Context: It is easy to see that for $p, v \in \mathbb H = \mathbb R^4$, the usual inner product is given by $\langle p,v\rangle = \mathrm{Re} (\bar p v)$. From this we may conclude that $\langle (p^1, p^2), (v^1, v^2)\rangle = \mathrm {Re} (\bar p^1 v^1 + \bar p^2 v^2)$, for any $(p^1, p^2),(v^1,v^2) \in \mathbb H^2= \mathbb R^8$.
Question:
$\mathrm {Re} (\overline{p^1 a} v^1 + \overline{ p^2a} v^2) = 0$ for every $a \in \mathrm {Im} (\mathbb H)$ iff $\bar p^1 v^1 + \bar p^2 v^2$ is real, i.e., iff$\mathrm{Im}(\bar p^1 v^1 + \bar p^2 v^2) = 0$.
Attempt 1: I'm failing miserably at seeing that this is true. So far, just managed the obvious part
$$\begin{align}\overline{p^1 a} v^1 + \overline{ p^2a} v^2 & = \bar a \bar p^1 v^1 + \bar a\bar p^2 v^2 \\&= \bar a (\bar p^1 v^1 + \bar p^2 v^2)\end{align}$$ Thus $0 = \mathrm {Re} (\overline{p^1 a} v^1 + \overline{ p^2a} v^2) = \mathrm {Re} (\bar a(\bar p^1 v^1 + \bar p^2 v^2))$, which is obviously true since $\bar a \in \mathrm {Im}$. But it doesn't give much about $(\bar p^1 v^1 + \bar p^2 v^2)$. Assuming on the other hand that $\mathrm{Im} (\bar p^1 v^1 + \bar p^2 v^2) = 0$ would only yield that $\mathrm{Im} (\overline{p^1 a} v^1 + \overline{ p^2a} v^2 )= 0 $, which again doesn't say anything about the real part.
Maybe there is some property of the quaternions that correlates the real and imaginary parts that I don't know about.
Attemp 2: Writing $q = \bar p^1 v^1 + \bar p^2 v^2$. As in Attempt 1, we have that $0 = \mathrm {Re} (\overline{p^1 a} v^1 + \overline{ p^2a} v^2) = \mathrm{Re}(\bar a q)$, where we know that $a \in \mathrm{Im} \mathbb H$ and $q \in \mathbb H$. Thus $$\langle a, q\rangle = 0 , \ \ \forall a \implies \mathrm{Im}(q) = 0$$ since $\langle \cdot , \cdot \rangle$ is non degenerate.
Any ideas? Maybe Attempt 2 is the way to go?