1. ## Singularities and residues

Find the isolated singularities (if any) of the following function. For each isolated singularity, describe its nature; that is is it removeable or a pole (and of what order) or essential? In each case, calculate the residue of the function at the singularity.

a) $\pi cot(\pi z) - 1/z$
I got a removable singularity at z=0.
but how do I exactly work out the residue for this?

b) $z^{-\frac{1}{2}}$
I got none. because you cannot have a complex number which is multiplied to a power of a half. I'm not quite sure about this one...

c) $sin(z)sin(\frac{1}{z})$
I got a removable singularity at z=0. Not sure about this one either.... These are some hard functions that I got...

2. for a), should I expand it to Lauren series... and then see what coefficients I get?

3. Originally Posted by lpd
Find the isolated singularities (if any) of the following function. For each isolated singularity, describe its nature; that is is it removeable or a pole (and of what order) or essential? In each case, calculate the residue of the function at the singularity.

a) $\pi cot(\pi z) - 1/z$
I got a removable singularity at z=0.
but how do I exactly work out the residue for this?

b) $z^{-\frac{1}{2}}$
I got none. because you cannot have a complex number which is multiplied to a power of a half. I'm not quite sure about this one...

c) $sin(z)sin(\frac{1}{z})$
I got a removable singularity at z=0. Not sure about this one either.... These are some hard functions that I got...

a) The residue at a removable singularity is zero. Your job is to think about why ....

b) The function is $f(z) = \frac{1}{\sqrt{z}}$ ....

c) I suggest you try getting the first few terms of the series around z = 0. Is there a 1/z term ....?

4. Originally Posted by mr fantastic

c) I suggest you try getting the first few terms of the series around z = 0. Is there a 1/z term ....?
theres no 1/z term when i computed the lauren series. does that imply that $res[sin(z)sin(1/z),0]=0$?

5. Originally Posted by lpd
theres no 1/z term when i computed the lauren series. does that imply that $res[sin(z)sin(1/z),0]=0$?
Yes.