Question 1021503
Let z = x+iy
==> {{{(abs(z))^2 - 2con(z)+iz = x^2+y^2-2(x-iy)+i(x+iy) = x^2+y^2-2x+2iy+ix-y}}}

= {{{(x^2+y^2-2x-y)+i(x+2y)}}}
Since this is supposed to be equal to 2i, it follows that

x+2y = 2 and {{{x^2+y^2-2x-y = 0}}}

Putting x = 2-2y into {{{x^2+y^2-2x-y = 0}}}, we get

{{{4(1-y)^2+y^2-4(1-y) - y = 0}}}

Simplifying this and solving for y (you should be able to do the algebra!), we get

y = 0 or y=1.
The corresponding x-values are x = 2 or x = 0 respectively..

Therefore there are two complex numbers satisfying the original equation namely

{{{z[1] = 2}}}, and {{{z[2] = i}}}.