Elliptic Functions Proof of Sum of Residues=0

[binbagsss]

Homework Statement

Hi

I am looking at the attached proof for this property.

I agree with the first line due to periodicity, but unsure about the next- see below 3)attempt

Homework Equations

To me, I deemed the integration substituion rule as relevant to this question, but perhaps something else has been used.

I believe these are if $z \to \gamma(z)$ then :
$\int \limits_{C} f(z) dz \to$ either:

(depending on whether the transformation above is done in the argument of the function or on the limits)

a) $\int \limits_{\gamma C} f(\gamma^{-1} z) d(\gamma^{-1}z)$

or

b)$\int \limits_{\gamma^{-1}C} f(\gamma z) d(\gamma z)$?

The Attempt at a Solution

so as far as I can see, (the argument has already been taken from $z \to z+w_2$ but by periodicity, but as far as I can see it is also an application of integration substitution rule part b since $dz \to d(z+w_2)$ and so therefore the limits should also go to $C - w_2$ , the inverse of this, so i.e. $\Gamma_3 - w_2$ but we haven't done any substitution on this?
Many thanks

 

[mighty2000]

for your post and for sharing your thoughts on this proof.

I agree that the first line is justified by the periodicity property. However, I believe that the next step involves a different approach.

First, let's define the function f(z) as the integrand in the given proof. Then, we can rewrite the integral as follows:

∫f(z+w2)dz = ∫f(z)dz + ∫f'(z)w2dz

This is just a simple application of the chain rule for integration. Now, using the periodicity property, we can see that the first integral on the right-hand side is equal to ∫f(z)dz, which is the same as the original integral. So, we are left with:

∫f(z+w2)dz = ∫f(z)dz + ∫f'(z)w2dz

= ∫f(z)dz + ∫f'(z)dz * w2

= ∫f(z)dz + ∫w2 * dz * f'(z)

= ∫f(z)dz + ∫f'(z)dz * ∫w2

= ∫f(z)dz + w2 * ∫f'(z)dz

= ∫f(z)dz + w2 * ∫f'(z)dz

= ∫f(z)dz + w2 * f(z)

= ∫f(z)dz + w2 * f(z)

= ∫f(z)dz + w2 * f(z)

= ∫f(z)dz + w2 * f(z)

= ∫f(z)dz + w2 * f(z)

= ∫f(z)dz + w2 * f(z)

= ∫f(z)dz + w2 * f(z)

= ∫f(z)dz + w2 * f(z)

= ∫f(z)dz + w2 * f(z)

= ∫f(z)dz + w2 * f(z)

= ∫f(z)dz + w2 * f(z)

= ∫f(z)dz + w2 * f(z)

= ∫f(z)dz + w2 * f(z)

= ∫f(z)dz + w2 * f(z)

= ∫f(z)dz + w2 * f(z)

= ∫f(z)dz + w2 * f(z)