Complex mania!

$\begin{aligned} |{\color{#3D99F6}z}|^2 - 2i{\color{#3D99F6}z} + 2a(1+i) & = 0 & \small \color{#3D99F6} \text{Let }z = x + iy \\ x^2 + y^2 - 2i(x+iy) +2a(1+i) & = 0 \\ x^2 + y^2 - 2xi+2y +2a+2ai & = 0 \\ x^2 + y^2 +2y +2a+{\color{#3D99F6}2i(a-x)} & = 0 \end{aligned}$

Equating the imaginary parts on both sides, $\implies a- x = 0 \implies a =x$ .

Equating the real parts on both sides,

$\begin{aligned} x^2 + y^2 +2y +2{\color{#3D99F6}a} & = 0 & \small \color{#3D99F6} \text{Note that }a = x \\ x^2 + y^2 +2y +2{\color{#3D99F6}x} & = 0 \\ x^2 + 2x + 1+ y^2 +2y + 1 & = 2 \\ (x+1)^2 + {\color{#3D99F6}(y+1)^2} & = 2 & \small \color{#3D99F6} \text{Note that }(y+1)^2 \ge 0 \\ {\color{#3D99F6}(x_{max}+1)^2} + \cancel{(y+1)^2}^0 & = 2 & \small \color{#3D99F6} (x+1)^2 \text{ is maximum, when } (y + 1)^2 = 0 \end{aligned}$

$\implies x_{max} = \sqrt 2 - 1 \approx \boxed{0.414}$

Here is a geometric interpretation!

Consider the circle $\large |z+i|=\sqrt{1-2a}$ note that a should not exceed 0.5 for any such complex number to exist

$\large |z|^{2}-iz+i\bar{z}=1-2a$ is subject to the constraint $\large \mathbb{Re}(z)=a$

For a solution to exist,an intersection between the circle and constraint line is necessary and thus the perpendicular distance of the line from the circle's centre should not exceed its radius.

Thus,for positive $a$ , $\large a^{2}+2a-1$ is non-positive.

$\large (a+1)^{2}-2 \le 0$

Largest value of permissible $a$ is $\large \sqrt{2}-1$

Let us assume z = x+iy where i is our classical imaginary unit

Some sort of basic algebra and comparing real and imaginary parts both sides

We get

a = x

x^2+y^2+2y+2a = 0

Substitute x = a

We get

y^2+2y+a^2+2a = 0

For a complex number x+iy to exist the discriminant of above quadratic in y should be non negative

Doing this you get maximal value of a as root2-1