- The problem statement, all variables and given/known data
Let ${w_1,w_2} $ be a basis for $\Omega$ the period lattice.
Use $\zeta (z+ w_{i})=\zeta(z)+ n_i$ , $i=1,2$ ; $ m \in N$ for the Weierstrass zeta function to show that
$\sigma ( z + mw_i )=(-1)^m \exp^{(mn_i(z+mw_i/2))}\sigma(z)$
- Relevant equations
To note that $\sigma(z)$ is an odd function.
$\zeta(z)=\frac{\sigma'(z)}{\sigma(z)}$
- The attempt at a solution
I am pretty close but messing up with not getting $(-1)^m$
From $\zeta (z+ w_{i})=\zeta(z)+ n_i$ I get $\zeta (z+ mw_{i})=\zeta(z)+ mn_i$ . $\frac{d}{dz} log \sigma(z+mw_i) = \frac{d}{dz} log \sigma(z) + mn_i$, and $ \sigma(z+mw_i) = \sigma(z)\exp{mn_i z} A $ , $A$ a constant of integration.
And now to determine $A$ I use the oddness of $\sigma(z)$ (odd as in odd function..) by setting $z=\frac{-mw_i}{2}$: $\sigma(\frac{mw_i}{2})=\sigma(\frac{-mw_i}{2})A\exp^{-m^2n_iw_i/2}$ $\implies A= - \exp^{\frac{m^2n_iw_i}{2}}$ $ \sigma(z+mw_i) = - \sigma(z)\exp{mn_i (z+mw_i/2)} $ So I have a minus sign rather than $(-1)^m$ could someone please tell me what I have done wrong? Many thanks in advance.
You made a simple mistake. In your step when you set $z=-mw_i/2$ you neglected to note that $\sigma(z)$ has zeros at all integer multiples of $w_i$, and so when you divided out by $\sigma(\frac{mw_i}{2})$ and $m$ is even then you are dividing by zero. When $m$ is odd, you just get $-1$ which is $(-1)^m$ when $m$ is odd.