# Removable Singularity

1. Nov 15, 2014

### eyesontheball1

How exactly would one show that [z^(c-1)]/[exp(z)-1]has a removable singularity at z=0? I tried using the methods introduced in my complex analysis book, but nothing seemed to work. Thanks!

2. Nov 15, 2014

### Staff: Mentor

That will depend on c.
You can determine the order of the pole for 1/(exp(z)-1) and then see what happens with the numerator.
Alternatively, you can just consider the limit.

3. Nov 15, 2014

### eyesontheball1

I'm reading that [z^(c-1)]/[exp(z)-1] has a removable singularity at z = 0 whenever Re(c) > 1, but if c = 3/2, then Re(c) > 1, and [z^(c-1)]/[exp(z)-1] = [z^(1/2)]/[exp(z)-1], which is of the indeterminate form of 0/0 at z=0, but then after applying L'Hopital's rule, this gives (1/2)[z^(-1/2)]/[exp(z)], which approaches infinity as z approaches 0. Then every subsequent application of L'Hopital's rule results in the same limit of infinity. How can it then be the case that this function has a removable singularity at z=0 when c = 3/2? My understanding was that removable singularities are removable precisely because after applying L'Hopital's rule a finite number of times, a finite limit is eventually reached, meaning the function is essentially analytic at the singularity. I'm not too familiar with complex analysis and I just started reading from my text and noticed this problem, which I haven't been able to completely understand. It's kind of an unique example compared to the rest of the examples given in my book. Can you please elaborate on this in detail?

4. Nov 16, 2014

### Staff: Mentor

[z^(1/2)]/[exp(z)-1] is not even continuous if you ignore z=0, clearly it cannot be continuous including z=0.

5. Jan 17, 2015

### Svein

A removable singularity for f(z) at z=0 means that z*f(z) is finite and defined. Thus, we must check (z*z^(c-1))/[exp(z)-1] = z^c/(exp(z)-1) at z=0. Using L'Hôpital, we get c*z^(c-1)/exp(z) = 0/1 (as long as Re(c) > 1). If c=1, the numerator is 1 for all z and the singularity is not removable