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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 :

JEE · Math · previous-year question

  1. A.straight line whose slope is $${3 \over 2}$$
  2. B.straight line whose slope is $$-{2 \over 3}$$
  3. C.circle whose diameter is $${{\sqrt 5 } \over 2}$$correct
  4. D.circle whose centre is at $$\left( { - {1 \over 2}, - {3 \over 2}} \right)$$

Answer

C. circle whose diameter is $${{\sqrt 5 } \over 2}$$

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 $$ \times $$ $${{\sqrt 5 } \over 4}$$ = $${{\sqrt 5 } \over 2}$$

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