If $$\operatorname{Re} \left( {\frac{{z + 4}}{{2z - i}}} \right) = \frac{1}{2}$$    then $$z$$ is represented by a point lying on

Solution :
$$\eqalign{ & \frac{{z + 4}}{{2z - i}} = \frac{{\left( {x + 4} \right) + iy}}{{2x + i\left( {2y - 1} \right)}} = \frac{{\left\{ {\left( {x + 4} \right) + iy} \right\}\left\{ {2x - i\left( {2y - 1} \right)} \right\}}}{{\left\{ {2x + i\left( {2y - 1} \right)} \right\}\left\{ {2x - i\left( {2y - 1} \right)} \right\}}} \cr & \frac{{z + 4}}{{2z - i}} = \frac{{2x\left( {x + 4} \right) + y\left( {2y - 1} \right) + i\left\{ {2xy - \left( {x + 4} \right)\left( {2y - 1} \right)} \right\}}}{{4{x^2} + {{\left( {2y - 1} \right)}^2}}} \cr & \operatorname{Re} \left( {\frac{{z + 4}}{{2z - i}}} \right) = \frac{1}{2} \cr & \Rightarrow \,\,\frac{{2x\left( {x + 4} \right) + y\left( {2y - 1} \right)}}{{4{x^2} + {{\left( {2y - 1} \right)}^2}}} = \frac{1}{2} \cr} $$
$$ \Rightarrow {\kern 1pt} \,\,16x + 2y = 1,$$    which represents a straight line.