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

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

(A) straight line whose slope is $\frac{3}{2}$

(C) straight line whose slope is $-\frac{2}{3}$

(D) circle whose centre is at $\left(-\frac{1}{2},-\frac{3}{2}\right)$

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Complex Numbers

Re (\frac{z - 1}{2 z + i}) = 1

, where

z = x + i y

, then the point

(x, y)

lies on a:

Answer: (B)

Put

z = x + i y

\frac{z - 1}{2 z + i}

and set its real part to

1

. Clearing the denominator leads to

2 x^{2} + 2 y^{2} + 2 x + 3 y + 1 = 0

, i.e.

x^{2} + y^{2} + x + \frac{3}{2} y + \frac{1}{2} = 0

, a circle centred at

(- \frac{1}{2}, - \frac{3}{4})

with radius

\frac{1}{4} + \frac{9}{16} - \frac{1}{2} = \frac{5}{4}

. Its diameter is

\frac{5}{2}

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Single correct option