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