# How do I express a unit quaternion in exponential form? [closed]

Let $$t,u,v$$ lie in the interval $$(-\pi, \pi]$$

If we assume that $$\cos(t)\cos(u)\cos(v) + \sin(t)\cos(u)\cos(v)i + \sin(u)\cos(v)j + \sin(v)k = e^z$$ such that $$z$$ is also a quaternion, what does z equal and why?

## closed as off-topic by José Carlos Santos, Adrian Keister, Lee David Chung Lin, metamorphy, GibbsJan 23 at 21:05

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The link provided by @MattiP shows this:

If $$\mathbf{v} \in \mathbb{H}_P$$ is an imaginary quaternion, putting $$\theta=|\mathbf{v}|$$ we have: $$e^\mathbf{v}= \cos\theta + \mathbf{v}\;\dfrac{\sin \theta}{\theta}$$

$$\newcommand{bv}{{\mathbf v}}$$ In your case, we want to find $$\bv$$ (which you've called "z"), but we know the right hand side. In particular, we can let $$\theta = \cos^{-1} (\cos(t)\cos(u)\cos(v))$$ and then $$\cos \theta$$ will be equal to the real part of your quaternion, as needed.

To determine $$\bv$$, we multiply the vector part of your quaternion by $$\frac{\theta}{\sin \theta}$$ to get $$\bv = \frac{\theta}{\sin \theta} \left( \sin(t)\cos(u)\cos(v)i + \sin(u)\cos(v)j + \sin(v)k \right)$$

In this expression, $$\sin \theta$$ can be simplified a little, because it's a sine of an arc-cosine, but it's probably not worth doing.

You can probably also get to all this by using Rodrigues' formula, but I doubt it's any simpler.

• Thank you John. – Wire Bowl Jan 23 at 11:18