The number of solutions of the equation ${\mathrm{z}}^2-\mathrm{z}-|\mathrm{z}{|}^2+\frac{64}{|\mathrm{z}{|}^5}=0$ is (where z = x + iy $\mathrm{x},\mathrm{y}\in \mathrm{R}$  and $\left.\mathrm{x}\ne \frac{1}{2}\right)$

Here ${\mathrm{z}}^2-\mathrm{z}=\left|{\mathrm{z}}^2\right|-\frac{64}{|\mathrm{z}{|}^5}----\left(\mathrm{i}\right)$

RHS is real number$\Rightarrow$${\mathrm{z}}^2-\mathrm{z} \mathrm{is} \mathrm{purely} \mathrm{real} \mathrm{number}$
$\Rightarrow {\mathrm{z}}^2-\mathrm{z}={\overline{\mathrm{z}}}^2-\overline{\mathrm{z}}\left(\because {\mathrm{z}}^2-\mathrm{z}\right. \mathrm{is} \mathrm{purely} \mathrm{real} \mathrm{number})$
$\Rightarrow (\mathrm{z}-\overline{\mathrm{z}})(\mathrm{z}+\overline{\mathrm{z}}-1)=0\Rightarrow \mathrm{z}=\overline{\mathrm{z}}\text{ as }\mathrm{z}+\overline{\mathrm{z}}=1$
is not possible $\left(\because \mathrm{x}\ne \frac{1}{2}\right)\Rightarrow \mathrm{z}=\mathrm{x}$
$\therefore$Equation (i) given as ${\mathrm{x}}^2-\mathrm{x}-\left|{\mathrm{x}}^2\right|+\frac{64}{|\mathrm{x}{|}^5}=0$
$\Rightarrow \mathrm{x} = 2 \mathrm{only} \mathrm{one} \mathrm{solution}$