Show the general form of a line in $\Bbb{C}$ is of the form
$$a\overline{z}+ \overline{a}z+C = 0$$
where $0 \neq a \in \Bbb{C}$ and $C \in \Bbb{R}$.
Do we start with setting $z=x+iy$, $a=u+iv$? A bit confused on this one.
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1https://math.stackexchange.com/questions/2344227/equation-of-a-line-in-complex-plane – Anne Bauval Sep 03 '22 at 22:25
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What's the definition of a "line"? – fleablood Sep 04 '22 at 01:27
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@fleablood y = mx+b? – homosapien Sep 04 '22 at 01:39
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Can I assume there's a line $ax+by+c=0$ in the reals and build off that? @fleablood – homosapien Sep 04 '22 at 01:40
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You could definitely start that way.
Using your notation described above, note that $z + \bar z = 2x$ and $\bar z-z = -2iy$. Then
\begin{align*} &a\bar z + \bar az + C = 0\\ \iff & u\bar z+ iv\bar z + uz -ivz+C=0\\ \iff & u(z+\bar z) +iv(\bar z-z) + C=0\\ \iff & 2ux -iv(2iy) + C=0\\ \iff & ux + vy = D, \end{align*} where the last equation is a line in $\mathbb R^2$ written in standard form.
Philip Speegle
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I got that too, but I ended up using the fact that there's a line in reals of form $ax+by+c=0$ where $x+iy=z$ then using the real and imaginary parts I substituted. – homosapien Sep 04 '22 at 01:41