If $c\ge 0$ , then the equation ${\left|z\right|}^2-2iz+2c\left(1+i\right)=0$  (z is complex) has

Let$z=x+iy$  . Then $\left(x^2+y^2\right)-2i\left(x+iy\right)+2c\left(1+i\right)=0$
Therefore

$x^2+y^2+2y+i\left(2c-2x\right)+2c=0$  $\underset{ }{\overset{ }{\to }}\left(1\right)$

$x^2+y^2+2y+2c=0$  $\underset{ }{\overset{ }{\to }}\left(2\right)$
  and  $2c-2x=0orx=c$
Substituting   in Eq. (2), we get that
   $c^2+y^2+2y+2c=0$    $\underset{ }{\overset{ }{\to }}\left(3\right)$
Equation (3) has solutions if $4-4\left(c^2+2c\right)\ge 0$ , that is $1-c^2-2c\ge 0$  . Therefore

${\left(c+1\right)}^2\le 2or-\sqrt{2}\le c+1\le \sqrt{2}$${\left(c+1\right)}^2\le 2or-\sqrt{2}\le c+1\le \sqrt{2}$
It is given that. Therefore  .$0\le c\le \sqrt{2}-1$
(i) If $c<\sqrt{2}-1$  ¸ then  $z=c+\left(-1\pm \sqrt{1-2c-c^2}\right)i$.
(ii) If  $c=\sqrt{2}-1$, then $z=\left(\sqrt{2}-1\right)-i$ .
(iii) If $c>\sqrt{2}-1$ , the equation has no solutions.