If the locus of $z \in \mathrm{C}$, such that
$\operatorname{Re}\left(\frac{z-1}{2 z+\mathrm{i}}\right)+\operatorname{Re}\left(\frac{\bar{z}-1}{2 \bar{z}-\mathrm{i}}\right)=2$
is a circle of radius $r$ and center $(a, b)$ then $\frac{15 a b}{r^2}$ is equal to :

Question

Aicharya · Accepted Answer

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

If the locus of $z \in \mathrm{C}$, such that $\operatorname{Re}\left(\frac{z-1}{2…

Solution