Solving a complex quadratic-like equation 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 $z$.
 A: Let's generalize slightly. 
Consider 
$$a z \bar{z} + b z\bar{w} + cw\bar{z} + dw\bar{w},$$
which is the "quadratic form"ish thing associated to the sesquilinear form
on $\newcommand{\CC}{\Bbb{C}}\CC^2\times\CC^2$ given by
$$B((z_1,w_1),(z_2,w_2))=az_1\bar{z_2} + bz_1\bar{w_2} + cw_1\bar{z_2} + dw_1\bar{w_2}.$$ Your equation is $B((z,1),(z,1))=0$.
Now if $B$ is positive definite, since $(z,1)\ne 0$ we'd have $\operatorname{Re} B((z,1),(z,1)) > 0$, so we certainly couldn't have $B((z,1),(z,1))=0$. Thus it suffices to find a positive definite
sesquilinear form such that its matrix w.r.t. the standard basis has no zero entries. Here's one given by its matrix:
$$
\begin{pmatrix}
1&\frac{1}{2}\\\frac{1}{2}& 1
\end{pmatrix},
$$
which is positive definite because $1 > 0$ and $1-\frac{1}{4} > 0$.
Edit: In response to the comment below.
EDIT 3: The below argument is wrong. It's too late at night for me to correct it right now. The key problem is that I was careless about positive and negative definite, which for general complex matrices are matrices $A$ such that $\Re x^*Ax  > 0$ for all nonzero $x$, since $x^*Ax$ is not necessarily real. I'll have to see if I can fix the argument later. I just need to record this note for future readers.
Note that the same argument shows that if $B$ is negative definite, your equation has no solutions.
Then if $B$ is not positive definite or negative definite, there is always a nonzero vector $v$ such that $B(v,v)=0$. 
Proof: suppose not, then because $B$ is not positive definite, there is a nonzero $v$ such that $B(v,v)<0$, and since $B$ is not negative definite, there is a nonzero $u$ such that $B(u,u)>0$. Then $\CC^2\setminus\{0\}$ is path connected, so choose a path from $u$ to $v$ not going through 0. By the intermediate value theorem, at some point $w$ on this path $B(w,w)=0$ contradiction.
Let the vector $v$ with $B(v,v)=0$ be $(z,w)$. If $w\ne 0$, then $B((z/w,1),(z/w,1))=0$ solving your equation.
Thus the only issue is that this vector $v$ with $B(v,v)=0$ might be of the form $(z,0)$. 
I will continue to think about how to deal with this issue.
Edit 2:
Note that the set of nonpositive-definite, nonnegative-definite sesquilinear forms that are bad (in the sense of having no solution for your equation) are those such that $B((z,w),(z,w))=0$ if and only if $w=0$. Hopefully we can characterize these. Actually note that if $B((z,0),(z,0))=0$, for $z\ne 0$, then $a\bar{z}z = 0$, so $a=0$ contrary to your assumption.
Hence your equation has a solution if and only if the corresponding matrix  $$\begin{pmatrix} a & b \\ c& d \end{pmatrix}$$ is not positive definite and not negative definite. I.e. if and only if its Hermitian part:
$$\begin{pmatrix} \operatorname{Re} a & \frac{b+\bar{c}}{2} \\ 
\frac{\bar{b}+c}{2} & \operatorname{Re} d \\
\end{pmatrix}
$$ 
is not positive definite and not negative definite. Assuming $a=1$, and using $b'$, $c'$, $d'$, the equation has a solution if and only if 
$$\begin{pmatrix} 1 & \frac{b'+\bar{c'}}{2} \\ 
\frac{\bar{b'}+c'}{2} & \operatorname{Re} d' \\
\end{pmatrix}
$$ 
is not positive definite and is not negative definite.
Now in fact, from the criterion for positive definiteness/negative definiteness given here, since the upper left corner of this matrix is 1, the matrix is not negative definite, and it is positive definite if and only if its determinant is positive, so your equation has a solution if and only if 
$$
\operatorname{Re} d' - \frac{(b'+\bar{c'})(\bar{b'}+c')}{4} 
=
\operatorname{Re} d' - \frac{\|b'+\bar{c'}\|^2}{4} 
\le 0.$$
A: The equation wil not always have solutions, as pointed out already.
The following gives a necessary condition for solutions to exist. Consider WLOG the case $\,a=1\,$, then taking the complex conjugates on both sides gives $\,|z|^2+\bar b \bar z+ \bar c z+\bar d=0\,$. Eliminating $\bar z\,$ between the latter and the original equation results in:
$$
(\bar b - c) |z|^2+(|b|^2-|c|^2) z + \bar b d - c \bar d = 0 \quad \iff \quad (|b|^2-|c|^2) z = (c - \bar b) |z|^2 + c \bar d - \bar b d
$$
Taking the squared modulus on both sides:
$$
(|b|^2-|c|^2)^2 |z|^2 = \big((c - \bar b) |z|^2 + c \bar d - \bar b d\big)\big((\bar c - b) |z|^2 + \bar c d - b \bar d\big)
$$
The latter is a quadratic in $\,|z|^2\,$, and a necessary condition for solutions to exist is that the quadratic must have at least one real positive root. This can be easily verified for particular values of $\,b, c, d\,$, but writing the condition in the general case in terms of arbitrary $\,b, c, d\,$ is not pretty.
Also, the condition is not necessarily sufficient: if a real positive root does exist, then its square root gives $\,|z|\,$, which can then be substituted back into the original equation to get $\,z\,$, but it still remains to be verified that the resulting $\,z\,$ does in fact satisfy the equation.
A: Take $a=b=c=1,$ then $d = 47.$ With real $x,y$ and $z = x + iy,$ this is
$$  x^2 + y^2 + 2x + 47 = 0, $$
$$ (x+1)^2 + y^2 + 46 = 0 $$
A: If we let $b'=c'=d'=1$, then $z=\alpha+\beta i$ gives$$|z|^2+b'z+c'\bar{z}+d'=(\alpha^2+\beta^2)+(\alpha+\beta i)+(\alpha-\beta i)+1=\color{red}{(\alpha+1)^2+\beta^2=0}$$ which is impossible even if $\alpha,\beta\ge0$.
A: (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.
