square roots

The unit quaternions are square roots of either unity or of minus one. Are there any others? According to Eq. 1, to get a root of unity needs

$$\begin{eqnarray*} ( s + {\bf u}) ^2 & = & s^2 + 2 s {\bf u}+ ( {\bf u}. {\bf u}) \\ & = & {\bf 1}, \end{eqnarray*}$$

leading to two mutually exclusive alternatives

$$\begin{eqnarray*} s^2 = {\bf 1}& & {\bf u}= 0 \\ ( {\bf u}. {\bf u}) = {\bf 1}& & s = 0. \end{eqnarray*}$$

Besides the two expected scalar roots, any vector of unit norm fills the bill, infinitely many in all. A second glance at the derivation shows that any vector, the square of whose norm is $-1$ (such as ${\bf i}$) is a root of $-{\bf 1}$, but that there are no (real) scalar roots.

As for square roots in general, the square root of any scalar follows the same line of reasoning with the exception that everything is scaled by the positive square root of the scalar when it has one. The general requirement for

$$\begin{eqnarray*} s + {\bf u}& = & \surd(t {\bf 1}+ {\bf v}) \end{eqnarray*}$$

would be

$$\begin{eqnarray*} ( s + {\bf u}) ^2 & = & s^2 + 2 s {\bf u}+ ( {\bf u}. {\bf u}) \\ & = & t {\bf 1}+ {\bf v}, \end{eqnarray*}$$

which would require in succession

$$\begin{eqnarray*} {\bf u}& = & \frac{{\bf v}}{2s} \\ s^2 + \frac{({\bf v}\cdo... ...\\ s^2 & = & \frac{1}{2}(t\pm\surd(t^2-({\bf v}\cdot{\bf v})) \end{eqnarray*}$$

If $s$ were $0$ then ${\bf v}$ would have to be zero, and only a scalar could have a vector square root. But non-vector quaternions can have quaternion roots, of which there would appear to be exactly four possible values for $s$, not all necessarily real. For example, $\surd{\bf i}= \pm({\bf 1}+{\bf i})/\surd2$.