Solving for moduli in the product of matrices of complex polar numbers when the angles of those numbers must be specific values. I have a 1-row matrix of complex numbers in polar form which represent the difference of 3 other numbers I am solving for.
Essentially, I am starting with the RHS of this equation:
$$AB=C$$
where $A=\pmatrix{a&b&c}$ and $C = \pmatrix{(a-b)&(b-c)&(c-a)}$, hence:
$$\pmatrix{a&b&c} \pmatrix{1&0&-1\\-1&1&0\\0&-1&1} = \pmatrix{(a-b)&(b-c)&(c-a)}$$
Note that the elements of the matrix on the RHS are all complex numbers which sum to zero.
I need to work back and solve for $A$ knowing that $a$, $b$, and $c$ are complex numbers whose angles are known and so will be of the form:
$$\pmatrix{r_{a}e^{i0\pi}&r_{b}e^{i\frac{2}{3}\pi}&r_{c}e^{i\frac{4}{3}\pi}}$$
So far, I have used a method where I right-multiplied both sides by the pseudoinverse of $$\pmatrix{1&0&-1\\-1&1&0\\0&-1&1}$$
Under the assumption that
$$ABB^{-1}=A=CB^{-1}$$
where $B^{-1} = \frac{1}{3}\pmatrix{1&-1&0\\0&1&-1\\-1&0&1}$
However, when I use this with actual values, I don't get a solution which has the right angles for my complex numbers.
Is there a technique to solve for the moduli which will guarantee that that the angles are at 0, $\frac{2}{3}\pi$, and $\frac{4}{3}\pi$?
 A: Here's how I would think about it.  Given a value of $a$, that uniquely determines $b$ and $c$ by $b=a-(a-b)$ and $c=a+(c-a)$.  Given that $a$ is supposed to have a certain polar form, that restricts $a$ to lie on some line (or really just a ray if you require $r_a\geq 0$).  The possible values of $b$ then just lie on a translated version of that line (subtract $a-b$).  But the required polar form of $b$ gives a different line that $b$ must also lie on.  So there is at most one choice of $b$ that can work: just solve for the intersection point of the two lines it must lie on.  You can then test whether the value of $c$ you get from this also has the correct polar form.
Alternatively, if you want to use the machinery of linear algebra here, you should work over $\mathbb{R}$ instead of over $\mathbb{C}$, so instead of using $a,b,c$ as components of the vector you're looking for you would use the real numbers $r_a,r_b,r_c$.  You can then still express the real and imaginary parts of $a-b,b-c,$ and $c-a$ as certain $\mathbb{R}$-linear combinations of $r_a,r_b,$ and $r_c$.  (So $B$ would become a $3\times 6$ matrix, whose columns compute the real and imaginary parts of $a-b,b-c$ and $c-a$ in terms of $r_a,r_b,$ and $r_c$.  Actually, you could get away with just using a $3\times 4$ matrix, since you don't need to care about $c-a$: it will always be equal to $-(a-b)-(b-c)$.)
