Let $q \in S^3$. Therefore $q$ can be represented as $q=\cos(\alpha/2) + \sin(\alpha/2)u$ for some $\alpha \in \mathbb{R}$ and some $u \in S^3$ with it's real part zero. Recall that the quaternions with real part zero can be identified with vectors in $\mathbb{R}^3$. Define a function $T_q: \mathbb{R^3} \rightarrow \mathbb{R^3}$ by
$T_q(v) = qvq^{-1}$ for $v \in \mathbb{R^3}$.
I need to show that $T_q(v)$ has real part zero by calculating $T_q(v)+ \overline{T_q(v)}$.
Essentially this means that $T_q(v)+ \overline{T_q(v)}= 0$.
What I have done so far:
$$ T_q(v)+ \overline{T_q(v)} = qvq^{-1} + \overline{qvq^{-1}}\\ =r(qw\overline{q} - \overline{q}wq) $$
since $q^{-1}=\overline{q}$ and $v \in \mathbb{R}^3$ implies $\overline{v} = -rw$ when identified as a quaternion, $w \in S^3$ and $r\in \mathbb{R}$ is the length of $v$. Hence
$$ qwq^{-1} - \overline{q}wq = (\cos(\alpha/2) + u\sin(\alpha/2))\;w\;(\cos(\alpha/2) + u\sin(\alpha/2)) \\ \qquad \qquad \qquad - (\cos(\alpha/2) - u\sin(\alpha/2))\;w\;(\cos(\alpha/2) + u\sin(\alpha/2))\\ = (uw-wu)\frac{1}{2}\sin(\alpha) $$
Now since $u,w \in \mathbb{H}$ each have real part zero, $uw = -(u\cdot w) + (u \times w)$ (LHS of this equation is multiplication in $\mathbb{H}$ and RHS it is the dot and cross product on $\mathbb{R}^3$). Hence,
$$ uw-wu = 2(u \times w) $$
And this is where I get stuck.
Am I going the wrong way or am I missing a trick? Please help
Ok, I think you've gone wrong fairly early on. What it looks like you've missed, is that fact the conjugate isn't quite multiplicative in the quaternions. Instead, we have: $$ \overline{zw} = \overline{w} \space \overline{z} $$
Hence $ qvq^{-1} + \overline{qvq^{-1}} = qv\overline{q} + \overline{qv \overline{q}} = qv\overline{q} + q \space \overline{qv} = qv\overline{q} + q \space \overline{v} \space \overline{q} = qv\overline{q} + q \space (-v) \space \overline{q} = q(v-v)\overline{q} $