I would like to clasify the singularities of the function
$$ f(z) = \frac{1}{z^2+2z+1} $$
I tried to make the Laurent expansion on $ z = -1$, but at this point the function is not holomorphic and I do not know how to calculate the coefficents because it is not posible to use the Cauchy's Integral Formula for the derivatives on points where the function is not holomorphic. The formula says that if you have a function $f$, holomorphic on an open set $\Omega$, and $\bar{D}(c,R)\subset \Omega$, for all $z_0\in \Omega$ you have
$$ f^{(k} (z_0) = \frac{k!}{2\pi i}\int_{\partial D(c;r)} \frac{f(\zeta)}{(\zeta-z_0)^{k+1}}d\zeta $$
And the coefficients are given by $$ a_n = \frac{1}{2\pi i}\int_{\gamma_r} \frac{f(\zeta)}{(\zeta+1)^{k+1}}d\zeta $$ where $\gamma_r$ is the disc $\gamma_r(t) = c + r\exp{it}$ with $ r>0$. Note that $f$ is holomorphic on $\mathbb C\backslash{\{-1\}}$.
Thanks.
Note $z^2+2z+1=(z+1)^2$. So $$f(z)=\frac{1}{(z+1)^2}$$ This is the Laurent series expansion about $z=-1$. I.e. if $f(z)=\sum_{-\infty}^\infty a_n (z+1)^n$, then $$a_n=\begin{cases}1&n=-2\\0&\text{else}\end{cases}$$ Can you now classify the singularity?