There’s something neat about an identity or result that seems completely unexpected, and this one is an especially nice one:

$$ e^{2\pi \sin( i \ln(\phi) }= -1$$

(where $\phi$ is the golden ratio.)

It’s one of those that just begs, “prove me!” So, here goes!

I’d start with the identity $2\sin(x) \equiv \frac{e^{ix} – e^{-ix}}{i}$. In particular, the power on the left hand side is $2\pi \sin( i \ln(\phi) ) = \frac{\pi}{i} \left( e^{-\ln(\phi)} – e^{\ln(\phi)} \right)$.

We’ve got $e$s and $\ln$s mixed up in the brackets, which is a Clue if ever I saw one: since $e^{\ln(x)} = x$, the…

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