<s>[INST] Define the term $a_n$ in Laurent's series expansion $\sum_{n=-\infty}^{\infty} a_n(z-a)^n$ of $f$ centered on $z=a.$... 
Define the term $a_n$ in Laurent's series expansion $\sum_{n=-\infty}^{\infty} a_n(z-a)^n$ of $f$ centered on $z=a.$ classify the singularity $ z=a.$


a)$f(z)=\frac{z^2+1}{z(z+1)}, a=0.$


b)$f(z)=\frac{1}{z(z-1)}$ for $a=0.1.$

we have the following formula $a_n = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}}dz$
So I need to do
a) $a_n = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}}dz =  \frac{1}{2 \pi i} \int_{\gamma} \frac{\frac{z^2+1}{z(z+1)}}{z}dz$ ?
b)$a=0 \to a_n = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}}dz =  \frac{1}{2 \pi i} \int_{\gamma} \frac{\frac{1}{z(z-1)}}{z}dz$?
$a=1 \to a_n = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}}dz =  \frac{1}{2 \pi i} \int_{\gamma} \frac{\frac{1}{z(z-1)}}{z-1}dz$?
If it is that, how do I solve this integral?
Thanks.
 [/INST] You don't need that formula.
a) For each $z\in\Bbb C$ with $|z|<1$, you have\begin{align}\frac{z^2+1}{z+1}&=z-1+\frac2{z+1}\\&=z-1+2\sum_{n=0}^\infty(-1)^nz^n\\&=1-z+2\sum_{n=2}^\infty(-1)^nz^n.\end{align}Therefore, if $z\ne0$,$$\frac{z^2+1}{z(z+1)}=\frac1z-1+2\sum_{n=1}^\infty(-1)^{n+1}z^n.$$So, $0$ is a simple pole.
b) Near $0.1$, $\frac1{z(z-1)}$ is the quotient of two analytic functions, and therefore it is an analytic function. It follows that $0.1$ is a removable singularity of $f$.
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