Question

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

• A. circle whosecentre is at $\left(-\frac{1}{2},-\frac{3}{2}\right).$
• 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

Answer: (C)

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

Put $z=x+iy$

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

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

$\Rightarrow 2x^2+2y^2+2x+3y+1=0$

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

$\Rightarrow$ locus is a circle whose

Centre is $\left(-\frac{1}{2},-\frac{3}{4}\right)$ and radius $\frac{\sqrt{5}}{4}$

$\Rightarrow$ diameter $=\frac{\sqrt{5}}{2}$

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

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

$\Rightarrow 2x^2+2y^2+2x+3y+1=0$

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

$\Rightarrow$ locus is a circle whose

Centre is $\left(-\frac{1}{2},-\frac{3}{4}\right)$ and radius $\frac{\sqrt{5}}{4}$
$\Rightarrow$ diameter $=\frac{\sqrt{5}}{2}$