# Formula for quaternion exponentiation

I'm having trouble understanding the polar representation of quaternions. That is, any quaternion $z=a+ib+jc+kd=a+\mathbf{v}$ can be expressed in polar form as: $$z = |z|\left(\cos \theta +\mathbf{n}\sin \theta \right)$$ with: $$\begin{cases} & |z|= \sqrt{a^2+b^2+c^2+d^2}\\ & \cos \theta=\dfrac{a}{|z|} \qquad \sin \theta=\dfrac{|\mathbf{v}|}{|z|}\\ & \mathbf{n}=\dfrac{\mathbf{v}}{|z|\sin \theta}=\dfrac{\mathbf{v}}{|\mathbf v|} \end{cases}$$

In particular, I'm having trouble with the formula for the exponentiation of a quaternion in this form: $$z^n = |z|^n\left(\cos n\theta +\mathbf{n}\sin n\theta \right)$$ I have tried exponentiating an arbitrary quaternion $a+ib+jc+kd$ by a power of 2 in hopes of understanding what exactly is going on here, but I can't seem to obtain the $\sin 2\theta$ and $\cos 2\theta$ terms. I understand why such a $\theta$ exists such that you can assign $$\cos \theta=\dfrac{a}{|z|} \qquad \sin \theta=\dfrac{|\mathbf{v}|}{|z|}$$ However, defined in this way, I don't know what the expressions $\sin n \theta$ and $\cos n \theta$ would represent, especially in the context of quaternions. If anyone could shed some light on this I would be extremely appreciative!

The easy way to derive de Moivre's formula for quaternions is exactly the same as the easy way to derive it from complex numbers -- with the exponential function.

It can be shown that the power series expansion $e^{z} = \sum_{i=0}^\infty \frac{z^i}{i!}$ converges for any quaternion $z$. Then we can identify the odd and even parts exactly the same as with complex numbers to get $e^{z} = e^{a+b\epsilon} = e^ae^{b\epsilon} = e^a(\cos(b)+\epsilon\sin(b))=c+d\epsilon$ where $\epsilon$ is a unit imaginary quaternion.

So once you've proven the assertions I've made above, then clearly $$\omega^n = (c+d\epsilon)^n = (e^{z})^n = e^{nz} = e^{na+nb\epsilon} = (e^{a})^n(\cos(nb)+\epsilon\sin(nb)) = \lvert\omega\rvert^n(\cos(nb)+\epsilon\sin(nb)),\quad \forall n\in\Bbb Z$$

So basically, go back and look at the sections in your complex analysis text on this and use the same methods (though being careful to note that in general quaterions don't commute) to derive this.

Edit: Here's how to verify it explicitly in the case $n=2$.

\begin{align}\big(\lvert z\rvert(\cos(\theta)+n\sin(\theta))\big)^2 &= \lvert z\rvert^2\big(\cos^2(\theta)+2n\sin(\theta)\cos(\theta)-\sin^2(\theta)\big)\\ &=\lvert z\rvert^2\left([\cos^2(\theta)-\sin^2(\theta)]+n[2\sin(\theta)\cos(\theta)]\right) \\ &= \lvert z\rvert^2\big(\cos(2\theta)+n\sin(2\theta)\big)\end{align}

• Okay cool, I hadn't even considered using the exponential form of quaternions although it seems obvious now. Before I accept your answer, I'm wondering if there is a way to perform this computation (for a small value of n, say n=2) without first converting to the exponential form. Really my question is: given $\cos \theta$, $\sin \theta$ as defined, is there a simple representation of $\cos 2\theta$, $\sin 2\theta$ also in terms of the coefficients & norm of $z$? Thanks! – Will Jan 5 '16 at 21:50
• @Will See my edit. – user137731 Jan 5 '16 at 22:00
• haha I was just attempting this but it seems you are faster than me! Thanks! – Will Jan 5 '16 at 22:03
• No problem. :-) – user137731 Jan 5 '16 at 22:04