# Elliptic functions, removable singularity, limits,

1. Apr 11, 2017

### binbagsss

1. The problem statement, all variables and given/known data

b) The solution seems a bit vague is the idea here, what this comment is saying, that since this is a simple zero the form of $lim_{z\to a} f_a(z) (z-a)=0$ since, crudely, it is of the form $\frac{0.0}{0}$.
Compared to the point $z=-a$ where I have $lim_{z \to -a} f_a(z) (z+a) = \frac{-2\psi'(a).0}{0}$, here can you immediately conclude that the limit is not zero so it is not a removable singularity, or does further work need to be done such as l'hopitals rules or computer the fourier coefficients via the contour integral and show the laurent series has only non-zero positive coneffiecients or something along those lines.i.e Crudely speaking, in general can you say the zeros are of the same order on the numerator and denominator (at $z=-a$ compared to $z=a$) and so 'cancel' so you can generally rule out a removable singularity in this case or not?

c) Is the idea here simply that the holomorphic part of a laurent series of a complex function about $z=c$ $\to 0$ as $z\to c$?

2. Relevant equations

see above.

3. The attempt at a solution

see above.

2. Apr 16, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.