Question

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

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

Answer 1

The correct option is C

A circle whose diameter is $\frac{\sqrt{5}}{2}$

Explanation for the correct option:

Step 1: Simplification of the given relation

Substitute $z=x+iy$ in the given relation, we have

$\begin{array}{rcl}\frac{\left(z-1\right)}{\left(2z+i\right)}& =& \frac{\left(x+iy-1\right)}{\left(2\left(x+iy\right)+i\right)}\\ & =& \frac{\left(x+iy-1\right)}{2x+i\left(2y+1\right)}\end{array}$

Now, multiply the numerator and the denominator of the RHS by the conjugate of $2x+i\left(2y+1\right)$ that will be $2x-i\left(2y+1\right)$, we get

$\begin{array}{rcl}\frac{\left(z-1\right)}{\left(2z+1\right)}& =& \frac{(x-1)+iy}{2x+i\left(2y+1\right)}\times \frac{2x-i\left(2y+1\right)}{2x-i\left(2y+1\right)}\\ & =& \frac{2x(x-1)+i2xy-i(x-1)(2y+1)+y(2y+1)}{{\left(2x\right)}^2+{\left(2y+1\right)}^2}\\ & =& \frac{2x(x-1)+y(2y+1)+i\left(2xy-(x-1)(2y+1)\right)}{{\left(2x\right)}^2+{\left(2y+1\right)}^2}\end{array}$

Thus, the real part of the given relation is $\frac{2x(x-1)+y(2y+1)}{{\left(2x\right)}^2+{(2y+1)}^2}$

Step 2: Formation of the equation to determine the correct option

Since, $\mathrm{Re}\left[\frac{\left(z-1\right)}{\left(2z+i\right)}\right]=1$

$\begin{array}{c}\therefore \frac{2x(x-1)+y(2y+1)}{{\left(2x\right)}^2+{(2y+1)}^2}=1\\ \Rightarrow 2x^2-2x+2y^2+y=4x^2+4y^2+4y+1\\ \Rightarrow 2x^2+2y^2+2x+3y+1=0\\ \Rightarrow x^2+y^2+x+\frac{3}{2}y+\frac{1}{2}=0\end{array}$

As the general form of the equation of the circle is $x^2+y^2+2gx+2fy+c=0$. Then after comparing, we have

$g=\frac{1}{2},f=\frac{3}{4},\text{and}c=\frac{1}{2}$

From this, the center of the circle is $\left(\frac{1}{2},\frac{3}{4}\right)$

We know, that the radius of the circle is given by $\mathit{r}\mathbf{=}\sqrt{{\mathbf{g}}^{\mathbf{2}}\mathbf{+}{\mathbf{f}}^{\mathbf{2}}\mathbf{-}\mathbf{c}}$,

$\begin{array}{rcl}r& =& \sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{3}{4}\right)}^2-\frac{1}{2}}\\ & =& \sqrt{\frac{1}{4}+\frac{9}{16}-\frac{1}{2}}\\ & =& \sqrt{\frac{4+9-8}{16}}\\ & =& \sqrt{\frac{5}{16}}\\ & =& \frac{\sqrt{5}}{4}\text{units}\end{array}$

Thus, the diameter of the circle would be

$\begin{array}{rcl}d& =& 2r\\ & =& \frac{\sqrt{5}}{2}\text{units}\end{array}$

Hence, the correct option is (C).