Answered: "What is the complex form through five points..."

Jakobe Norton

Answered question

2022-04-19

What is the complex form through five points of a conic section in the complex plane?

Answer & Explanation

cm1mmeboulbes21e1

Beginner2022-04-20Added 8 answers

In complex plane, we use z and $\overset{―}{z}$ as bases. Hence,
$0 = F (z, \bar{z}) \equiv \det (\begin{matrix} (z + \bar{z})^{2} & z^{2} - {\bar{z}}^{2} & (z - \bar{z})^{2} & z & \bar{z} & 1 \\ (z_{1} + {\bar{z}}_{1})^{2} & z_{1}^{2} - {\bar{z}}_{1}^{2} & (z_{1} - {\bar{z}}_{1})^{2} & z_{1} & {\bar{z}}_{1} & 1 \\ (z_{2} + {\bar{z}}_{2})^{2} & z_{2}^{2} - {\bar{z}}_{2}^{2} & (z_{2} - {\bar{z}}_{2})^{2} & z_{2} & {\bar{z}}_{2} & 1 \\ (z_{3} + {\bar{z}}_{3})^{2} & z_{3}^{2} - {\bar{z}}_{3}^{2} & (z_{3} - {\bar{z}}_{3})^{2} & z_{3} & {\bar{z}}_{3} & 1 \\ (z_{4} + {\bar{z}}_{4})^{2} & z_{4}^{2} - {\bar{z}}_{4}^{2} & (z_{4} - {\bar{z}}_{4})^{2} & z_{4} & {\bar{z}}_{4} & 1 \\ (z_{5} + {\bar{z}}_{5})^{2} & z_{5}^{2} - {\bar{z}}_{5}^{2} & (z_{5} - {\bar{z}}_{5})^{2} & z_{5} & {\bar{z}}_{5} & 1 \end{matrix})$
and may further rearrange into $\overset{―}{u} z^{2} + p z \overset{―}{z} + u {\overset{―}{z}}^{2} + \overset{―}{v} z + v \overset{―}{z} + q = 0$
with p, $q \in R$ .

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