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

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

(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 power can become $\frac{\pi}{i} \left( \frac{1}{\phi} - \phi \right)$

Now for some golden ratio jiggery-pokery. Since $\phi = \frac{\sqrt{5} + 1}{2}$, by definition, $\frac{1}{\phi} = \frac{2}{\sqrt{5}+1}$, which we can sort out using a conjugate trick: $\frac{2}{\sqrt{5}+1}\times \frac{\sqrt{5}-1}{\sqrt{5}-1} = \frac{\sqrt{5}-1}{2}$. We've got $\frac{1}{\phi} - \phi$ in our final bracket, so that's $\frac{\sqrt{5}-1}{2} - \frac{\sqrt{5} +1}{2}$, which becomes.... just -1. That's simpler!

Our power is now $\frac{\pi}{i} \left(-1\right)$, which is $\pi i$.

And, oh look, here's our friend Leonhard Euler dropping in for a visit! The left hand side is now $e^{\pi i}$, which (as any fule no) is -1. With a tombstone!

* Edited 2016-05-31 to fix a missing bracket. Thanks to @chrishazell72 for pointing it out.

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.