Inspired by the fruitful answer to this question regarding numerically solving polynomial equations in terms of simpler fields (in that case representing real numbers as fractions of integers), I though I would try to take it a step further. Knowing the matrix representation of complex numbers:

$$z = a+bi \rightarrow \left[\begin{array}{rr}a&b\\-b&a\end{array}\right], a,b \in\mathbb{R}, z \in \mathbb{C}$$ If we further restrict $a,b \in \mathbb{Z}$ we will be able to do integer arithmetic with complex "integers".

Then next step is to use block multiplication property

Recalling the Companion matrix structure from answer to previous question

$$p(z) = z^n+a_{n-1}z^{n-1}+\cdots + a_1z + a_0 \hspace{0.5cm}\text{ then } \hspace{0.5cm} {\bf C}(p) = \left[\begin{array}{ll}0& -a_0\\{\bf I}&-{\bf a}_{1:n-1} \end{array}\right]$$

Now use $\otimes$ (Kronecker product) with ${\bf I}_2$ to represent these matrix elements and instead of considering $n \times 1$ column vectors $\bf v$, consider $2n \times 2$ matrices which can be written like $${\bf v} = {\bf a} \otimes \left[\begin{array}{rr}1&0\\0&1\end{array}\right]+ {\bf b}\otimes\left[\begin{array}{rr}0&1\\-1&0\end{array}\right]$$ where ${\bf a} \in \mathbb{Z}^n$ is real part, and ${\bf b} \in \mathbb{Z}^n$ imaginary part.

Can we prove that we will get properties for which ensure the procedure in this answer to work to find solutions expressed as the quotient between two complex integers: $({\bf C}(p)^N{\bf v})_1$, $({\bf C}(p)^N{\bf v})_n$?

Extended question: If this is possible to do for complex numbers, is there any theoretical limit as to what kinds of numbers will be approximal by this approach? Will it always work to do this with the Kronecker product as long as we have some matrix representation of the numbers?


I have tried this for some low degree polynomials with real coefficients and the "eigenvalue mover" mentioned in the other question. And it seems to work for second and third degrees. 15 iterations:

$$\left[\begin{array}{rr} 10^{-7}&1.9999999\\ -1.9999999&-10^{-8} \end{array}\right] \text{ appr root of } p(z) = z^2+4 = 0$$ which is very close to the correct $2i$ with a $+i {\bf I}$ perturbation. And
$$\left[\begin{array}{rr} -10^{-6}&-2.236067\\ 2.236067&10^{-6} \end{array}\right]\text{ appr root of } p(z) = z^2+5 = 0$$ which is very close to the correct $-\sqrt{5}i$ with a $-i {\bf I}$ perturbation. So this seems to work for complex polynomials. Would still welcome some theoretical explanation as to why it works.


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