Find the number of solutions of

Let $z=x+i y$.

Then,

$|z|=\sqrt{x^{2}+y^{2}}$

$\therefore z^{2}+|z|^{2}=0$

$\Rightarrow(x+i y)^{2}+\left(\sqrt{x^{2}+y^{2}}\right)^{2}=0$

$\Rightarrow x^{2}+i^{2} y^{2}+2 i x y+x^{2}+y^{2}=0$

$\Rightarrow x^{2}-y^{2}+2 i x y+x^{2}+y^{2}=0$

$\Rightarrow 2 x^{2}+2 i x y=0$

$\Rightarrow 2 x(x+i y)=0$

$\Rightarrow x=0$ or $x+i y=0$

$\Rightarrow x=0$ or $z=0$

For $x=0, z=0+i y$

Thus, there are infinitely many solutions of the form $z=0+i y, y \in \Re$.