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From a complex number, $z=x+iy$ and its conjugate there is a simple way to evaluate the real part and imaginary part, \begin{equation} \text{Re}(z)=\frac{z+z^*}{2} \end{equation} and \begin{equation} \text{Im}(z)=\frac{-i(z-z^*)}{2} \end{equation} I was wondering for a quaternion, $q=w+ix+jy+kz$, there is a similarly simple way to get Im($q$), or indeed any specific component?

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The imaginary axis in $\mathbb{C}$ is spanned by $\{i\}$, so it makes sense for $\mathrm{Im}()$ to extract the coefficient of $i$ from the expression $a+bi$ (with $a,b\in\mathbb{R}$). But in $\mathbb{H}$, the 3D real subspace of imaginary quaternions is not spanned by a single element ($\{\mathbf{i},\mathbf{j},\mathbf{k}\}$ is an orthonormal basis), the $\mathrm{Im}()$ operator cannot extract a single real number, instead of just gives the entire imaginary part, as a (purely) imaginary quaternion. That is, if $a$ is a real number and $\mathbf{b}$ is a vector (i.e. pure imaginary quaternion) then we define

$$ \mathrm{Re}(a+\mathbf{b})=a, \qquad \mathrm{Im}(a+\mathbf{b})=\mathbf{b}. $$

A quaternion $q$ and its conjugate $\bar{q}$ are then given by

$$ q=\mathrm{Re}(q)+\mathrm{Im}(q), \qquad \bar{q}=\mathrm{Re}(q)-\mathrm{Im}(q). $$

We can then solve

$$ \mathrm{Re}(q)=\tfrac{1}{2}\big(q+\bar{q}\big), \qquad \mathrm{Im}(q)=\tfrac{1}{2}\big(q-\bar{q}\big). $$

You can also get a specific component, but for many purposes $\mathbf{i},\mathbf{j},\mathbf{k}$ is an arbitrary basis. So say we want the $\mathbf{u}$-component, where $\mathbf{u}$ is any unit vector. If $\mathbf{v}$ is another vector, then we have the formula $\mathbf{uv}=-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$, which means the $\mathbf{u}$-component of $\mathbf{v}$ is given by

$$ \|\mathrm{proj}_\mathbf{u}\mathbf{v}\|=-\tfrac{1}{2}\big(\mathbf{uv}+\mathbf{vu}\big). $$

To extract the $\mathbf{u}$-component of an arbitrary quaternion $q$, then, we can set $\mathbf{v}=\mathrm{Im}(q)$ in the above formula. More generally, $\langle x,y\rangle=\mathrm{Re}(\bar{x}y)$ is the usual inner product on $\mathbb{H}$, so the projection of a quaternion $q$ onto a unit quaternion $p$ is given by $\mathrm{proj}_pq=\mathrm{Re}(\bar{p}q)p$, which means the $p$-component of $q$ is simply $\mathrm{Re}(\bar{p}q)$.

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The conjugate can be defined purely in terms of quaternion addition and multiplication.

$q^*=-{1\over 2}(q+iqi+jqj+kqk)$

Then as mentioned in other answers:

$Re(q)={1\over 2}(q+q^*)$

$Im(q)={1\over 2}(q-q^*)$

Individual components can also be extracted.

If $q=w+ix+jy+kz$, then:

$iqi=i^2w+i^3x+ijiy+ikiz=-w-ix+jy+kz$

so:

${1\over 2}(q-iqi)=w+ix$

Hence:

${1\over 2}(q^*-iq^*i)=w-ix$

And we can subtract to find $x$. Similarly for $y$ and $z$ with $jqj$ and $kqk$.

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