Im trying to express the following quaternion in polar form (axis-angle)


Would this be the resultant ?

$$\cos \frac{θ}{2} +\sin \frac{θ}{2} (i-2j+k)$$


2 Answers 2


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).

  • $\begingroup$ If a quaternion a + bi + cj + dk (containing 4 values) is converted into r*e^nθ (with only 2 values), isn't some information is lost? Is this one to one? Is this reversible? $\endgroup$ Jan 20, 2022 at 14:38
  • $\begingroup$ Sorry, now I understand the formula better, so ignore above comment. My doubt is, is this function one-one and onto? $\endgroup$ Jan 21, 2022 at 4:01
  • $\begingroup$ Why not just write $\tan\theta=\frac {|\mathbf v|}a$? $\endgroup$ Nov 4, 2023 at 0:23
  • $\begingroup$ Because for the same value of $\tan \theta$ we can have two opposite values of $\cos \theta$ $\endgroup$ Nov 4, 2023 at 8:25

In general if you have a number $z=a+b_1i+b_2j+b_3k$, you can write it as $$ z = A\cos\theta + A\sin\theta(\cos\varphi_1 i+\cos\varphi_2 j+\cos\varphi_3 k) $$ where $A = \sqrt{a^2+b_1^2+b_2^2+b_3^2}$, and $\cos\theta = a/A$. And if you define $B=\sqrt{A^2-a^2}$, then $\cos\varphi_i = b_i/B$. Clearly $\cos^2\varphi_1+\cos^2\varphi_2+\cos^2\varphi_3=1$. A quick calculation shows $$(\cos\varphi_1 i+\cos\varphi_2 j+\cos\varphi_3 k)^2 = -1$$ This means that in fact you can also write $$z = A\exp\Big[(\cos\varphi_1 i+\cos\varphi_2 j+\cos\varphi_3 k)\theta\Big]$$ This is one possibility of polar coordinates for quarternions. One can do other things too.

  • $\begingroup$ So is my answer from above an acceptable answer. Or should i say the correct answer ? $\endgroup$
    – JekasG
    Oct 25, 2015 at 7:29
  • $\begingroup$ I doubt it is correct but then again depends on what's you definition of $\theta$? $\endgroup$
    – Hamed
    Oct 25, 2015 at 7:34
  • $\begingroup$ So how would i work out the answer ? And what is the correct answer ? $\endgroup$
    – JekasG
    Oct 25, 2015 at 8:19

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