Why the derivative of the logarithm of a theta function is not an elliptic function?

Question

Let's consider the theta functions periodicity conditions $$\vartheta\left(z+1\right) =\vartheta\left(z\right),$$ $$\vartheta\left(z+\tau\right) =e^{-\pi i\tau-2\pi iz}\vartheta\left(z\right),$$ Those imply conditions on the derivatives of theta functions $$\vartheta'\left(z+1\right) =\vartheta'\left(z\right),$$ $$\vartheta'\left(z+\tau\right) =e^{-\pi i\tau-2\pi iz}\vartheta'\left(z\right),$$ Therefore if I take the derivative of the logarithm I should have an elliptic function since $$f(z)=\frac{d}{dz}log\left( \vartheta\left(z\right)\right) =\frac{\vartheta'\left(z\right)} {\vartheta\left(z\right)}$$ then we have $$f(z+\tau)=f(z)=f(z+1).$$ Anyway this can't be since it has only one pole in the fundamental period, so what am I missing?

Answer 1

From $$\vartheta(z+\tau) =e^{-\pi i\tau-2\pi iz}\vartheta(z)$$ you get $$\vartheta'(z+\tau) =e^{-\pi i\tau-2\pi iz}\vartheta'(z) -2\pi i e^{-\pi i\tau-2\pi iz}\vartheta(z)$$ and so $$\frac{\vartheta'(z+\tau)}{\vartheta(z+\tau)} =\frac{\vartheta'(z)}{{\vartheta(z)}} -2\pi i.$$ Therefore $\vartheta'/\vartheta$ is not an elliptic function but $(\vartheta'/\vartheta)'$ is (essentially the Weierstrass $\wp$-function).