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I want to find the largest open set where I can analytically continue $\frac{\log(1-z^2)}{z^2}$.

Attempt The numerator is analytic in $\mathbb{C}\setminus[-\infty,-1]\cup[1,\infty]$ (by linking the branch points $\pm 1$ and $\infty$)and the denumerator non-zero outside zero, so by intersecting:

$$U=\mathbb{C}\setminus[-\infty,-1]\cup[1,\infty].$$

However, taking them together I think there is a larger set i.e.: $\mathbb{C}\setminus [-1,1]$

How to show that $\infty$ is not a branch point for $\dfrac{\log(1-z^2)}{z^2}$? Any hints.

As $z\to \infty$ we set $\frac{1}{z}$ and take $z\to 0$ then:

$$r^{2}e^{i2\theta}(log(z-1)+log(z+1)-2log(z))$$

so by making a small circle around 0, the first two terms are fixed but the last one is shifted. So infinity is actually a branch point.

Any mistakes?

user133100
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1 Answers1

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To analyze behaviour at $\infty$, set $z = 1/\zeta$. Then you are studying the neighborhood of $0$ of \begin{align*} \zeta^2 \log(1-\zeta^{-2}) &= \zeta^2 \log(\zeta^{-2}(\zeta^{2}-1)) \\ &= \zeta^2 \left( \log(\zeta + 1) + \log(\zeta - 1) - 2 \log(\zeta) \right) \text{.} \end{align*} Put a small loop around $0$. The $\zeta^2$, $\log(\zeta + 1)$, and $\log(\zeta - 1)$ do not cause any branch difficulties. However, $-2 \log(\zeta)$ moves by $-4\pi \mathrm{i}$, so $\infty$ is a branch point and there is a branch cut terminating at $\infty$. (It's other end is one of the other two branch points or is a point on your cut, $[-1,1]$.)

Page 11 here has much nicer diagrams than I can make. (The document is Dr. R. Rosales's Branch Points and Branch Cuts (18.04, MIT), from 11 October 1999.)

In slightly more detail, $\oint -2 \zeta^2 \log \zeta \,\mathrm{d}\zeta$ where the path is a small circle of radius $r \in (0,1)$ centered at $0$ is $\frac{4\pi\mathrm{i}}{3} \mathrm{e}^{3r} \rightarrow \frac{4\pi\mathrm{i}}{3}$ as $r \rightarrow 0$, so $\zeta = 0$ is a branch point. Therefore $z = \infty$ is a branch point. (I did that integral "by memory" and I don't recall if I've dropped a minus sign. Either way, you get a branch point.)

Eric Towers
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