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Suppose f holomorphic and injective on $C\setminus \{0\}$. Then $0$ is an order one pole or a removable singularity and the continuation of $f$ is injective.

I'm reading Pole of order $\ge 2 \; \Rightarrow \;$ not injective and confused about the last sentence. Why are the zeros distinct when $C$ is small enough? Also, he used Rouche theorem to get the number of roots of $g-\alpha$, right?

As for the removable case, I don't know how to show it's continuation is injective.

Fluffy Skye
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  • When a zero has multiplicity the derivative is $0$. The derivative is an analytic function. If its zeros accumulate then it would be the function zero. If the derivative is constant equal to zero, the original function would be constant, at least locally. – logarithm May 03 '19 at 02:55
  • Around a pole of order 2 or higher the function behaves locally like $\frac{1}{z^n}, n\ge 2$ and by inspection it cannot be injective in a small neighborhood of zero; for an essential singularity, it's even worse as any small neighborhood has dense image in the plane - actually much more is true but this result is elementary; at a removable singularity, if $f(0)=f(a), a\ne 0$, then a small neighborhood of zero and of $a$ have the same image , or maybe easier to prove, have images intersecting in an open non empty set, contradicting the previous injectivity outside $0$ – Conrad May 03 '19 at 02:55
  • @Conrad so by open mapping theorem, the image intersect in an open non empty set, right? – Fluffy Skye May 03 '19 at 03:02
  • @logarithm I'm confused. In the linked solution, it says $g'$ has isolated zeros and thus not constantly zero? I'm not understanding the logic here. – Fluffy Skye May 03 '19 at 03:04
  • yes, pretty much so ($f(a)=f(0)$ and open mapping means indeed that a small neighborhood of $a$ is sent into a small neighborhood of $f(a)$ and same with $0$ but two open neighborhoods of the same point have open non-empty intersection - as noted with a little more work you can actually make them equal but is not needed) – Conrad May 03 '19 at 03:05
  • @FluffySkye Read the part where I said "If its zeros accumulate". That is how a proof by contradiction looks like. Assumption, implication, contradiction, and conclude that the assumption is false. – logarithm May 03 '19 at 03:08
  • @logarithm So for the linked problem, if $g'$ is constantly zero, then $g$ is locally constant zero and thus not injective, is that what you're saying? – Fluffy Skye May 03 '19 at 03:18
  • If $g'$ is constant $0$ then $g$ is locally constant, not necessarily zero. But the important part of what I said is that this implies that the points where $g$ has multiplicity, larger than $1$, don't accumulate. – logarithm May 03 '19 at 03:22
  • @logarithm So you're saying since $g$ has multiplicity, its derivative is $0$ at those multiplicity points, such points can't accumulate because that would lead to $g$ being constant, right? – Fluffy Skye May 03 '19 at 03:42
  • @logarithm But what does this have to do with roots of $g-\alpha$ are distinct? – Fluffy Skye May 03 '19 at 03:42

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