Identifying singularities of f and classifying them

  • #1
Hi guys, just wanting to know if I'm doing this right. [tex] f(z) = \frac{z}{(z^2 + 4) (z^2+1/4)} [/tex]
Singularities of f(z) are when [tex] (z^2 + 4), (z^2 + 1/4) = 0 [/tex]

In this case, the singularities are [tex] \pm2i , \pm\frac{i}{2} [/tex]

Lets call these singularities [itex] s [/itex] and [itex] s [/itex] is a simple pole if [itex] \lim_{z \to s} \frac{z}{(z^2 + 4) (z^2+1/4)} [/itex] exists.

I got all these singularities to be simple poles, correct or incorrect?
 

Answers and Replies

  • #2
2,797
602
A simple pole is a pole of order one which means it appears in the laurent expansion as [itex] (z-z_0)^{-1} [/itex]
For your functions:

[itex]
f(z)=\frac{z}{(z+2i)(z-2i)(z+\frac{i}{2})(z-\frac{i}{2})}=\frac{2}{15}[-\frac{1}{z+2i}-\frac{1}{z-2i}+\frac{1}{z+\frac{i}{2}}+\frac{1}{z-\frac{i}{2}}]
[/itex]

Where I have used the method of partial fractions.
 
  • #3
A simple pole is a pole of order one which means it appears in the laurent expansion as [itex] (z-z_0)^{-1} [/itex]
For your functions:

[itex]
f(z)=\frac{z}{(z+2i)(z-2i)(z+\frac{i}{2})(z-\frac{i}{2})}=\frac{2}{15}[-\frac{1}{z+2i}-\frac{1}{z-2i}+\frac{1}{z+\frac{i}{2}}+\frac{1}{z-\frac{i}{2}}]
[/itex]

Where I have used the method of partial fractions.

Thanks, but on my course we don't use any sort of expansions so I'm having difficulty understanding what the expansion actually tells me, I just wanted to know if those singularities I have are just simple poles or poles of higher order.
 
  • #4
Office_Shredder
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Science Advisor
Gold Member
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Your definition of a simple pole is incorrect. A pole s of f(z) called removable if
[tex] \lim_{z \to s} f(z) [/tex]
exists, and it is a simple pole if
[tex] \lim_{x\to s} (z-s) f(z) [/tex]
exists and is not equal to zero (if it was zero then you would have a removable singularity).

You can easily check by just canceling the numerator and denominator of the simple pole limit that the poles you have are simple
 

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