Elliptic functions, removable singularity, limits,

[binbagsss]

Homework Statement

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please see attached.

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 $z\to c$?

Homework Equations

see above.

The Attempt at a Solution

[/B]
see above.

 

[mighty2000]

Hello,

Thank you for your post. Based on the information provided, it seems like you are asking for clarification on the concepts of removable singularities and Laurent series. I will try my best to address your questions.

To start, a removable singularity is a type of singularity that can be "removed" by defining the function at that point. In other words, if we have a function f(z) with a removable singularity at z=a, we can redefine the function at z=a to be continuous and the limit as z approaches a will exist. In the case of a simple zero, as you mentioned, this can be seen by looking at the limit as z approaches a of f(z)(z-a). This limit will be zero, indicating that there is no singularity at z=a.

In contrast, for a singularity at z=-a, the limit as z approaches -a of f(z)(z+a) will not be zero. This indicates that there is a singularity at z=-a and it is not removable. In order to determine the type of singularity at z=-a, further work such as using L'Hopital's rule or computing the Laurent series may be necessary.

In general, it is not safe to assume that the zeros of the numerator and denominator will "cancel out" and result in a removable singularity. Each case must be analyzed individually.

As for your question about the holomorphic part of a Laurent series, the idea is that the holomorphic part of a Laurent series is the part of the series with non-negative powers of (z-c). This part of the series will approach zero as z approaches c, indicating that the function is holomorphic at z=c.

I hope this helps clarify the concepts of removable singularities and Laurent series. Please let me know if you have any further questions.