Locate and Classify Singularities

[jjangub]

Homework Statement

Locate and classify the singularities of the following functions
a) f(z) = 1 / (z^3*(z^2+1))
b) f(z) = (1 - e^z)/z
c) f(z) = 1 / (1-e^z(^2))
d) f(z) = z / (e^(1/z))

Homework Equations

The Attempt at a Solution

I am not sure what I need to do when it asks me to locate and classify the singularities. I tried this way.
a) f(z) has singularities when z = i or -i and singularities are fifth order poles.
(z^3*(z^2+1) = z^5 + z^3
b) f(z) has essential singularity at 0 because f(z) is defined in the whole complex plane except for z = 0.
c) f(z) has essential singularity at 0 because f(z) is defined in the whole complex plane except for z = 0.
d) since we know e^(1/z) is essential singularity at 0, f(z) = z / (e^(1/z)) is same.

Did I do right?
Please tell me if I did something wrong.
Thank you.

 

[jackmell]

Need to write those in quotient form. It's easier to talk about the singularities that way.

(a)\quad \frac{1}{z^3(z^2+1)}

Ok, that one has a third-order pole at zero and simple poles at \pm i right?

(b)\quad \frac{1-e^z}{z}

That has a limit as z\to 0 so has a removable singularity at zero.

(c)\quad \frac{1}{1-e^{z^2}}

That one is tricky because e^u is entire so reaches every value, except maybe one, infinitely often so can you solve multiwise:

e^{z^2}=1

I mean:

z^2=\log(1)

z=\sqrt{\ln|1|+i(0+2n\pi)}

z=\sqrt{2n\pi i},\quad n=0,\pm 1,\pm2,\cdots

So (c) has a pole at those values of z. Keep in mind the square root is double-valued also, so the poles are along the two rays with arguments \pi/4 and -3\pi /4. What are the orders of those poles? Can you figure out what the order of the one at zero is if I write it as:

-\frac{1}{\sum_{n=1}^{\infty} \frac{z^{2n}}{n!}}

And (d) is the canonical form of an essential singularity (e^{1/z}). You can write the taylor series for e^{w} and then substitute w=1/z, z\ne 0[/tex] and obtain a Laurent series with a non-terminating singular part (the part with powers of z in the denominator). Also, e^u is entire and non-polynomial so then has an essential singularity at infinity (Picard) which means e^{1/z}[/tex] has one at zero.

 

[jjangub]

I understand most of them excpet c).
Could you explain again about c)?
Thank you.