This question may have been asked before..
what is the complex form through five points of a conic section in the complex plane?
Thanks for any hints.
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2What did you try? – Brien Navarro Jan 24 '22 at 01:49
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The real point matrix is known but not mentioned as it may not be of direct use here. With any hint I would try improving my down voted question. – Narasimham Jan 30 '22 at 08:15
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In complex plane, we use $z$ and $\bar z$ as bases. Hence,
$$0=F(z,\bar z) \equiv \det \begin{pmatrix} (z+\bar z)^2 & z^2-\bar z^2 & (z-\bar z)^2 & z & \bar z & 1 \\ (z_1+\bar z_1)^2 & z_1^2-\bar z_1^2 & (z_1-\bar z_1)^2 & z_1 & \bar z_1 & 1 \\ (z_2+\bar z_2)^2 & z_2^2-\bar z_2^2 & (z_2-\bar z_2)^2 & z_2 & \bar z_2 & 1 \\ (z_3+\bar z_3)^2 & z_3^2-\bar z_3^2 & (z_3-\bar z_3)^2 & z_3 & \bar z_3 & 1 \\ (z_4+\bar z_4)^2 & z_4^2-\bar z_4^2 & (z_4-\bar z_4)^2 & z_4 & \bar z_4 & 1 \\ (z_5+\bar z_5)^2 & z_5^2-\bar z_5^2 & (z_5-\bar z_5)^2 & z_5 & \bar z_5 & 1 \\ \end{pmatrix}$$
and may further rearrange into
$$\bar u z^2+pz\bar z+u\bar z^2+\bar v z+v\bar z+q=0$$
with $p,q\in \mathbb{R}$.
Ng Chung Tak
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This is mostly wrong, since the source-pdf in the link needs the coefficients of $z^2$ and $\bar{z}^2$ to be conjugate as well as those of $z$ and $\bar{z},$ and the coefficients of $z\bar{z}$ and $1$ to be real. In fact, it's a real equation with imaginary part $0$, dressed up as a complex equation. Your "equation" in most cases becomes two; one from the real part and one from the imaginary part. You'd be better off doing the determinant calculation as usual in ${\Bbb R}^2$ and retrofitting to $\bar{u}z^2+pz\bar{z}+u\bar{z}^2+\bar{v}z+v\bar{z}+q=0, p,q\in {\Bbb R}.$ – Jan-Magnus Økland Jan 30 '22 at 07:53
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@Jan-MagnusØkland, You're right and I've jumped to conclusion carelessly. I've revised my answer finally. – Ng Chung Tak Jan 30 '22 at 08:15