Why does the real part of quaternion conjugation with a pure quaternion stay 0?

Question

This answer stated a few facts which I don't understand completely:

The "real part" (the $w$) of the product of two quaternions $pq$ is the same as the "real part of $qp.$

So, what happens when I take a unit quaternion $q$ and a "pure" quaternion $v,$ and calculate $$ p = q^\ast v q.$$

Well, we have $$\parallel p \parallel = 1 \cdot \parallel v > \parallel \cdot 1 = \parallel v \parallel $$

But as to the "real part," we begin with $$ \Re v = 0,$$

then $$ \Re q^\ast (v q) = \Re (v q) q^\ast = \Re v (q q^\ast) = \Re v = 0. $$

My question is not short enough to comment it there, so here it goes as its own question: Why is the real part of $q^*(vq)$ equal to the real part of $(vq)q^*$?
The other steps in the last line seem to be simple enough.

Ultimately, I'm trying to figure out under what preconditions the quaternion multiplication and/or conjugation keep the real part zero. I was unable to find any better resource than the linked answer.

In short:

1. How to get the equality $\Re {q^*(vq)} = \Re (vq)q^*$ ? ($q^*$ denotes the conjugate of $q$, which is equal to $q^{-1}$ for a unit quaternion)

2. Which of the preconditions stated in the quote are neccessary for the implication $\Re v = 0 \Rightarrow \Re (qvq^{-1}) = 0$ ?

Answer 1

1. Since, for any two quaternions $p$ and $q$, you have $\operatorname{Re}(pq)=\operatorname{Re}(qp)$, then, in particular$$\operatorname{Re}(q^*vq)=\operatorname{Re}(vqq^*).$$
2. Nothing is needed; it is always true that $\operatorname{Re}(v)=0\implies\operatorname{Re}(q^{-1}vq)=0$ . And I saw no preconditions in the passage that you quote.