Define $f(z)=(z^{2}+1)^{1/2}$. My argument is as follows: if we want to find out if $z=\infty$ is a branch point of this function, we define $\zeta=\frac{1}{z}$. From this, $f(\zeta(z))=(\frac{1}{\zeta^2}+1)^{1/2}=(\frac{1}{\zeta^2})^{1/2}(1+\zeta^2)^{1/2}$. Near $\zeta=0$, $f(\zeta)\rightarrow\infty$, so it has a branch point. This means $f(z)$ has a branch point at $z=\infty$. However, this was marked wrong, and I don't know why. Where does my argument fail, and why?
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Are you familiar with the different branches of log(z)? – Omry Feb 22 '17 at 17:28
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You create a singularity articially by taking the inverse. – Feb 22 '17 at 17:28
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@Omry Yes I am. – John Doe Feb 22 '17 at 17:32
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@YvesDaoust, Could you elaborate? – John Doe Feb 22 '17 at 17:33
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No, it's not a branch point, because the local solutions are $\pm \zeta^{-1} A(\zeta)$ where $A(\zeta)$ is analytic at $\zeta=0$. Since the exponent $-1$ is an integer, one sees that $\zeta=0$ is not a branch point. – Mark Feb 22 '17 at 17:44
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Ah yes, that's what I was missing - so the exponent being an integer means that $z^n=(re^{i\theta})^n$ is continuous when jumping from $2\pi$ to $0$, and so is not a branch point, right? – John Doe Feb 22 '17 at 17:52
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No, it's not a branch point, because the local solutions are $\pm\xi^{−1}A(\xi)$ where $A(\xi)$ is analytic at $\xi=0$. Since the exponent $−1$ is an integer, one sees that $\xi=0$ is not a branch point.
John Doe
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@rasmus the solutions near $\zeta=0$. There may be other solutions, but we don't care about them here. – John Doe Jun 03 '21 at 18:26