Question

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

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

Answer 1

The correct option is C circle whose diameter is $\frac{\sqrt{5}}{2}$.

$z=x+iy\frac{x+iy-1}{2x+2iy+i}=\frac{(x-1)+iy}{2x+i(2y+1)}\left(\frac{2x-i(2y+1)}{2x-i(2y+1)}\right)\Rightarrow \frac{2x(x-1)+y(2y+1)}{4x^2+(2y+1{)}^2}=1\Rightarrow 2x^2+2y^2-2x+y=4x^2+4y^2+4y+1\Rightarrow x^2+y^2+x+\frac{3}{2}y+\frac{1}{2}=0$

Circle's centre will be $(-\frac{1}{2},-\frac{3}{4})$

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

Diameter $=\frac{\sqrt{5}}{2}$