The solution of the equation $\left|z\right|-z=1+2i$ is

$\left|z\right|-z=1+2i$ 

Let $z=x+iy,$ therefore $|x+iy|-(x+iy)=1+2i$

Equating real and imaginary parts, we get 

$\sqrt{x^2+y^2}-x=1$ and $y=-2 \Rightarrow x=\frac{3}{2}$

Hence complex number $z=\frac{3}{2}-2i.$

Trick : Since $\left|\frac{{\displaystyle 3}}{{\displaystyle 2}}-2i\right|-\left(\frac{{\displaystyle 3}}{{\displaystyle 2}}-2i\right)$ 

$=\sqrt{\frac{9}{4}+4}-\frac{3}{2}+2i=\frac{5}{2}-\frac{3}{2}+2i=1+2i$