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Locate and Classify Singularities

  • Thread starter jjangub
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  • #1
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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.
 

Answers and Replies

  • #2
1,796
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Need to write those in quotient form. It's easier to talk about the singularities that way.

[tex](a)\quad \frac{1}{z^3(z^2+1)}[/tex]

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

[tex](b)\quad \frac{1-e^z}{z}[/tex]

That has a limit as [itex]z\to 0[/itex] so has a removable singularity at zero.

[tex](c)\quad \frac{1}{1-e^{z^2}}[/tex]

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

[tex]e^{z^2}=1[/tex]

I mean:

[tex]z^2=\log(1)[/tex]

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

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

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 [itex]\pi/4[/itex] and [itex]-3\pi /4[/itex]. 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:

[tex]-\frac{1}{\sum_{n=1}^{\infty} \frac{z^{2n}}{n!}}[/tex]

And (d) is the canonical form of an essential singularity ([itex]e^{1/z}[/itex]). You can write the taylor series for [itex]e^{w}[/itex] and then substitute [itex]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, [itex]e^u[/itex] is entire and non-polynomial so then has an essential singularity at infinity (Picard) which means [itex]e^{1/z}[/tex] has one at zero.
 
Last edited:
  • #3
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I understand most of them excpet c).
Could you explain again about c)?
Thank you.
 
Last edited:

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