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 :

${\mathop{\rm Re}\nolimits} \left( {{{z - 1} \over {2z + i}}} \right) = 1$ <br><br>Put z = x + iy <br><br>$\therefore$ $${\mathop{\rm Re}\nolimits} \left( {{{\left( {x + iy} \right) - 1} \over {2\left( {x + iy} \right) + i}}} \right) = 1$$ <br><br>$\Rightarrow$ $${\mathop{\rm Re}\nolimits} \left( {\left( {{{\left( {x - 1} \right) + iy} \over {2x + i\left( {2y + 1} \right)}}} \right)\left( {{{2x - i\left( {2y + 1} \right)} \over {2x - i\left( {2y + 1} \right)}}} \right)} \right) = 1$$ <br><br>$\Rightarrow$ $${\mathop{\rm Re}\nolimits} \left( {{{\left\{ {\left( {x - 1} \right) + iy} \right\}\left\{ {2x - i\left( {2y + 1} \right)} \right\}} \over {4{x^2} + {{\left( {2y + 1} \right)}^2}}}} \right) = 1$$ <br><br>Real part of this equation is = 1 <br><br>$\therefore$ $${{2x\left( {x - 1} \right) + y\left( {2y + 1} \right)} \over {4{x^2} + {{\left( {2y + 1} \right)}^2}}}$$ = 1 <br><br>$\Rightarrow$ 2x² + 2y² +2x + 3y + 1 = 0 <br><br>$\Rightarrow$ x² + y² +x + ${3 \over 2}$y + ${1 \over 2}$ = 0 <br><br>This is an equation of circle. <br><br>$\therefore$ Locus is a circle whose <br><br>center is $\left( { - {1 \over 2}, - {3 \over 4}} \right)$ and radius ${{\sqrt 5 } \over 4}$ <br><br>$\therefore$ Diameter = 2 $\times$ ${{\sqrt 5 } \over 4}$ = ${{\sqrt 5 } \over 2}$