Given : $|z|-z=1+2 i$
If $z=x+i y$, then th is equation reduces to
$$
\begin{array}{l}
|x+i y|-(x+i y)=1+2 i \\
\Rightarrow\left(\sqrt{x^{2}+y^{2}}-x\right)+(-i y)=1+2 i
\end{array}
$$
On comparing real and imaginary parts of both sides of this equation, we get
$$
\begin{array}{l}
\sqrt{x^{2}+y^{2}}-x=1 \\
\Rightarrow \sqrt{x^{2}+y^{2}}=1+x \Rightarrow x^{2}+y^{2}=(1+x)^{2} \\
\Rightarrow x^{2}+y^{2}=1+x^{2}+2 x \\
\Rightarrow y^{2}=1+2 x...(i) \\
\text { and }-y=2 \\
\Rightarrow y=-2
\end{array}
$$
Putting this value in eq. (i), we get
$$
\begin{array}{l}
(-2)^{2}=1+2 x \\
\Rightarrow 2 x=3 \Rightarrow x=\frac{3}{2} \\
\therefore z=x+i y=\frac{3}{2}-2 i
\end{array}
$$