Mathematics - Complex Number Question with Solution | TestHub

We have,

$Z=\frac{2z-3i}{4z+2i}$

$\Rightarrow Z=\frac{2x+i\left(2y-3\right)}{4x+i\left(4y+2\right)}$

$\Rightarrow Z=\frac{2x+i\left(2y-3\right)}{4x+i\left(4y+2\right)}\times \frac{4x-i\left(4y+2\right)}{4x-i\left(4y+2\right)}$

$\Rightarrow Z=\frac{8x^2+\left(4y+2\right)\left(2y-3\right)+i\left[4x\left(2y-3\right)-\left(8xy+4x\right)\right]}{16x^2+{\left(4y+2\right)}^2}$

Since, $Z$ is purely real, so $\frac{4x\left(2y-3\right)-\left(8xy+4x\right)}{16x^2+{\left(4y+2\right)}^2}=0$

$\Rightarrow -12x+8xy-\left(8xy+4x\right)=0$

$\Rightarrow x=0$

And,

$4x+i\left(4y+2\right)\ne 0$

$\Rightarrow i\left(4y+2\right)\ne 0$

$\Rightarrow y\ne -\frac{1}{2}$

Hence this is the required option.