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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?

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    Perhaps-sadly, no. Already the square roots of $-1$ form a spherical continuum (as you can check directly). It makes geometric sense, but it's not as discretized as one might have hoped on aesthetic grounds. – 2017-01-03

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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 $\{x + y{\bf w}: {\bf w} \in \mathbb R^3,\; \|{\bf w}\|=1\}$.

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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 $c=d=0$.