I assume the equation is
[tex]2z=|z|+2i[/tex]
then let [tex]z=x+iy[/tex].
[tex]2x+2iy=\sqrt{x^2+y^2}+2i[/tex]
[tex]2x+2i(y-1)=\sqrt{x^2+y^2}[/tex]
The right side is real-valued and positive, and only on the left side do we have an imaginary term, which tells us
[tex]2(y-1)=0\implies y=1[/tex]
Then
[tex]2x=\sqrt{x^2+1}[/tex]
[tex]4x^2=x^2+1[/tex]
[tex]3x^2=1[/tex]
[tex]x=\pm\dfrac1{\sqrt3}[/tex]
but again we omit the negative solution, so that
[tex]\boxed{z=\dfrac1{\sqrt3}+i}[/tex]
