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.

  • $\begingroup$ 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. $\endgroup$ – logarithm May 3 at 2:55
  • $\begingroup$ 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 $0$ $\endgroup$ – Conrad May 3 at 2:55
  • $\begingroup$ @Conrad so by open mapping theorem, the image intersect in an open non empty set, right? $\endgroup$ – Fluffy Skye May 3 at 3:02
  • $\begingroup$ @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. $\endgroup$ – Fluffy Skye May 3 at 3:04
  • $\begingroup$ 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) $\endgroup$ – Conrad May 3 at 3:05

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