Pole order of $\frac{z-1}{z^4(z+1)}$ at $z_0=-1$

Question

Determine the type of singularity of $z_0$ of $f(z)=\frac{z^2-1}{z^6+2z^5+z^4}\:z_0=-1$.

I can simplify the function:

$f(z)=\frac{z^2-1}{z^6+2z^5+z^4}=\frac{z-1}{z^4(z+1)}$

$\lim_{z\to -1}\frac{z-1}{z^4(z+1)} =\infty$

However I am not seeing how to compute the pole order. It should be 1 according to the solution.

Question:

How should I compute the pole order of $\frac{z-1}{z^4(z+1)}$ at $z_0=-1$?

Thanks in advance!

Answer 1

Since $\frac{z-1}{z^4}$ is an analytic function defined in the neighborhood of $-1$ of which $-1$ is not a zero, its Taylor series centered at $-1$ is of the type$$a_0+a_1(z+1)+a_2(z+1)^2+\cdots,$$with $a_0\neq0$. Therefore, near $0$ you have$$\frac{z-1}{z^4(z+1)}=\frac{a_0}{z+1}+a_1+a_2(z+1)+\cdots,$$from which it follows that the order of the pole is indeed $1$.

Answer 2

After we know that $\;z=-1\;$ is a pole, observe that

$$\lim_{z\to-1}(z+1)f(z)=\lim_{z\to-1}\frac{z-1}{z^4}=\frac{-2}1=-2$$

and thus $\;z=-1\;$ is a simple pole or pole of multiplicity one.

You can read more about this here: https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Pole_(complex_analysis).html