A complex number z satisfies the equation 

$|z{|}^2-2iz+2c(1+i)=0,$ where c is real. The values of c for which the above equation has no solution can be given by

Let $z=x+iy,$ then the equation is$c\in [-1-\sqrt{2},-1+\sqrt{2}]$ 

    $\begin{array}{l}{x}^2+y^2-2i(x+iy)+2c(1+i)=0\\ \Rightarrow \left(x^2+y^2+2y+2c\right)+i(2c-2x)=0\end{array}$

    $\Rightarrow x^2+y^2+2y+2c=0$ and $x=c$

   $\begin{array}{l}\Rightarrow c^2+y^2+2y+2c=0\\ \Rightarrow y=-1\pm \sqrt{1-2c-c^2}\\ \because y\in \mathbf{R}\Rightarrow 1-2c-c^2\ge 0\\ \Rightarrow c^2+2c-1\le 0\Rightarrow -1-\sqrt{2}\le c\le -1+\sqrt{2}\end{array}$

$\therefore$  The equation has a solution, if 

$c\in [-1-\sqrt{2},-1+\sqrt{2}]$ and the solution is given by

$z=c+i\left(-1\pm \sqrt{1-2c-c^2}\right)$

The equation has no solution, if 

$c\in (-\mathrm{\infty },-1-\sqrt{2})\cup (-1+\sqrt{2},\mathrm{\infty })$