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I have a system of two equations that need to be satisfied simultaneously, where $A,B,x,y\in\mathbb{C}$; they have the form: \begin{eqnarray} Ax=-yB\nonumber\\ B\bar{x}=-\bar{y}A \end{eqnarray} where $\bar{z}$ represents the complex conjugate of a complex number $z$. We aim to solve for $A,B$, more explicitly, the ratio between both, without making any assumptions on the form of $A,B,x,y$.

We can obtain by dividing them that: \begin{eqnarray} \left(\frac{A}{B}\right)^{2}=\frac{\bar{x}y}{x\bar{y}}=e^{i2\theta}, \end{eqnarray} where $\theta$ is a phase angle that will generally be given by: \begin{eqnarray} \theta = \arctan\left(\frac{\text{Im}(\bar{x}y)}{\text{Re}(\bar{x}y)}\right). \end{eqnarray} Is this solution a valid one? Otherwise, I assume the system has no "non-trivial" solution. The only important ratio to determine is $A/B$.

Zarathustra
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    The system always has a solution, since $(0,0)$ is always a solution... – 5xum Jan 13 '23 at 12:33
  • We aim to solve for $A/B$. If $A=\bar{B}$, then the second equation is just the first equation conjugated. But a priori, $A$ need not be $\bar{B}$, so I would say one just solves taking into account the two equations, as I wrote above. The important question is how would one, in general, go on solving such system of simultaneous equations, without assuming anything in the form of $A,B$, – Zarathustra Jan 13 '23 at 13:05
  • As pointed by 5xum, $(A,B)=(0,0)$ is always a solution. So you can "aim to solve for $A,B$" but not for "the ratio between both" (which is moreover not "more explicit"). – Anne Bauval Jan 13 '23 at 13:21

1 Answers1

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The system has non-zero solutions $(A,B)\in\Bbb C^2$ iff $|x|=|y|.$

Assuming $x\ne0$ and $y=xe^{i\theta},$ the solutions are simply $(-e^{i\theta}B,B)$ ($B\in\Bbb C$).

Your formulation is not correct because it would also allow $(+e^{i\theta}B,B).$

Anne Bauval
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  • The solution still makes an assumption on $x,y$, that $|x|=|y|$. But then, what is wrong with the approach above of dividing both equations to obtain $(A/B)^{2}$?. – Zarathustra Jan 13 '23 at 13:09
  • The answer assumes $|x|=|y|$ only after having solved the system when $|x|\ne|y|.$ – Anne Bauval Jan 13 '23 at 13:12
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    2 things are wrong with the "approach above". 1: dividing is not legit because $B=0$ for at least one of the solutions. 2: squaring introduces fake "solutions", as pointed at the end of my answer. – Anne Bauval Jan 13 '23 at 13:17
  • So $A/B=-e^{i\theta}$, with $\theta=\arctan(\text{Im}(-y/x)/\text{Re}(-y/x))$, correct? – Zarathustra Jan 13 '23 at 13:22
  • $\theta$ is not that arctan (which is defined only $\bmod\pi$ and would be the same with $-y$ replaced by $y$). $\theta$ is defined $\bmod{2\pi}$ (if $|x|=|y|\ne0$) by $y=xe^{i\theta}.$ – Anne Bauval Jan 13 '23 at 13:27
  • But $|x|\neq |y|$ in general. So, for $|x|\neq |y|$, the only solution is the trivial one, $(0,0)$? In other words, we only have a solution $(-e^{i\theta}B,B)$ (as in the response) if and only if $|x|=|y|$, which is in general not the case. – Zarathustra Jan 13 '23 at 13:31
  • Yes, this obvious remark was my 1st sentence. Beware also of the trivial case $x=y=0,$ so your "if and only if" is not completely correct. The solution(S) $(-e^{i\theta}B,B)$ are the only one(S) iff $|x|=|y|\ne0,$ as in the answer. – Anne Bauval Jan 13 '23 at 13:32