I've met this formula and I need to demonstrate that it is purely imaginary (it has no real part).
$\frac{1}{2}\log(-\exp(i2\pi q))$, //for a real "input" q.
As I don't know much about maths, what I've tried untill now was to prove it by applying Euler's formula, but getting a real cosine there didn't take me to anywhere. Any help? Thanks!
Use the definition of complex logarithm: \begin{equation*} \log(z) = \ln|z| + i \text{arg}(z) = \ln(z) + i \Big( \text{Arg}(z) + 2 \pi \mathbb{Z} \Big) \end{equation*} (which in fact is a set of infinite numbers), with $\ln|z|$ the real logarithm of the complex number $z$, and $\text{Arg}(z)$ its principal argument, i.e. the angle $\theta \in [0,2\pi]$ when $z$ is written in polar coordinates as $z = r e^{i \theta}$.
For example, taking $z = 2i$, its module is 2, and principal argument $\frac{\pi}{2}$, hence \begin{equation*} \log(2i) = \ln(2) + i\Big( \frac{\pi}{2} + 2 \pi \mathbb{Z}\Big) \end{equation*}
Using this, you can attack now your problem.