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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?

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closed as off-topic by José Carlos Santos, Adrian Keister, Lee David Chung Lin, metamorphy, Gibbs Jan 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.

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  • $\begingroup$ Thank you John. $\endgroup$ – Wire Bowl Jan 23 at 11:18

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