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How to solve the equation $$ a z - b \bar{z}^2 - c z^2 \bar{z} = 0 $$ for $z \in \mathbb{C}$ where $a, b, c \in \mathbb{C}$ are known coeffients. I want to solve for z. I started with substituting $z = r \, \mathrm{e}^{i \theta}$ to arrive at $$ a - b \, r \, \mathrm{e}^{-i 3 \theta} - c \, r^2 = 0 $$ but I'm not sure how to continue from here.

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Hint: using that $\,z\bar z = |z|^2\,$ the equation can be written as:

$$ \begin{align} a z - b \bar{z}^2 - c z |z|^2 = 0 \\ \iff\;\; \left(a - c|z|^2\right) z = b \bar z^2 \tag{1} \end{align} $$

Taking conjugates on both sides:

$$ \left(\bar a - \bar c|z|^2\right) \bar z = \bar b z^2 \tag{2} $$

Multiplying $(1) \cdot (2)\,$:

$$ \left(a - c|z|^2\right) \left(\bar a - \bar c|z|^2\right) |z|^2 = |b|^2 |z|^4 \\ \iff\;\;|z|^2 \cdot \left(|c|^2|z|^4-\left(|b|^2+a \bar c + \bar a c\right)|z|^2+|a|^2\right) = 0 $$

The latter gives the obvious root $\,z=0\,$, and the second factor is a biquadratic in $\,|z|\,$ with real coefficients. Any real positive roots can then be substituted back in $(1)$ and solved for $z$.

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