- #1
binbagsss
- 1,259
- 11
Homework Statement
I am wanting to show that
##lim_{z\to\infty} f(z)=c## does not exist for ##c \in C##, ##C## the complex plane, where ##f## is non-constant periodic meromorphic function. (elliptic)
Homework Equations
The Attempt at a Solution
So I want to proove this is not true
.
Beginning with the definition:
For any ##\epsilon >0## can find a ##R## s.t ##|z|>R \implies |f(z)-c|<\epsilon## (1)
Now let ##w## be a period.
Then we also have
##|f(z+mw)-c|<\epsilon##(2), for ##m## some integer
Now I attach the solution:
Questions:
- I thought that (2) would hold automatically, from (1) by periodicty, without additionally needing to have ##|z+mw |>R ##
- Also this would hold following from (1), if not using periodicity, but simply looking at ##|z+mw |>R ##, for ##z## large since a shift by some constant is insignificant? Therefore in the solution, I do not understand the need to specify both, by periodicity and include this condition ##|z+mw |>R ##
- Finally I do not understand the conclusion itself, perhaps due to being unclear on the above points- I understand the goal is to show that if this limit exists it implies that ##f## is holomoprhic (since I then have the theorem that an elliptic function holomorphic is constant, and so proof by contradiction).
I do not understand the conclusion that
##|f(z)-c|=|f(z+mw)-c|<1## (3) implies the ##f(z)## is a bounded holomorphic function.
My interpretation is that since we can vary ##m## to cover a range of values of ##f(z)## which are bounded this conclusion is made? So ##m## generating a lot of points? However isn't this only for large ##z## ? We have only showed bounded for large ##z##? (When ##|z|>R## is satisfied)? Or am I barking up the wrong tree as to why we conclude ##f(z)## is constant from (3)?
Many thanks in advance