if Re
Question:

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

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

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

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

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

Correct Option: , 4

Solution:

$\because \quad z=x+i y$

$\left(\frac{z-1}{2 z+i}\right)=\frac{(x-1)+i y}{2(x+i y)+i}$

$=\frac{(x-1)+i y}{2 x+(2 y+1) i} \times \frac{2 x-(2 y+1) i}{2 x-(2 y+1) i}$

$\operatorname{Re}\left(\frac{z+1}{2 z+i}\right)=\frac{2 x(x-1)+y(2 y+1)}{(2 x)^{2}+(2 y+1)^{2}}=1$

$\Rightarrow\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{3}{4}\right)^{2}=\left(\frac{\sqrt{5}}{4}\right)^{2}$