Mathematics - Complex Number Question with Solution | TestHub

$\begin{array}{rl}& \mathrm{Re}\left(\frac{z-1}{2z+i}\right)+\mathrm{Re}\left(\frac{\bar{z}-1}{2\bar{z}-i}\right)=2\\ & \text{ Here, }\frac{z-1}{2z+i}=\left(\frac{\overline{\bar{z}-1}}{2\bar{z}-i}\right)=2\\ & =\mathrm{Re}\left(\frac{z-1}{2z+i}\right)+\mathrm{Re}\left(\frac{\mathrm{z}-1}{2z+i}\right)=2\\ & =2\mathrm{Re}\left(\frac{z-1}{2z+1}\right)=2\Rightarrow \mathrm{Re}\left(\frac{z-1}{2z+i}\right)=1\end{array}$
Let $z=x+iy$
$\begin{array}{rl}& \mathrm{Re}\left(\frac{(x-1)+iy}{2x+i(2y+1)}\right)=1\Rightarrow \mathrm{Re}\left[\frac{((x-1)+iy)(2x-i(y+1)}{(2x+i(2y+1)(2x-i(2y+1))}\right]=1\\ & \Rightarrow \frac{2x(x-1)+y(2y+1)}{4x^2+(2y+1{)}^2}=1\\ & \Rightarrow 2x^2-2x+2y^2+y=4x^2+4y^2+1+4y\\ & \Rightarrow 2x^2+2y^2+3y+2x+1=0\\ & \Rightarrow x^2+y^2+x+\frac{3}{2}y+\frac{1}{2}=0\\ & \text{ centre }=\left(\frac{-1}{2},\frac{-3}{4}\right),r=\sqrt{\frac{1}{4}+\frac{9}{16}-\frac{1}{2}}=\frac{\sqrt{5}}{4}\\ & a=\frac{-1}{2},b=\frac{-3}{4},r^2=\frac{5}{16}\\ & 15\frac{ab}{r^2}=15\times \left(\frac{-1}{2}\right)\times \left(\frac{-3}{4}\right)\times \frac{16}{5}=18\end{array}$