Solve the equation

Let $z=x+i y$.

Then,

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

$\therefore|z|=z+1+2 i$

$\Rightarrow \sqrt{x^{2}+y^{2}}=(x+i y)+1+2 i$

$\Rightarrow \sqrt{x^{2}+y^{2}}=(x+1)+i(y+2)$

$\Rightarrow \sqrt{x^{2}+y^{2}}=(x+1)$ and $y+2=0$

$\Rightarrow x^{2}+y^{2}=(x+1)^{2}$ and $y=-2$

$\Rightarrow x^{2}+y^{2}=x^{2}+1+2 x$ and $y=-2$

$\Rightarrow y^{2}=2 x+1$ and $y=-2$

$\Rightarrow 4=2 x+1$ and $y=-2$

$\Rightarrow 2 x=3$ and $y=-2$

$\Rightarrow x=\frac{3}{2}$ and $y=-2$

$\therefore z=x+i y=\frac{3}{2}-2 i$

Thus, $z=\frac{3}{2}-2 i$