JEE MAIN - Mathematics (2020 - 7th January Morning Slot - No. 2)
If $${\mathop{\rm Re}\nolimits} \left( {{{z - 1} \over {2z + i}}} \right) = 1$$, where z = x + iy, then the point (x, y) lies on a :
straight line whose slope is $${3 \over 2}$$
straight line whose slope is $$-{2 \over 3}$$
circle whose diameter is $${{\sqrt 5 } \over 2}$$
circle whose centre is at $$\left( { - {1 \over 2}, - {3 \over 2}} \right)$$
Explanation
$${\mathop{\rm Re}\nolimits} \left( {{{z - 1} \over {2z + i}}} \right) = 1$$
Put z = x + iy
$$ \therefore $$ $${\mathop{\rm Re}\nolimits} \left( {{{\left( {x + iy} \right) - 1} \over {2\left( {x + iy} \right) + i}}} \right) = 1$$
$$ \Rightarrow $$ $${\mathop{\rm Re}\nolimits} \left( {\left( {{{\left( {x - 1} \right) + iy} \over {2x + i\left( {2y + 1} \right)}}} \right)\left( {{{2x - i\left( {2y + 1} \right)} \over {2x - i\left( {2y + 1} \right)}}} \right)} \right) = 1$$
$$ \Rightarrow $$ $${\mathop{\rm Re}\nolimits} \left( {{{\left\{ {\left( {x - 1} \right) + iy} \right\}\left\{ {2x - i\left( {2y + 1} \right)} \right\}} \over {4{x^2} + {{\left( {2y + 1} \right)}^2}}}} \right) = 1$$
Real part of this equation is = 1
$$ \therefore $$ $${{2x\left( {x - 1} \right) + y\left( {2y + 1} \right)} \over {4{x^2} + {{\left( {2y + 1} \right)}^2}}}$$ = 1
$$ \Rightarrow $$ 2x2 + 2y2 +2x + 3y + 1 = 0
$$ \Rightarrow $$ x2 + y2 +x + $${3 \over 2}$$y + $${1 \over 2}$$ = 0
This is an equation of circle.
$$ \therefore $$ Locus is a circle whose
center is $$\left( { - {1 \over 2}, - {3 \over 4}} \right)$$ and radius $${{\sqrt 5 } \over 4}$$
$$ \therefore $$ Diameter = 2 $$ \times $$ $${{\sqrt 5 } \over 4}$$ = $${{\sqrt 5 } \over 2}$$
Put z = x + iy
$$ \therefore $$ $${\mathop{\rm Re}\nolimits} \left( {{{\left( {x + iy} \right) - 1} \over {2\left( {x + iy} \right) + i}}} \right) = 1$$
$$ \Rightarrow $$ $${\mathop{\rm Re}\nolimits} \left( {\left( {{{\left( {x - 1} \right) + iy} \over {2x + i\left( {2y + 1} \right)}}} \right)\left( {{{2x - i\left( {2y + 1} \right)} \over {2x - i\left( {2y + 1} \right)}}} \right)} \right) = 1$$
$$ \Rightarrow $$ $${\mathop{\rm Re}\nolimits} \left( {{{\left\{ {\left( {x - 1} \right) + iy} \right\}\left\{ {2x - i\left( {2y + 1} \right)} \right\}} \over {4{x^2} + {{\left( {2y + 1} \right)}^2}}}} \right) = 1$$
Real part of this equation is = 1
$$ \therefore $$ $${{2x\left( {x - 1} \right) + y\left( {2y + 1} \right)} \over {4{x^2} + {{\left( {2y + 1} \right)}^2}}}$$ = 1
$$ \Rightarrow $$ 2x2 + 2y2 +2x + 3y + 1 = 0
$$ \Rightarrow $$ x2 + y2 +x + $${3 \over 2}$$y + $${1 \over 2}$$ = 0
This is an equation of circle.
$$ \therefore $$ Locus is a circle whose
center is $$\left( { - {1 \over 2}, - {3 \over 4}} \right)$$ and radius $${{\sqrt 5 } \over 4}$$
$$ \therefore $$ Diameter = 2 $$ \times $$ $${{\sqrt 5 } \over 4}$$ = $${{\sqrt 5 } \over 2}$$
Comments (0)
