {y_2}} \over {{x_1} - {x_2}}} = {1 \over {\sqrt 3 }}$$
$$ \therefore $$ $${y_1} + {y_2} = 2\sqrt 3 $$
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