The total number of poles of an elliptic function in $P_0$ is always $\geq$ 2 I'm trying to follow this proof from Stein & Shakarchi "Complex Analysis". The statement of the theorem is in the title of the question. The Proof is as follows:
Suppose that $f$ (an elliptic function) has no poles on the boundary $\partial P_0$ where $P_0 =\{ z \in \mathbf{B} : z = a + b\tau \text{ where } 0\leq a < 1 \text{  and  } 0 \leq b < 1 \}$ is the fundamental parallelogram. By residue theorem
$\int_{\partial P_0} f(z) dz = 2\pi i \sum res(f)$.
The integral is zero due to periodicity of $f$, implying $f$ must hae at least two poles in $P_0$.
My question is: why does this imply that $f$ must have atleast 2 poles? 

We know $f$ can't have zero poles, since then it'd be bounded on the fundamental parallelogram, so bounded on the whole plane by periodicity and hence constant by Liouville's theorem.
And $f$ can't have one pole (counted with multiplicity), since if it did its residue there would be nonzero and so $2\pi i\sum {\rm Res}(f)$ would just be a single nonzero term, contradicting the fact that this equals $\oint_{\partial P_0}f(z)dz$ which is zero by periodicity.
