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.

  • $\begingroup$ Funny thing about English is "a set of infinite numbers" is extremely different from "an infinite set of numbers." I'm sure the latter is what you meant. $\endgroup$ – Matt Samuel Jul 16 '16 at 23:50

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