Show that if $\lvert a \rvert \neq 1$, then the equation $\overline{z}^2 = az^2+bz+c$ has only a discrete number of solutions.

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I knew the proof for this at some point, but I'm having trouble piecing it back together. At least, I think the proof I'm thinking of was for this result, or a result which implied this result. The proof involved some integrals of $e^x$, but the details escape me... I have been searching my notes and will continue to do so, but I would appreciate any help in the meantime.

Although I had attempted to recreate the proof mentioned above, as it escaped me, I tried one alternate approach. Expand the expressions by the substitution $z = x+iy$ and separate the real and complex parts (one should also accordingly expand the coefficients as, for instance, $a = a_1 + a_2 i$). You then have two curves, the real and imaginary parts of the curve, each of which are of degree two. The resultants of the curve, then, are of degree 4 with respect to each variable, but from here it would be a matter of showing that the curves are not the same if $\lvert a \rvert \neq 1$. I've used a similar approach in the past for these kinds of proofs, but I'm stuck at this step.