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)

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

Answer 2

Let $z = x+iy$. The given condition is $Re(\frac{z-1}{2z+i})=1$. Substituting $z = x+iy$: $\frac{z-1}{2z+i} = \frac{(x-1)+iy}{2x+i(2y+1)}$ Multiply by the conjugate of the denominator: $\frac{((x-1)+iy)(2x-i(2y+1))}{(2x)^2+(2y+1)^2} = \frac{(x-1)(2x) + y(2y+1) + i[y(2x) - (x-1)(2y+1)]}{4x^2+(2y+1)^2}$ The real part is $\frac{2x^2-2x+2y^2+y}{4x^2+(2y+1)^2}$. Given $Re(\frac{z-1}{2z+i})=1$: $\frac{2x^2-2x+2y^2+y}{4x^2+4y^2+4y+1} = 1$ $2x^2-2x+2y^2+y = 4x^2+4y^2+4y+1$ $2x^2+2x+2y^2+3y+1 = 0$ Dividing by 2: $x^2+x+y^2+\frac{3}{2}y+\frac{1}{2} = 0$ Completing the square: $\left(x+\frac{1}{2}\right)^2 - \frac{1}{4} + \left(y+\frac{3}{4}\right)^2 - \frac{9}{16} + \frac{1}{2} = 0$ $\left(x+\frac{1}{2}\right)^2 + \left(y+\frac{3}{4}\right)^2 = \frac{1}{4} + \frac{9}{16} - \frac{1}{2} = \frac{4+9-8}{16} = \frac{5}{16}$ This is the equation of a circle with center $(-\frac{1}{2}, -\frac{3}{4})$ and radius $r = \sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4}$. The diameter is $2r = 2 \times \frac{\sqrt{5}}{4} = \frac{\sqrt{5}}{2}$. The locus is a circle with diameter $\frac{\sqrt{5}}{2}$.