Solving Removable Singularity in [z^(c-1)]/[exp(z)-1] at z=0

[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!

 

[mfb]

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.

 

[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?

 

[mfb]

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

 

[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