Let $D \subseteq \mathbb{C}$ be open and $f : D \rightarrow \mathbb{C}$ meromorphic with a pole of order $\ge 2$ in $a \in D$. Then $f$ is not injective.

Is there an easy proof to this? This is not homework; it comes from user8268's answer in entire 1-1 function.


If $f$ has a pole of order $m$ at $a$, then (after removing the removable singularity) $g = 1/f$ has a zero of order $m$ there. Let $C$ be a small circle (oriented positively) around $a$. For $\alpha \notin g(C)$, the number of zeros (counted by multiplicity) of $g - \alpha$ inside $C$ is $\dfrac{1}{2\pi i} \oint_C \dfrac{g'(z)}{g(z)}\ dz$, and this is continuous (and therefore constant) in a neighbourhood of $\alpha = 0$, with value $m$ at $\alpha$. But the zeros of $g'$ are isolated, so the $m$ zeros of $g-\alpha$ are all distinct if $C$ is small enough.


Here's one way. You can just assume $f$ is just a holomorphic function with a zero of order 2 at $a$, at least locally. (If there is a pole/zero of a different order, modify the approach appropriately, as shown below.) Now, the "big cool theorem" is that if a function has a zero of order 2, by choosing an appropriate holomorphic coordinate system, we can actually assume it's just $z^2$. (We've assumed $a=0$.) To build this chart, write $f(z) = z^2h(z)$ with $h(0) \neq 0$ and holomorphic. Then the function $g(z) = z\sqrt{h(z)}$ is well-defined and holomorphic in a neighborhood of zero and $g(z)^2 = f(z)$. And now for the cool part. Taking derivatives, we find $g'(0) = \sqrt{h(0)} \neq 0$, and so by the inverse function theorem $g$ is actually an invertible holomorphic function taking $0$ to $0$. So, changing coordinates with $w = g(z)$, we have $w^2 = f(g^{-1}(w))$. This rewrites $f$ in the nice form.

Since $z^2$ is not 1-1 and $g$ is a bijection, $f$ is not 1-1. The upshot to this approach is that you now locally understand the behavior of all meromorphic functions! (Up to a nice chart.)

In general, if you write $f(z) = z^kh(z)$, where $h$ holomorphic with $h(0) \neq 0$, then $g(z) = zh(z)^\frac{1}{k}$ gives a well-defined holomorphic function in a neighborhood of zero with $g'(z) = \frac{1}{k}zh(z)^{\frac{1}{k}-1}h'(z) + h(z)^\frac{1}{k}$, and so $g'(0) \neq 0$. So we can perform the change of coordinates with $w = g(z)$, as in the above case. So for a pole or zero of any order $\geq 2$, we can represent $f$ in a coordinate system in which it is obviously not injective.

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    $\begingroup$ I don't understand how $z \mapsto 1/z$ can work like you say. What if $f(z) = z + 1/z^2$, where we still have a pole? $\endgroup$ – user108717 Nov 14 '13 at 7:01
  • $\begingroup$ Right, you can totally have something like that happen. You don't want to do the inversion map, but with modifications it will still work. For instance, your example can be factored as $f(z) = \frac{1}{z^2}(1 + z^3)$. Now, we take $h(z) = 1 + z^3$. Then $g(z) = \frac{1}{z}\sqrt{h(z)}$. This, instead, represents $f$ locally as the pole $\frac{1}{z^2}$. I've edited the original post to refer to this comment. $\endgroup$ – Zach L. Nov 14 '13 at 7:10
  • $\begingroup$ OK, I understand this now $\endgroup$ – user108717 Nov 14 '13 at 7:57

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