How to find complex numbers in form z=x+yi which satisfies :
Let $\displaystyle z=x+iy$ be a solution of this equation, then:
$\displaystyle (x+iy)(x-iy) -2i = 2i(x-iy) \Rightarrow (x-iy)(x+iy-2i)=2i$
Now,
$\displaystyle (x-iy)(x+i(y-2)) = x^2-2xi+y^2-2$
Substitute back and get: $\displaystyle x^2+y^2-2-2xi = 2i \rightarrow -2x = 2 \Rightarrow x=-1$
$\displaystyle x^2+y^2=2\Rightarrow y^2=1 \Rightarrow y=\pm 1$
So our solutions are:
$\displaystyle z_1=-1+i$
$\displaystyle z_2=\bar{z_1} = -1-i$