Roots of quaternions in space

Question

Roots of complex numbers are known to take the shapes of regular polygons when represented on the Argand Diagram. This happens due to de Moivre's Theorem. I would like to know whether the same can be observed in the case of quaternions. Would roots of quaternions, when represented in $\mathbb{R^3}$, take the shapes of regular polyhedra?

Answer 1

Somewhat more generally: for any unit vector ${\bf v} \in \mathbb R^3$, identified as the quaternion $v_1 {\bf i} + v_2 {\bf j} + v_3 {\bf k}$, let $V$ be the 2-dimensional subspace of quaternions of the form $x + y \bf v$, for $x, y \in \mathbb R$. This subspace is closed under multiplication, since ${\bf v}^2 = - v_1^2 - v_2^2 - v_3^2 = -1$. In particular every polynomial $P$ with real coefficients maps $V$ into itself, and $P(x + y {\bf v}) = \alpha + \beta {\bf v}$ if and only if $P(x + y i) = \alpha + \beta i$ in the complex numbers.

The result is that the roots of the polynomial equation $P(q) = \alpha + \beta {\bf v}$ (where $q$ is a quaternion variable) are not terribly interesting: if the complex roots of $P(z) = \alpha + \beta i$ are $r_j = x_j + y_j i$, then if $\beta \ne 0$ they are the corresponding points $x_j + y_j {\bf v}$ in the plane $V$, while if $\beta = 0$ they form spheres of the form $\{x + y{\bf w}: {\bf w} \in \mathbb R^3,\; \|{\bf w}\|=1\}$.

Answer 2

Let's solve $(a+bi+cj+dk)^2=-1$. Expanding partially, $$ a^2 +2a(bi+cj+dk) + (bi+cj+dk)^2 = -1. $$ Expanding the last bracket gives $$ (bi+cj+dk)^2 = -b^2-c^2-d^2 + bc(ij+ji)+cd(jk+kj)+db(ki+ik) = -b^2-c^2-d^2, $$ since the units $i,j,k$ anticommute, so we have $$ a^2 - b^2 -c^2-d^2 +2a(bi+cj+dk) = -1 $$ Equating coefficients, we find $a=0$ (or $b=c=d=0$, which doesn't work) and then $$ b^2+c^2+d^2 = 1, $$ which agrees with what we know about $i^2=j^2=k^2=-1$. So no, we never get polyhedra, but instead, similar methods will give us the intersection of various four-dimensional surfaces. It so happens that the complex case $\pm i$ comes from the two poles of this sphere when we intersect it with $c=d=0$.