TITLE: How can i express a quaternion in polar form?
QUESTION [5 upvotes]: Im trying to express the following quaternion in polar form (axis-angle)
$a=1+i-2j+k$
Would this be the resultant ?
$$\cos \frac{θ}{2} +\sin \frac{θ}{2} (i-2j+k)$$

REPLY [4 votes]: To find the polar (exponential) form of a quaternion we have to start from the definition of the exponential of a quaternion. ( see :Exponential Function of Quaternion - Derivation)
For a quaternion $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k} = a+\mathbf{v}$ this is given by:
$$
e^z = e^{a+\mathbf{v}}=e^a \left(  \cos  |\mathbf{v}| +\dfrac{\mathbf{v}}{|\mathbf{v}|} \,\sin |\mathbf{v}| \right)
$$
Now note that we have obviously:
$$
z=a+\mathbf{v}= |z| \left( \dfrac{a}{|z|}+\dfrac{|\mathbf{v}|}{|z|} \dfrac{\mathbf{v}}{|\mathbf{v}|} \right)
$$
end, since: 
$$
a^2+|\mathbf{v}|^2=|z|^2
$$
we have:
$$
\left(\dfrac{a}{|z|}\right)^2+ \left( \dfrac{|\mathbf{v}|}{|z|}\right)^2 =1
$$
so there exists  $0\le\theta<2\pi$ such that:
$$
\dfrac{a}{|z|}=\cos \theta \qquad \dfrac{|\mathbf{v}|}{|z|}=\sin \theta
$$
and every quaternion  $z=a+ib+jc+kd=a+\mathbf{v}$ can be expressed in exponential (polar) form as:
$$
z = |z|\left(\cos \theta +\mathbf{n}\sin \theta \right)=|z| e^ {\mathbf{n} \theta}= e^{\ln |z| + \mathbf{n} \theta}
$$
with:
$$
\begin{cases}
& |z|= \sqrt{a^2+b^2+c^2+d^2}\\
& \cos \theta=\dfrac{a}{|z|} \qquad \sin \theta=\dfrac{|\mathbf{v}|}{|z|}\\
& \mathbf{n}=\dfrac{\mathbf{v}}{|z|\sin \theta}=\dfrac{\mathbf{v}}{|\mathbf v|}
\end{cases}
$$
Now you can apply this general result to your case (note that your answer is wrong and I don't see where it come from).