Suppose $a\in \mathbf{R}$ and the equation $z+a\left|z\right|+$$2i=0$ has no solution in$C,$ then $\alpha$ satisfies the relation.

$z=-a\left|z\right|-2i$

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} |z{|}^2=a^2|z{|}^2+4\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} |z{|}^2\left(1-a^2\right)=4\end{array}$

This equation has no solution if  $1-a^2\le 0\text{ or }\left|a\right|\ge 1$

For $\left|a\right|<1,\left|z\right|=\frac{2}{\sqrt{1-a^2}}$ and 

         $z=\frac{2a}{\sqrt{1-a^2}}-2i$