Square roots of quaternions

Question

In class we saw that $-1$ has infinitely many square roots in the ring of quaternions. Is it possible to compute the square roots of a given nonreal quaternion? What do they look like?

Answer 1

With non-real quaternions we can always find two square roots.

You can write any quaternion in the form $$ q=a+b\vec{u}, $$ where $a$ and $b$ are real, and $\vec{u}$ is a unit vector. You probably know that as a quaternion $\vec{u}^2=-1$. Therefore we can treat $\vec{u}$ as if it were the usual imaginary unit $i$ of the complex numbers. So we can use the usual techniques of finding square roots of complex numbers.

Note that with non-real quaternions we only get two square roots. This is because the square of the above quaternion is $$ q^2=(a^2-b^2)+2ab\vec{u}. $$ For this to be non-real, we need both $a$ and $b$ to be non-zero. So whenever $q_1^2=q_2$ for some quaternions $q_1,q_2$ where $q_2\notin\mathbb{R}$, they must both lie in the same plane, i.e. they must be linear combinations of $1$ and the same unit vector $\vec{u}$. This means that $q_1$ and $q_2$ must belong to the same copy of $\mathbb{C}=\mathbb{R}\oplus\mathbb{R}\vec{u}$. As any complex number has only two square roots, the same applies to non-real quaternions because of this.