Pole of order $\ge 2 \; \Rightarrow \;$ not injective 
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.
 A: 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.
A: 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.
