Elliptic functions proof f(z)-c has N zeros, N the order

[binbagsss]

Homework Statement

proof of theorem

Homework Equations

The Attempt at a Solution

Hi,
I have a couple of questions on the attached proof and theorem

1) On the last line, how is it we go from the order of the zeros = the number of zeros, or is it's meaning the number of zeros counted with multiplicity. - Am I correct in thinking that:

order of pole/zero = number of poles/zeros * multiplicity at each point ?

2) The zero expansion considered is fine since we are considering zeros of $f(z)-c$. But for expansion about a pole of $f(z)$, near that pole, how is it we can say that that same expansion holds for $f(z)-c$? Is this not making some assumptions on $c$ being small?

Many thank in advance.

 

[mighty2000]

Hello,
Thank you for your questions. Let me address them one by one:

1) Yes, you are correct. The order of a pole/zero is equal to the number of poles/zeros multiplied by their respective multiplicities. This means that a pole/zero with multiplicity n will be counted n times in the order, while a simple pole/zero (with multiplicity 1) will only be counted once.

2) You are correct that we are making an assumption on c being small in order for the same expansion to hold for f(z)-c. This is because when we expand about a pole of f(z), we are essentially expanding about the singularity of the function. However, when we subtract a constant c from the function, we are essentially shifting the singularity to a different point. If c is small enough, this shift will not affect the expansion and we can still use the same expansion as before. However, if c is too large, the singularity will be shifted too far and the expansion will no longer hold. Therefore, we must make the assumption that c is small enough for the expansion to hold.

I hope this helps clarify your questions. Let me know if you have any further concerns. Thank you for your interest in the proof and theorem.