Identifying singularities of f and classifying them

[SALAAH_BEDDIAF]

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?

 

[ShayanJ]

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.

 

[SALAAH_BEDDIAF]

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.

 

[Office_Shredder]

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

 

[blue_raver22]

I cannot confirm the accuracy of your calculations without seeing your work. However, the general approach of identifying and classifying singularities is correct. It is important to note that in this case, the singularities are not just simple poles, but also include a double pole at z=0. This is because the denominator can be factored as (z+2i)(z-2i)(z+1/2i)(z-1/2i), indicating a double pole at z=0.

Furthermore, it is important to consider the behavior of the function near the singularities. For example, at z=0, the function approaches infinity, indicating a removable singularity. At z=2i and z=1/2i, the function approaches a finite value, indicating simple poles. At z=-2i and z=-1/2i, the function approaches negative infinity, indicating essential singularities.

In summary, while your approach is generally correct, it is important to also consider the behavior of the function near the singularities in order to fully classify them.