The Hopf map is given by the projection $\pi: \mathbb{S}^3 \to \mathbb{S}^2$, and:
$\pi: z \mapsto zi\bar{z}$,
where $z \in \mathbb{S}^3$ and $i \in \mathbb{H}$ is a unit quaternion.
Show that the level set passing through a point $z \in \mathbb{S}^3$ is given by $\{e^{i \theta}z: \theta \in \mathbb{R}\}$.
[Hint: Show that $\pi(z) = \pi(w)$ if and only if $\bar{w} z$ commutes with $i$].
I can prove the hint easily enough, but I've no idea how it helps! If anyone can explain this, and how it leads to the final answer, that'd be great. If anyone has any ideas, I'd be very grateful.
Let $q = a + bi + cj + dk \in \mathbb{H}$. Then
$$iq = i(a + bi + cj + dk) = ai - b + cij + dik = ai - b + ck -dj$$
and
$$qi = (a + bi + cj + dk)i = ai - b + cji + dki = ai - b - ck + dj.$$
Therefore, $iq = qi$ if and only if $c = d = 0$. That is, the quaternions $q$ which commute with $i$ are of the form $q = a + ib$.
As $\bar{w}z$ commutes with $i$, $\bar{w}z = a + ib = re^{i\theta}$ for some $r \in [0, \infty)$ and $\theta \in \mathbb{R}$. As $\bar{w}z \in S^3$, $\|\bar{w}z\| = 1$ so $r = 1$. Hence, $\bar{w}z = e^{i\theta}$ so $z = we^{i\theta}$ and we have
$$\pi^{-1}(w) = \{z \in S^3 \mid \pi(z) = \pi(w)\} = \{z \in S^3 \mid z = we^{i\theta}, \theta \in \mathbb{R}\} = \{we^{i\theta} \mid \theta \in \mathbb{R}\}.$$