# Identifying singularities of f and classifying them

1. Nov 29, 2013

### SALAAH_BEDDIAF

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

In this case, the singularities are $$\pm2i , \pm\frac{i}{2}$$

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

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

2. Nov 29, 2013

### ShayanJ

A simple pole is a pole of order one which means it appears in the laurent expansion as $(z-z_0)^{-1}$

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

Where I have used the method of partial fractions.

3. Nov 29, 2013

### SALAAH_BEDDIAF

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. Nov 29, 2013

### Office_Shredder

Staff Emeritus
Your definition of a simple pole is incorrect. A pole s of f(z) called removable if
$$\lim_{z \to s} f(z)$$
exists, and it is a simple pole if
$$\lim_{x\to s} (z-s) f(z)$$
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