QJEEM20Maeb54: JEE Main 2020 Mathematics - Class 11 JEEM20Maeb54 Solution: If , where z = x + iy, then the point (x, y) lies on a :

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

{\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}

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