Sum of squares of modulus of all the complex numbers z satisfying $\overline{\mathrm{z}}={\mathrm{iz}}^2+{\mathrm{z}}^2-\mathrm{z}$ is equal to ________.

$\begin{array}{l}\overline{\mathrm{z}}={\mathrm{iz}}^2+{\mathrm{z}}^2-\mathrm{z}\\ \Rightarrow \mathrm{z}+\overline{\mathrm{z}}={\mathrm{iz}}^2+{\mathrm{z}}^2\\ \text{ Consider }\mathrm{z}=\mathrm{x}+\mathrm{iy}\\ 2\mathrm{x}=(\mathrm{i}+1)\left({\mathrm{x}}^2-{\mathrm{y}}^2+2\mathrm{xyi}\right)\\ \Rightarrow 2\mathrm{x}={\mathrm{x}}^2-{\mathrm{y}}^2-2\mathrm{xy}\text{ and }{\mathrm{x}}^2-{\mathrm{y}}^2+2\mathrm{xy}=0\\ \Rightarrow 2\mathrm{x}=-4\mathrm{xy}\end{array}$

$\Rightarrow \mathrm{x}=0\text{ or }\mathrm{y}=\frac{-1}{2}$

Case 1 : $\mathrm{x}=0\Rightarrow \mathrm{y}=0\text{ here }\mathrm{z}=0$

Case 2 : $\mathrm{y}=\frac{-1}{2}$

$\begin{array}{l}\Rightarrow 4{\mathrm{x}}^2-4\mathrm{x}-1=0\\ (2\mathrm{x}-1{)}^2=2\\ 2\mathrm{x}-1=\pm \sqrt{2}\\ \mathrm{x}=\frac{1\pm \sqrt{2}}{2}\end{array}$

Here $\mathrm{z}=\frac{1+\sqrt{2}}{2}-\frac{\mathrm{i}}{2}\text{ or }\mathrm{z}=\frac{1-\sqrt{2}}{2}-\frac{\mathrm{i}}{2}$

Sum of squares of modulus of z 

$=0+\frac{(1+\sqrt{2}{)}^2+1}{4}+\frac{(1-\sqrt{2}{)}^2+1}{4}=\frac{8}{4}=2$