Question

If $Re(\frac{z-1}{2z+i}) = 1$, where $z = x + iy$, then the point (x, y) lies on a

• A. circle whose diameter is $\frac{\sqrt{5}}{2}$.
• B. straight line whose slope is $\frac{3}{2}$.
• C. circle whose centre is at $(-\frac{1}{2}, -\frac{3}{2})$.
• D. straight line whose slope is $-\frac{2}{3}$.

Answer 1

Answer: (A)

(1) circle whose diameter is $\frac{\sqrt{5}}{2}$.

Answer 2

Let $w = \frac{z-1}{2z+i}$. We are given $Re(w) = 1$. Substituting $z = x+iy$: $w = \frac{(x-1) + iy}{2x + i(2y+1)}$ Multiply by the conjugate of the denominator: $w = \frac{((x-1) + iy)(2x - i(2y+1))}{(2x)^2 + (2y+1)^2}$ The numerator is: $(2x^2 - 2x) - i(x-1)(2y+1) + i(2xy) + y(2y+1)$ $= (2x^2 - 2x + 2y^2 + y) + i(-2xy - x + 2y + 1 + 2xy)$ $= (2x^2 - 2x + 2y^2 + y) + i(-x + 2y + 1)$ The denominator is: $4x^2 + (4y^2 + 4y + 1) = 4x^2 + 4y^2 + 4y + 1$ So, $Re(w) = \frac{2x^2 - 2x + 2y^2 + y}{4x^2 + 4y^2 + 4y + 1}$. Given $Re(w) = 1$: $2x^2 - 2x + 2y^2 + y = 4x^2 + 4y^2 + 4y + 1$ $2x^2 + 2y^2 + 2x + 3y + 1 = 0$ Dividing by 2: $x^2 + y^2 + x + \frac{3}{2}y + \frac{1}{2} = 0$ This is the equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$. Here, $2g = 1 \implies g = \frac{1}{2}$, $2f = \frac{3}{2} \implies f = \frac{3}{4}$, $c = \frac{1}{2}$. The center is $(-g, -f) = (-\frac{1}{2}, -\frac{3}{4})$. The radius squared is $r^2 = g^2 + f^2 - c = (\frac{1}{2})^2 + (\frac{3}{4})^2 - \frac{1}{2} = \frac{1}{4} + \frac{9}{16} - \frac{1}{2} = \frac{4+9-8}{16} = \frac{5}{16}$. The radius is $r = \sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4}$. The diameter is $2r = 2 \times \frac{\sqrt{5}}{4} = \frac{\sqrt{5}}{2}$. Option (1) matches this result. Option (3) has an incorrect center. Options (2) and (4) describe lines, which is incorrect.