Editing Quaternion (section)

=== Square roots of arbitrary quaternions ===

Any quaternion <math>\mathbf q = (r,\, \vec{v})</math> (represented here in scalar–vector representation) has at least one square root <math>\sqrt{\mathbf q} = (x,\, \vec{y})</math> which solves the equation <math>\sqrt{\mathbf q}^{\,2} = (x,\, \vec{y})^2 = \mathbf q</math>. Looking at the scalar and vector parts in this equation separately yields two equations, which when solved gives the solutions

<math display=block>
\sqrt{\mathbf q}
= \sqrt{(r,\, \vec{v})}
= \pm\left(\sqrt{\tfrac12\bigl({\|\mathbf q\|+r}\bigr)},\ \frac{\vec{v}}{\|\vec{v}\|}\sqrt{\tfrac12\bigl({\|\mathbf q\|-r}\bigr)}\right),
</math>

where <math display="inline">\|\vec{v}\| = \sqrt{\vec{v}\cdot\vec{v}}=\sqrt{-\vec v\vphantom{v}^2}</math> is the norm of <math>\vec{v}</math> and <math display="inline">\|\mathbf q\| = \sqrt{\mathbf q^*\mathbf q} = \sqrt{r^2 + \|\vec{v}\|^2}</math> is the norm of <math>\mathbf q</math>. For any scalar quaternion <math>\mathbf q</math>, this equation provides the correct square roots if <math display="inline">\vec{v} / \|\vec{v}\|</math> is interpreted as an arbitrary unit vector.

Therefore, nonzero, non-scalar quaternions, or positive scalar quaternions, have exactly two roots, while 0 has exactly one root (0), and negative scalar quaternions have infinitely many roots, which are the vector quaternions located on <math>\{0\} \times S^2\bigl(\sqrt{-r}\bigr)</math>, i.e., where the scalar part is zero and the vector part is located on the [[n-sphere|2-sphere]] with radius <math>\sqrt{-r}</math>.