Question:
If $\operatorname{Re}\left(\frac{z-1}{2 z+i}\right)=1$, where $z=x+i y$, then the point $(x, y)$ lies on $a$ :
Correct Option: , 4
Solution:
$\because \quad z=x+i y$
$\left(\frac{z-1}{2 z+i}\right)=\frac{(x-1)+i y}{2(x+i y)+i}$
$=\frac{(x-1)+i y}{2 x+(2 y+1) i} \times \frac{2 x-(2 y+1) i}{2 x-(2 y+1) i}$
$\operatorname{Re}\left(\frac{z+1}{2 z+i}\right)=\frac{2 x(x-1)+y(2 y+1)}{(2 x)^{2}+(2 y+1)^{2}}=1$
$\Rightarrow\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{3}{4}\right)^{2}=\left(\frac{\sqrt{5}}{4}\right)^{2}$
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