1
$\begingroup$

I have an equation as follows:

$$\text{Imaginary}\left(\left( C_1 - \frac{C_2}{X} \right) \left(\overline{C_3 - \frac{0.75}{X}}\right)\right)=0$$

where the bar represents complex conjugation and $C_1$, $C_2$, $C_3$ are constants that are complex numbers, and $X$ is the complex number I am solving for. I have kept the constants general, but I can provide the numbers if required.

I am told that if you plot $\frac{C_2}{X}$ on the complex plane it produces a circle, but I am having trouble how to define it.

I can easily find one point on the circle by setting the left-most bracket expression to zero and solving for X. I can find another point on the circle by setting the right-most bracket expression to zero and solving for X. After that point, I am really stumped. I cannot define a unique circle with two points.

Any help would be greatly appreciated. I've been stuck on this for weeks.

$\endgroup$
2
  • $\begingroup$ "I have an equation" $\;-\;$ The equation has two solutions, so it represents just two points in the complex plane, not a circle and not any kind of curve. "I am told that if you plot $C_2/X$ on the complex plane it produces a circle" $\;-\;$ You must have misunderstood something. It is not possible to plot a complex function on the complex plane (just as you cannot plot a real function on the real axis). Besides, this has no relation to the equation you started with. $\endgroup$
    – dxiv
    Nov 7, 2021 at 3:32
  • $\begingroup$ @dxiv My apologies, I realised I bungled my original equation and forgot the Imaginary operator, which I have now added. I believe there are an infinite number of solutions to the equation. The expression in each nested bracket can be thought of as a vector. A solution is achieved if they have the same angle or are 180 degrees displaced. Sorry, pure math is obviously not my main field so I may be sometimes 'loose' in my terminology. $\endgroup$
    – razorfin
    Nov 7, 2021 at 4:27

1 Answer 1

2
$\begingroup$

(Not a complete answer, but too long for a comment.)

  • With $z=1/X$ and after folding the constants, the equation can be written as:

$$ \begin{align} && \text{Im} (z-a)(\overline{z}-b)=0 \\ &\iff &(z-a)(\overline{z}-b) \in \mathbb R \\ &\iff &(z-a)(\overline{z}-b)=\overline{(z-a)(\overline{z}-b)} \\ &\iff &(\overline{a}-b)z-(a-\overline{b})\overline{z}+ab-\overline{a}\overline{b}=0 \end{align} $$

  • See for example here why the latter is the equation of a line in the complex plane.

  • $X = 1/z$ is a particular case of the general Möbius transformation, which maps lines to (generalized) circles, see for example 1, 2, 3.


[ EDIT ] As suggested in a comment, putting it all together:

  • The $z$ line passes through points $a, \bar b$ and is unbounded, so its image via inversion is the $X$ circle passing through $1/a, 1/\bar b$ and the origin.
$\endgroup$
2
  • 1
    $\begingroup$ [+1] I think it is important to say that it is "the" equation of the line passing through $a$ and $b$. The OP can take a look at the interesting explanations here $\endgroup$
    – Jean Marie
    Nov 8, 2021 at 11:12
  • 1
    $\begingroup$ @JeanMarie Thanks. Edited the answer to include that. $\endgroup$
    – dxiv
    Nov 8, 2021 at 17:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .