Elliptic functions, diff eq, why proof on open disc holds for C

[binbagsss]

Homework Statement

Hi

I am looking at this derivation of differential equation satisfied by $\phi(z)$.

To start with, I know that such a disc $D$ described in the derivation can always be found because earlier in the lecture notes we proved that their exists an $inf=min \omega$ for $\omega \in \Omega/{0}$Following the derivation through I agree that $f(z)=0$, however, the trouble I’m having is why having that this proof holds on the disc $D$, extending it to the entire complex plane?

if a function is constant then its constant everywhere, but because we required convergence and $|z/\omega|<1$ haven't we only shown that this differential equation is satisfied for such a disc?Many thanks

Homework Equations

see above

The Attempt at a Solution

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see below

 

[pasmith]

Has the concept of "analytic extension" not been explained in the notes?

Because whenever you prove something for a holomorphic function in an open disc and immediately extend the result to some larger connected open set (such as the entire complex plane), it's analytic extension that is being used.

 

[binbagsss]

Has the concept of "analytic extension" not been explained in the notes?

Because whenever you prove something for a holomorphic function in an open disc and immediately extend the result to some larger connected open set (such as the entire complex plane), it's analytic extension that is being used.

Never heard of that term, nope.