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What would be the general form in the complex plane for the equation: $$ z\overline{z} + \beta z + \gamma \overline{z} + \delta=0 \quad ? $$ Straightforward is, that when $\gamma = \overline{\beta}$ and $\delta\in \mathbb{R}$ and $\delta<\overline{\beta}\beta$, it is a circle because it is then equivalent to $|z+\beta|^2=\overline{\beta}\beta - \delta$.

When $\gamma = \overline{\beta}$ and $\delta\notin \mathbb{R}$ or $\delta>\overline{\beta}\beta$ the solution is empty.

What if $\gamma \neq \overline{\beta}$? Does it become an ellipse e.g.?

By substituting $z\mapsto \frac{1}{z}$ one sees that the form of the equation is invariant under inversion.

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  • $\begingroup$ if you write out the coordinates using $z=x+iy,$ your variant still gives $x^2 + y^2 + Ax + B y + C = 0.$ It is still a circle or a point or empty. $\endgroup$
    – Will Jagy
    Commented Jan 13, 2016 at 1:49
  • $\begingroup$ It would be a contradiction with the form $|z+\beta|=R$. Your A and B are not real numbers anymore, unless $\gamma=\overline{\beta}$. $\endgroup$
    – Gerard
    Commented Jan 13, 2016 at 10:38

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We write $z = x + i y,$ $\beta = \beta_1 + i \beta_2,$ $\gamma = \gamma_1 + i \gamma_2,$ with all new quantities demanded real. Write out $z \bar{z} + \beta z + \gamma \bar{z} + \delta$ and identify the real and imaginary parts. These must both be zero. $$ x^2 + y^2 + (\beta_1 + \gamma_1)x + (\gamma_2 - \beta_2)y + \delta = 0, $$ $$ (\beta_2 + \gamma_2)x + (\beta_1 - \gamma_1)y = 0. $$ When $\gamma \neq \bar{\beta},$ at least one of the coefficients in the second equation is not zero, so we have defined a line in the $xy$ plane. The first equation still gives a circle or a point or the empty set.

All together, the possibilities are two points, one point, or the empty set.

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  • $\begingroup$ The first equation is a circle when the coefficients before $x$ and $y$ are equal and also the real part of $\delta$ must be smaller then the product of these coefficients. Still the empty set, or when it is a circle perhaps two points, is probably right. $\endgroup$
    – Gerard
    Commented Jan 14, 2016 at 8:31

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