Weierstrass zeta & sigma fncts, pseudo-periodicity identity

[binbagsss]

Homework Statement

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

$\sigma ( z + mw_i )=(-1)^m \exp^{(mn_i(z+mwi/2))}\sigma(z)$

Homework Equations

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To note that $\sigma(z)$ is an odd function.

$\zeta(z)=\frac{\sigma'(z)}{\sigma(z)}$

The Attempt at a Solution

[/B]
I am pretty close but messing up with not getting $(-1)^m$

From $\zeta (z+ w_{i})=\zeta(z)+ n_i$ I get $\zeta (z+ mw_{i})=\zeta(z)+ mn_i$ .
$\frac{d}{dz} \log \sigma(z+mw_i) = \frac{d}{dz} \log \sigma(z) + mn_i$
$\sigma(z+mw_i) = \sigma(z)\exp{mn_i z} A$ , $A$ 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.

 

[mighty2000]

It looks like you made a mistake when determining the value of A. When setting z = -mw_i/2, you should get:
sigma(-mw_i/2) = sigma(mw_i/2)Aexp(-m^2n_iw_i/2)
This gives you A = -exp(-m^2n_iw_i/2), which would give you the correct result of (-1)^m in the final expression.