I am reading the book Complex Analysis by Gamelin and I am having trouble understanding the following definition:

Let $p_j$ be a sequence of non zero complex numbers. We say that $\prod p_j$ converges if $p_j\to 1$ and $\sum \log{p_j}$ converges (Here $\log$ denotes de principal branch of the logarithm).

I don't understand why the expression $\sum \log{p_j}$ makes sense because the $p_j$ can be negative integers, say $p_1=-1$, in this case how we define $\log(-1)$? (the principal branch of the logarithm is not defined for negative integers)

  • 1
    $\begingroup$ Notice that, however, the $p_j$-s will eventually be all in a neighbourhood of $1$. Since the first few terms are inconsequent, you may just consider any tailing sum, without worrying too much to whether $\log(-1)$ has been defined as $i\pi$ or $-i\pi$. $\endgroup$ – Saucy O'Path Apr 19 at 21:02
  • $\begingroup$ You are right, thanks. $\endgroup$ – Sebitas Apr 19 at 21:11

The principal branch of the logarithm is defined by $z=re^{i\theta}\mapsto \log(r)+i\theta$, where we understand $\log(r)$ as a function on the positive reals and we force $\theta\in(-\pi,\pi]$. Since $-1=e^{i\pi}$ we find that $\log(-1)=i\pi$ (at least for the principal branch).

Thus, you can make sense of all negative integers, and you can then talk about the convergence of that sum.

  • $\begingroup$ The problem with that is that the logarithm you are defining is not continuous, because $\log(-1+0i)=i\pi$ while $\log(-1-i0)=-i\pi$ $\endgroup$ – Sebitas Apr 19 at 21:10
  • $\begingroup$ @Sebitas The principal branch of the logarithm isn't supposed to be a continuous function, so all is well. $\endgroup$ – ItsJustLogsBro Apr 19 at 21:11
  • $\begingroup$ Yes, now I understand a lot of things, thanks. Nice name btw $\endgroup$ – Sebitas Apr 19 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.