Why is it the case that infinity is also a branch point for log

Question

I thought about considering $L(z)=log|\frac{1}{z}|+i\theta$ and consider the behaviour as $z$ tends to $0$? All I am observing now is that the function tends to infinity with it. I am really stuck on how to proceed any further.

Answer 1

We recall a point is a branch point of a function $w(z)$ iff it is discontinuous upon traversing a small circuit around this point. We use polar coordinates and write $z=re^{i\theta}$ with $\theta=\theta_{p}+2\pi n$ and $0\leq \theta_{p}<2\pi, n\in\mathbb{Z}$. We then have \begin{align*} \log z=\log{r} + i\theta_{p}+2n\pi i\tag{1} \end{align*} where $\log r> 0$ and $n$ is an integer. For a given value of $z$, the function $\log{z}$ takes infinite values corresponding to $n\in\mathbb{Z}$.

• The key observation here is, when traversing a small circuit around $z=0$, we do not return to the original value. When starting for instance at $z=\varepsilon$ for real $\varepsilon >0$ and letting $n=0$ we start with $\theta_p=0$ and after returning to $z=\varepsilon, \theta_p=2\pi$ we have according to (1) $\log{z}=\log{r} + i\theta_{p}+2\pi i$. Traversing another circuit we have $\log{z}=\log{r} + i\theta_{p}+4\pi i$ and this continues for $n\in \mathbb{Z}$. This way we have $0$ identified as branch point of the function $\log{z}$.

• The point at $z=\infty$ is another branch point of $\log{z}$ which can be seen as follows: We consider $z=\frac{1}{t}$ and obtain \begin{align*} \log{\frac{1}{t}}=-\log{t}=-\left(\log|t|+i\arg{t}+2n\pi i\right) \qquad n\in\mathbb{Z}\tag{2} \end{align*} In (2) we see that $0$ is a branch point of $-\log {t}$ since we can argue in the same way as we did above in (1) for the branch point $0$ of $\log {z}$ .