It is important to realize that $\pm i$, $\pm j$, and $\pm k$ are not the only square roots of $-1$. Rather, any imaginary quaternion squares to a negative number, so it is only necessary to choose its norm to be 1 in order to get a square root of $-1$. The imaginary quaternions of norm 1 form a sphere; in the above notation, this is the set of points \begin{equation} q_2^2 + q_3^2 + q_4^2 = 1 \end{equation} (and $q_1=0$). Any such unit imaginary quaternion $u$ can be used to construct a complex subalgebra of $\HH$, which we will also denote by $\CC$, namely \begin{equation} \CC = \{a + b\,u\} \end{equation} with $a,b\in\RR$. Furthermore, we can use the identity (13) of §2.1. to write \begin{equation} e^{u\theta} = \cos\theta + u\sin\theta \end{equation} This means that any quaternion can be written in the form \begin{equation} q = r e^{u\theta} \end{equation} where \begin{equation} r=|q| \end{equation} and where $u$ denotes the direction of the imaginary part of $q$.

A useful strategy for solving problems such as the linear equations in the previous section, which involve both complex numbers and quaternions, is to break up the quaternions into a pair of complex numbers. Consider the following examples.

We define conjugation of one quaternion $q$ by another quaternion $p$ by $pqp^{-1}$. The norm of $p$ is irrelevant here, so we might as well assume that $|p|=1$, in which case $p^{-1}=\bar{p}$

• What is the result of conjugating a quaternion by $i$?

Write $q$ in terms of a pair of complex numbers via \begin{equation} q = q_\Cone + q_\Ctwo j \end{equation} Then $i$ commutes with the complex numbers $q_\Cone$ and $q_\Ctwo$, but anticommutes with $j$. Thus, \begin{equation} iq\bar\imath = i q_\Cone \bar\imath + i q_\Ctwo j \bar\imath = i q_\Cone \bar\imath - i q_\Ctwo \bar\imath j = q_\Cone - q_\Ctwo j \end{equation} Conjugation by $i$ therefore leaves the complex plane untouched, but yields a rotation by $\pi$ in the $jk$-plane. Analogous results would hold for conjugation by any other imaginary quaternionic unit, such as $u$.

• What is the result of conjugating a quaternion by $e^{i\theta}$?

Interchanging the roles of $i$ and $j$ in the previous discussion, conjugation by $j$ yields a rotation by $\pi$ in the $ki$-plane, so that \begin{equation} j e^{-i\theta} \bar\jmath = e^{i\theta} \end{equation} Multiplying both of these equations on the right by $j$ yields the important relation \begin{equation} j e^{-i\theta} = e^{i\theta} j \end{equation} Thus, \begin{align} e^{i\theta} q e^{-i\theta} &= e^{i\theta} q_\Cone e^{-i\theta} + e^{i\theta} q_\Ctwo j e^{-i\theta} \nonumber\\ &= e^{i\theta} q_\Cone e^{-i\theta} + e^{i\theta} q_\Ctwo e^{i\theta} j \nonumber\\ &= q_\Cone + q_\Ctwo e^{2i\theta} j \label{Hrot} \end{align} corresponding to a rotation by $2\theta$ in the $jk$-plane. We will return to such examples when discussing symmetry groups in Chapter 6.