'Even-order of zeros and poles' property of an even elliptic function While I am studying that every even elliptic function with periods 1 and τ is a rational function of Weierstrass function from 'Complex Analysis' by Stein and Shakarchi(p.271), I have exactly the same question as the link below. 
Orders of poles/zeros of an even elliptic function
But I still don't understand the answer and don't have enough reputation to add a comment for additional question to the answer.


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*The part that I don't understand is the first paragraph of the answer. I don't know why the equality in the first paragraph implies that the order of vanishing of f at  ω/2 is even. I can only get the information that f is symmetrical to z =  ω/2 in the parallelogram.   

*In the second paragraph, I also can't understand why the vanishing order of f at 0 is even. 
What happens if f satisfies 
f′(0)=f′′′(0)=…=0, and the vanishing order of f at 0 is odd? (f is not identically 0)
Thanks in advance for your answer. 

Let $f$ be any even holomorphic function. If its order of vanishing at zero
is $k$, then
$$f(z)=a_kz^k+a_{k+1}z^{k+1}+\cdots$$
near zero, with $a_k\ne0$. Then
$$f(-z)=(-1)^k(a_kz^k-a_{k+1}z^{k+1}+\cdots)$$
and for $f(z)=f(-z)$ we need $(-1)^ka_k=a_k$, that is $k$ must be even.
If $f$ is now an even elliptic function, and $\omega$ is a period, consider
$$g(z)=f(\omega/2+z).$$
Then
$$g(-z)=f(\omega/2-z)=f(z-\omega/2)=f(z+\omega/2)=g(z)$$
where at the second equality is evenness and the third is periodicity.
Then $g$ is even, and has even order at $0$. This means that $f$ has even order
at $\omega/2$.
