Solving the equation $|z|^2-2iz+2c(1+i)=0$ for $c\in \mathbb{R}^+\cup \{0\}$

From a given equation we get $$ 2i(z-c) = |z|^2+2c$$ so $z-c =ir$ where $r\in \mathbb{R}$, so $z=c+ri$. If we put this back to starting equation we get $$c^2+r^2+2r+2c=0$$ or $$(c+1)^2+(r+1)^2=2$$

So $z$ describes a minor arc between $0$ and $-2i$ on circle with center at $-1-i$ and radius $\sqrt{2}$.

Alt. hint:   write the equation and its conjugate as:

$$ \begin{align} 2iz = |z|^2 + 2c + 2ic \tag{1}\\ -2i \bar z = |z|^2 + 2c -2ic \end{align} $$

Multiplying the above:

$$ 4|z|^2 = \left(|z|^2+2c\right)^2+4c^2 \;\;\iff\;\;|z|^4+4(c-1)|z|^2+8c^2=0 \tag{2} $$

For the quadratic in $\,|z|^2\,$ to have (at least) a real positive root, the conditions are:

$$ \begin{cases} \begin{align} \frac{1}{4}\Delta = 4(c-1)^2-8c^2=4(-c^2-2c+1) \ge 0 \;\;&\iff\;\; -1-\sqrt{2} \le c \le -1 +\sqrt{2} \\ 4(c-1) \le 0 \;\;&\iff\;\; c \le 1 \end{align} \end{cases} $$

When those conditions are satisfied, solving $(2)$ gives $\,|z|^2 = 2(1-c) \pm \dfrac{1}{2}\sqrt{\Delta}\,$, and substituting this $\,|z|^2\,$ back in $(1)$ gives the solutions for $\,z\,$.

$$|z|^2-2iz+2c(1+i)=0 \tag 1$$

In the comments section, Ishan's asked to elaborate this hint :

" It seems that you lost one equation when considering only the sum of the two conjugate equations. You should not consider only the sum of both but also the difference. So the solutions are not all the points of the circle, but only the points which also satisfy the second equation."

This requires a too long explanation to be put in the comment section. So, I post it as an answer.

The two equations to which I refer are your equations :

$$|z_0|^2-2iz_0+2c(1+i)=0 \tag 2$$ $$|z_0|^2-\overline{2iz_0}+2c(\overline{1+i})=0 \tag 3$$

You add Eq.$(2)$ and Eq.$(3)$ , which gives : $$2|z_0|^2+2i(\bar z_0-z_0) +4c=0 \tag 4$$

I will not repeat your calculus which is correct. You proved that all the $z_0$ solutions of Eq.$(4)$ are located on a circle.

But all solutions of Eq.$(4)$ are not solutions of Eq.$(1)$.

In fact, the system of two equations $(2)$ and $(3)$ has been reduced to only one equation $(4)$. This is not equivalent. In order to have an equivalent system of two equations, one have to consider another linear combination of $(2)$ and $(3)$. For example in subtracting instead of adding. $$|z_0|^2-2iz_0+2c(1+i)-\left(|z_0|^2-\overline{2iz_0}+2c(\overline{1+i})\right)=0$$ After simplification : $$z_0+\overline{z_0}-2c=0 \tag 5$$ This shows that all the $z_0$ solutions of Eq.$(5)$ are located on the straight line $|z_0|=c$.

But all solutions of Eq.$(5)$ are not solutions of Eq.$(1)$.

On the other hand, the solutions of Eq.$(1)$ are solutions of both Eq.$(4)$ and $(5)$. This means that the solutions of Eq.$(1)$ are at the intersection of the circle and the straight line.

I suppose that you can continue. If I made no mistake, the solutions of Eq.$(1)$ are : $$z_0=c+i\left(-1\pm\left(2-(c+1)^2\right)^{1/2}\right)$$

Well, this is a boring method to solve Eq.$(1)$. It is much simpler to put $z_0=x+iy$ into $(1)$ and solve for $x,y$ , as it was already pointed out in some comments.