Elliptic function $f(z)=\frac{a_{-2}}{z^2}+a_0+a_1z+a_2z^2+\dots$ must be even 
Let $f:\mathbb{C}\longrightarrow \mathbb {C}$ be a nonconstant elliptic function such that $$f(z)=\frac{a_{-2}}{z^2}+a_0+a_1z+a_2z^2+\dots$$
How to prove that $f(z)$ is even.

Notice that $f'(z)+f'(-z)$ is entire and elliptic therefore constant.
Any help would be appreciated.
 A: Counterexample: $f(z) = \wp(z) + \wp'(z)/\wp(z)^2 = 1/z^2 - 2z + ...$
A: This is not a complete answer. 
First note that if $\omega$ is any period for $z\mapsto f(z)$ then $\omega$ is also a period for $z\mapsto f(-z)$, because $$f(-(z+\omega)) = f(-z-\omega) = f(-z)$$
for all $z$. (The set of periods of $f$ is invariant under negation.)
Now, $f$ is even iff $f(-z) = f(z)$ iff $f(z) - f(-z) = 0$.
Since $f(z) - f(-z)$ is a difference of elliptic functions, it is itself elliptic, 
with power series expansion around the origin given by
$$f(z) -\!f(-z) = 2a_1 z + 2a_3 z^3 + 2a_5 z^5 + \dotsb.$$
At first I thought this would tell us $f(z) - f(-z)$ was entire, but 
this only says that $f(z) -\!f(-z)$ has no pole at the origin.
If we knew, say, that the pole-set of $f$ was invariant under negation, then we'd be done.
Supposing someone shows that $f(z)-\!f(-z)$ has no poles, then we'll know that $f(z) -\!f(-z)$ is a constant. Which constant? $0$, by continuity of the power series about the origin: $f(z)-\!f(-z) \to 0$.
