If Re((z - 1)/(2z + i)) = 1, where z = x + iy, then the point (x,...

Let $z = x + iy$
\left(\dfrac{z - 1}{2z + i}\right) = \dfrac{\left(x - 1\right) + iy}{2\left(x + iy\right) + i} = \dfrac{\left(x - 1\right) + iy}{2x + \left(2y + 1\right)i} \times \dfrac{2x - \left(2y + 1\right)i}{2x - \left(2y + 1\right)i}$<br />$Re\left(\dfrac{z + 1}{2z + i}\right) = \dfrac{2x\left(x - 1\right) + y\left(2x + 1\right)}{\left(2{x}^{2}\right) + {\left(2y + 1\right)}^{2}} = 1$<br />$⇒ 2{x}^{2} + 2{y}^{2} - 2x + y = 4{x}^{2} + 4{y}^{2} + 4y + 1$<br />$⇒ 2{x}^{2} + 2{y}^{2} + 2x + 3y + 1 = 0$<br />$⇒ {x}^{2} + {y}^{2} + x + \dfrac{3}{2}y + \dfrac{1}{2} = 0$</p><p>Circle with center \left(- \dfrac{1}{2}, - \dfrac{3}{4}\right)$
$r = \sqrt{\dfrac{1}{4} + \dfrac{9}{16} - \dfrac{1}{2}} = \sqrt{\dfrac{4 + 9 - 8}{16}} = \dfrac{\sqrt{5}}{4} ⇒ d = \dfrac{\sqrt{5}}{2}$

where, $d =$diameter of the given circle.