If Re(z-12z+i)=1, where z=x+iy, then the point (x,y) lies on a:

Question

If $\operatorname{Re}\left(\dfrac{z-1}{2z+i}\right)=1$, 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 $$ \times $$ $${{\sqrt 5 } \over 4}$$ = $${{\sqrt 5 } \over 2}$$

Solutions · 1