TITLE: Solving a complex quadratic-like equation
QUESTION [6 upvotes]: Let $a,b,c,d$ are given non-zero complex numbers (i.e., constants). Then is it true that the equation 
$$a|z|^2+b z+c\bar{z}+d=0$$
which is equivalent to $$|z|^2+b'z+c'\bar{z}+d'=0$$
will always have a (at least one) solution for $z$? Or there is some necessary and sufficient condition that $a,b,c,d$ must satisfy to guarantee a solution in $z$? Notation: $\bar{z}$ is the complex conjugate of $z$.

REPLY [2 votes]: (Posting this separately since it is a different approach with different results, and it neither overlaps nor obsoletes my first answer.)
Assuming again WLOG that $\,a=1\,$, the following proves that:

the condition for $|z|^2+bz+c\bar{z}+d=0$ to have solutions depends solely on the compound expressions $\,\lambda=b-\bar c\,$ and $\,\mu=-bc+d\,$;
the necessary and sufficient condition for solutions to exist is $\,|\lambda|^2 \ge  2 \big(\,|\mu| + \operatorname{Re}(\mu) \,\big)\,$.

First off, the equation can be rewritten as:
$$
|z|^2+bz+c\bar{z}+d=0 \quad\iff\quad(z+c)(\bar z + b) - bc + d = 0 \tag{1}
$$
Then, the substitution $\,z+c=w\,$ and $\,\lambda=b-\bar c\,$, $\,\mu=-bc+d\,$ gives:
$$
|w|^2+\lambda w + \mu = 0 \quad\iff\quad \lambda w = -\left(|w|^2+ \mu\right) \tag{2}
$$
Taking the squared modulus on both sides:
$$
|\lambda|^2|w|^2=\left(|w|^2+ \mu\right)\left(|w|^2+ \bar \mu\right) \quad\iff\quad |w|^4 - \left(|\lambda|^2 - 2 \operatorname{Re}(\mu)\right)|w|^2 + |\mu|^2 = 0 \tag{3}
$$
The latter is a quadratic in $\,|w|^2\,$ with real coefficients and the product of the roots is $\,|\mu|^2 \ge 0 \,$, so if the roots are real then they have the same sign. Therefore, the conditions for real positive roots are:

non-negative discriminant: $\;\Delta \ge 0 \iff \left||\lambda|^2 - 2 \operatorname{Re}(\mu)\right| \ge 2 |\mu|\,$;
non-negative sum of roots: $\;|\lambda|^2 - 2 \operatorname{Re}(\mu) \ge 0\,$.

Combining the two inequalities gives the necessary condition $\,|\lambda|^2 \ge  2 \big(\,|\mu| + \operatorname{Re}(\mu) \,\big)\,$.
To prove that the condition is also sufficient, take $\,|w|^2=w_2\,$ to be a positive real solution of $\,(3)\,$:

If $\lambda \ne 0$, then substituting $\,|w|^2=w_2\,$ back in $\,(2)\,$ gives $\,w = - \dfrac{w_2+\mu}{\lambda}\,$:

it follows from $\,(3)\,$ that $\,|w|^2=w_2\,$, and
it follows from $\require{cancel}\,(2)\,$ that $\,|w|^2+\lambda w+\mu=\cancel{w_2} - \lambda\cdot \dfrac{\cancel{w_2}+\bcancel{\mu}}{\lambda}+\bcancel{\mu} = 0\,$.

Therefore $\,w = - \dfrac{w_2+\mu}{\lambda}\,$ satisfies the original equation.
If $\,\lambda=0\,$, then $\,|\lambda|^2 = 0 \ge  2 \big(\,|\mu| + \operatorname{Re}(\mu) \,\big) \iff \mu \in \mathbb{R}, \mu \le 0\,$. In this case the equation reduces to $\,|w|^2+\mu=0\,$ which has solutions with $\,|w| = \sqrt{-\mu}\,$, infinitely many if $\,\mu \lt 0\,$.

Thus $\,|\lambda|^2 \ge  2 \big(\,|\mu| + \operatorname{Re}(\mu) \,\big)\,$ is the necessary and sufficient condition for solutions to exist.