Having trouble classifying a singularity...
$f(z)=$$z^2-1\over z^6+2z^5+z^4$ with $z_0=0$ and $z_0=-1$
The $z_0=0$ is pretty simple, just need to put $z^4$ in evidence.
But $z_0=-1$ I can't seem to solve...
The problematic part is in how to solve the $1\over z^2+2z+1$ = $1\over \left(z+1\right)\left(z+1\right)$
If I multiply $f(z)$$\left(z+1\right)$ it still doesn't work.
How do I solve this?
Thanks in advance
On
It seems that your error is thinking that if $f$ has a pole at $c$, then has $f(z)(z-c)$ is always bounded (more precisely, analytic) near $c$. The correct characterization is that $f(z)(z-c)^k$ is bounded near $c$ for some $k\in \mathbb{Z}^+$.
As @ireallyjustdontknow says, note that $f(z)(z+1)^2$ is bounded near $z=-1$, since it behaves like $z^4$ there.
Solving $$g(z)=\frac{1}{z^2+2z+1}=\frac{1}{(z+1)^2}$$
Observe that $(z+1)g(z)$ is not bounded (near $-1$ (thanks Eric)), but $(z+1)^2g(z)$ is. Hence $z=-1$ is a pole of order $2$.