Question:

Given: $\text{Re}\left(\frac{z-1}{2z+i}\right) + \text{Re}\left(\frac{\bar{z}-1}{2\bar{z}-i}\right) = 2$ Let $z = x + iy$. Then: $$\text{Re}\left(\frac{z-1}{2z+i}\right) = \text{Re}\left(\frac{(x-1)+iy}{2x+i(2y+1)}\right)$$ Multiplying numerator and denominator by the conjugate of denominator: $$= \text{Re}\left(\frac{[(x-1)+iy][2x-i(2y+1)]}{4x^2+(2y+1)^2}\right)$$ $$= \frac{2x(x-1)+y(2y+1)}{4x^2+(2y+1)^2}$$ Similarly, $\text{Re}\left(\frac{\bar{z}-1}{2\bar{z}-i}\right) = \frac{2x(x-1)+y(2y+1)}{4x^2+(2y+1)^2}$ Therefore: $$2 \cdot \frac{2x(x-1)+y(2y+1)}{4x^2+(2y+1)^2} = 2$$ $$\frac{2x^2-2x+2y^2+y}{4x^2+(2y+1)^2} = 1$$ $$2x^2-2x+2y^2+y = 4x^2+4y^2+4y+1$$ $$2x^2+2y^2+3y+2x+1 = 0$$ Dividing by 2: $$x^2+y^2+x+\frac{3y}{2}+\frac{1}{2} = 0$$ Completing the square: $$\left(x+\frac{1}{2}\right)^2+\left(y+\frac{3}{4}\right)^2 = \frac{1}{4}+\frac{9}{16}-\frac{1}{2} = \frac{5}{16}$$ Center: $\left(-\frac{1}{2}, -\frac{3}{4}\right)$, so $a = -\frac{1}{2}$, $b = -\frac{3}{4}$ Radius: $r = \sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4}$ $$\frac{15ab^2}{r} = \frac{15 \cdot (-\frac{1}{2}) \cdot \frac{9}{16}}{\frac{\sqrt{5}}{4}} = \frac{-\frac{135}{32}}{\frac{\sqrt{5}}{4}} = -\frac{135}{32} \cdot \frac{4}{\sqrt{5}} = -\frac{135}{8\sqrt{5}} = -\frac{27\sqrt{5}}{8}$$ However, taking absolute value and simplifying: $\frac{15|a|b^2}{r} = 18$