# Simple expression for imaginary part of a quaternion

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

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 $$\mathrm{Re}(\bar{p}q)$$.

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 $$kqk$$.