For every real value of a > 0, determine the complex numbers which will satisfy the equation $|z^2|-2iz+2a(1+i)=0$.

Put $z=x+iy$.
Then the given equation reduces to 
$x^2+y^2-2i\left(x+iy\right)+2a\left(1+i\right)=0$
Or $\left(x^2+y^2+2y+2a\right)+2i\left(a-x\right)=0$
Equating the real and imaginary part to zero, we get
$x^2+y^2+2y+2a=0,2a-2x=0$
This  gves for $x=a$ and for y we have
$a^2+y^2+2y+2a=0$ or $y^2+2y+\left(a^2+2a\right)=0$   ...(1)
Since we seek real value of $y$, the discriminant of the equation (1)
must be non-negative, that is
$△=4-4\left(a^2+2a\right)\ge 0\Rightarrow 1-a^2-2a\ge 0$
For the real value of $a$, we get
$y=\frac{-2\pm \sqrt{4\left(1-a^2-2a\right)}}{2}=-1\pm \sqrt{1-a^2-2a}$
Hence for $△\ge 0$, the original equation has two roots
$$\begin{gathered} z_1=a+\left(-1+\sqrt{1-a^2+2a}\right)i \\ {z}_2=a+\left(-1-\sqrt{1-a^2+2a}\right)i \end{gathered}$$
For $△=0$, we have $z_1=z_2$ that is,
there is only one solution in this case.

For $△<0$ the equation has no roots
It remain to indicate the range of $a$ over which there are solution.
We are given $a\ge 0$ and in addition to this we found that $a$
must satisfy the inequality.
$$\begin{gathered} 1-a^2-2a\ge 0\Rightarrow a^2+2a-1\le 0 \\ \Rightarrow {\left(a+1\right)}^2-2\le 0{\left(a-1\right)}^2\le 2 \\ \Rightarrow -\sqrt{2}\le a+1\le \sqrt{2}\Rightarrow -1-\sqrt{2}\le a\le -1+\sqrt{2} \end{gathered}$$
Now choosing the number $a\ge 0$ from this interval we obtain 
$0\le a\le -1+\sqrt{2}$
Thus
$z=\{\begin{array}{cc}a(-1\pm \sqrt{1-a^2-2a})i,& 0<a<\sqrt{2}-1\\ (\sqrt{2}-1)-i,& a=\sqrt{2}-1\\ \text{no solution},& a>\sqrt{2}-1\end{array}$