There's a small detail in this proof on why $\sum_{k=1}^{\infty}\frac{1}{k^2} = \frac{\pi^2}{6}$ that I can't figure out

Question

http://www.maa.org/sites/default/files/pdf/upload_library/2/Kalman-2013.pdf

Here is a link to the article I have been reading. It's really interesting and easy to follow. What bothers me is a result the author uses. On page 45, the author goes on to say that $\ln(-1) = i\pi$ because $e^{i\pi} = -1$. Just take $\ln$ to both sides. Then he substitutes that result in and gets $\displaystyle\frac{\pi^2}{6}$. Makes sense. Here is my problem. Technically $e^{3i\pi}$ is also equal to $-1$ but if we were to substitute it in, we would not get the correct solution. So I guess my question is, does using other forms of $e^{i\pi}$ like $e^{3i\pi}$ work or is there a reason why you can only choose $e^{i\pi}$?

Answer 1

In general one can't take a logarithm of a negative number (This is at least the way we see it in calculus). One can extend the logarithm so that one can take the logarithm of a negative number; in fact one can take a logarithm of any non-zero complex number. But this depends on definition.

A standard way to define this is to say that $$ \log(z) = \ln\lvert z \rvert + i Arg(z). $$

With this definition $$ \log(-1) = \ln(1) + i(\pi) = i\pi. $$

Answer 2

The Taylor series $$\log(1+z) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}z^k}{k}$$ is only valid for the principal branch of the logarithm.

Answer 3

the first section of note is only the outlook of proof and in page 46 (Making the Proof Rigorous) the author has mentioned that the branch $r>0$ , $-\pi<\theta\le\pi$ has been used.