As stated in the question, I'm supposed to prove that there doesn't exist an analytic function on $\mathbb{C}$ s.t. $f(z)^2 = z^2 - 1$, and I'm not sure where to start with this, as we've talked about extending functions using Monodromy's Theorem, but not necessarily showing why it couldn't exist.
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You want $f(z) = \sqrt{z^2-1}.$ To show that this cannot be analytic, look at the argument of $f(z-1).$
Igor Rivin
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I did some digging and from this post https://math.stackexchange.com/questions/1747720/is-sqrtz-an-analytic-function they suggested showing the $arg(z)$ is discontinuous at $\pm \pi$. Was this kind of like what you were suggesting? I'm also not too sure how I'll find the $arg(f(z-1))$ since $f(z-1) = \sqrt{r^2e^{2i\theta} - 2re^{i\theta}}$ and I'm not sure how to reduce that. – Jill Apr 01 '21 at 21:52
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Such a function $f$ has a zero in $1$, whence $ord_1 f=m$ is a positive integer. Therefore, $$\mathrm{ord}_1 f^2=2m\neq 1= \mathrm{ord}_1 (z^2-1).$$
Pomponazzo
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